Probability & Statistics

1. Identify the sample space: $S = \{1, 2, 3, 4, 5, 6\}$, so the total number of outcomes is 6.
2. Identify the successful outcomes: the numbers greater than 4 are 5 and 6, so there are 2 successful outcomes.
3. Apply the probability ratio: $P = \frac{2}{6}$.

Answer: 1/3

1. Calculate the total number of marbles: $4 + 5 + 11 = 20$.
2. Identify the successful outcomes (not blue): these are the red and green marbles, so $4 + 5 = 9$.
3. Apply the probability ratio: $P = \frac{9}{20}$.

Answer: 0.45

1. List the total sample space: $HHH, HHT, HTH, THH, HTT, THT, TTH, TTT$. There are $2^3 = 8$ outcomes.
2. Count the outcomes with exactly two heads: $HHT, HTH, THH$. There are 3 such outcomes.
3. Apply the probability ratio: $P = \frac{3}{8}$.

Answer: 0.375

1. Identify the number of choices for each course: $n_1 = 6$, $n_2 = 10$, $n_3 = 5$.
2. Apply the multiplication principle: $6 \times 10 \times 5$.
3. Calculate $6 \times 10 = 60$.
4. Calculate $60 \times 5 = 300$.

Answer: 300

1. There are $26$ options for the first letter and $26$ for the second.
2. There are $10$ options for each of the three digits.
3. The total number of combinations is $26 \times 26 \times 10 \times 10 \times 10$.
4. Calculate $26^2 = 676$.
5. Calculate $676 \times 1000 = 676000$.

Answer: 676000

1. Slot 1: $10$ options (digits).
2. Slot 2: $26$ options (lowercase letters).
3. Slots 3, 4, and 5: Each has $10 + 26 = 36$ options.
4. Total = $10 \times 26 \times 36 \times 36 \times 36$.
5. Calculate $10 \times 26 = 260$.
6. Calculate $36^3 = 46656$.
7. Calculate $260 \times 46656 = 12130560$.

Answer: 12130560

1. Identify $n = 10$ and $r = 2$.
2. Use the permutation formula $P(10, 2) = \frac{10!}{(10-2)!}$.
3. Simplify to $\frac{10!}{8!} = 10 \times 9$.
4. Calculate $10 \times 9 = 90$.

Answer: 90

1. Since we are arranging all items in the set, $n = 6$ and $r = 6$.
2. The number of arrangements is $6!$.
3. Calculate $6 \times 5 \times 4 \times 3 \times 2 \times 1$.
4. Step by step: $30 \times 4 = 120$, $120 \times 3 = 360$, $360 \times 2 = 720$.

Answer: 720

1. Identify $n = 15$ and $r = 3$.
2. Use the permutation formula $P(15, 3) = \frac{15!}{(15-3)!}$.
3. Simplify to $\frac{15!}{12!} = 15 \times 14 \times 13$.
4. Calculate $15 \times 14 = 210$.
5. Calculate $210 \times 13 = 2730$.

Answer: 2730

1. Identify that order does not matter, so we use combinations with $n=12$ and $k=4$.
2. Apply the formula: $\binom{12}{4} = \frac{12!}{4!(12-4)!} = \frac{12!}{4!8!}$.
3. Simplify the fraction: $\frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1}$.
4. Calculate the numerator: $11880$.
5. Calculate the denominator: $24$.
6. Divide: $11880 / 24 = 495$.

Answer: 495

1. The order of numbers on a ticket does not matter, so we use $\binom{49}{6}$.
2. Set up the formula: $\frac{49!}{6!(49-6)!} = \frac{49!}{6!43!}$.
3. Expand the numerator: $49 \times 48 \times 47 \times 46 \times 45 \times 44$.
4. Divide by $6! = 720$.
5. The calculation results in $13,983,816$.

Answer: 13,983,816

1. First, calculate the ways to choose $3$ women from $8$: $\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.
2. Next, calculate the ways to choose $2$ men from $7$: $\binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21$.
3. Use the multiplication principle to combine these independent choices.
4. Total ways = $56 \times 21 = 1176$.

Answer: 1176

1. Step 1: Check if order matters. Since the roles are different, swapping the president and secretary creates a different outcome. Order matters.
2. Step 2: Check for repetition. One person cannot hold more than one office. Repetition is not allowed.
3. Step 3: Use the permutation formula $P(n, r)$ with $n=15$ and $r=3$.
4. Step 4: Compute $15 \times 14 \times 13$.

Answer: 2730

1. Step 1: Check if order matters. A pizza with pepperoni, mushrooms, and onions is the same as one with onions, mushrooms, and pepperoni. Order does not matter.
2. Step 2: Check for repetition. The problem states toppings cannot be repeated.
3. Step 3: Use the combination formula $\binom{n}{k}$ with $n=10$ and $k=3$.
4. Step 4: Compute $\frac{10 \times 9 \times 8}{3 \times 2 \times 1}$.

Answer: 120

1. Step 1: Check if order matters. In a security code, the sequence of letters is critical. Order matters.
2. Step 2: Check for repetition. There is no restriction saying letters cannot be repeated, so repetition is allowed.
3. Step 3: Since repetition is allowed, use the multiplication principle $n^r$.
4. Step 4: Compute $26^5$.

Answer: 11881376

1. Construct the triangle up to row 7.
2. Row 5 is $1, 5, 10, 10, 5, 1$.
3. Row 6 is $1, 6, 15, 20, 15, 6, 1$.
4. Row 7 is $1, 7, 21, 35, 35, 21, 7, 1$.
5. The value at position $k=3$ (the 4th number) is 35.

Answer: 35

1. Use the row sum formula $\sum_{k=0}^{n} \binom{n}{k} = 2^n$.
2. Substitute $n=9$ into the formula: $2^9$.
3. Calculate $2^9 = 512$.

Answer: 512

1. Pascal's Identity states $\binom{n+1}{k+1} = \binom{n}{k} + \binom{n}{k+1}$.
2. Substitute the given values: $28 + 56$.
3. Calculate the sum: $28 + 56 = 84$.

Answer: 84

1. Identify $a = x$, $b = 4$, and $n = 3$.
2. Use the binomial coefficients for $n=3$: $1, 3, 3, 1$.
3. Write the terms: $1(x^3)(4^0) + 3(x^2)(4^1) + 3(x^1)(4^2) + 1(x^0)(4^3)$.
4. Simplify the constants: $1 \cdot 1 = 1$, $3 \cdot 4 = 12$, $3 \cdot 16 = 48$, $1 \cdot 64 = 64$.

Answer: $x^3 + 12x^2 + 48x + 64$

1. Identify $a = 2x$, $b = -1$, and $n = 5$.
2. The $x^2$ term occurs when the power of $a$ is $2$, so $n-k = 2$, which means $k = 3$.
3. Apply the formula for the term: $\binom{5}{3} (2x)^2 (-1)^3$.
4. Calculate the parts: $\binom{5}{3} = 10$, $(2x)^2 = 4x^2$, and $(-1)^3 = -1$.
5. Multiply together: $10 \cdot 4x^2 \cdot (-1) = -40x^2$.

Answer: $-40$

1. Identify $a = 3x$, $b = 2y$, and $n = 4$.
2. Use coefficients $1, 4, 6, 4, 1$.
3. Term 1: $1(3x)^4(2y)^0 = 81x^4$.
4. Term 2: $4(3x)^3(2y)^1 = 4(27x^3)(2y) = 216x^3y$.
5. Term 3: $6(3x)^2(2y)^2 = 6(9x^2)(4y^2) = 216x^2y^2$.
6. Term 4: $4(3x)^1(2y)^3 = 4(3x)(8y^3) = 96xy^3$.
7. Term 5: $1(3x)^0(2y)^4 = 16y^4$.

Answer: $81x^4 + 216x^3y + 216x^2y^2 + 96xy^3 + 16y^4$

1. Identify $n = 10$ (coins) and $k = 4$ (pirates).
2. Use the stars and bars formula for non-negative integers: $\binom{n+k-1}{k-1}$.
3. Substitute the values: $\binom{10+4-1}{4-1} = \binom{13}{3}$.
4. Calculate $\frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 13 \times 2 \times 11 = 286$.

Answer: 286

1. Positive integer solutions mean each variable must be at least 1.
2. Use the 'no zeros' formula: $\binom{n-1}{k-1}$.
3. Here $n = 20$ and $k = 4$.
4. Calculate $\binom{20-1}{4-1} = \binom{19}{3} = \frac{19 \times 18 \times 17}{3 \times 2 \times 1} = 19 \times 3 \times 17 = 969$.

Answer: 969

1. Pre-allocate 2 marbles to jar 1 and 3 marbles to jar 2.
2. Remaining marbles: $15 - 2 - 3 = 10$.
3. Now distribute 10 identical marbles into 3 distinct jars with no further restrictions.
4. Use $\binom{n+k-1}{k-1}$ with $n = 10$ and $k = 3$.
5. Calculate $\binom{10+3-1}{3-1} = \binom{12}{2} = \frac{12 \times 11}{2} = 66$.

Answer: 66

1. Identify the sets: $|A| = 40$ and $|B| = 30$.
2. Identify the intersection: $|A \cap B| = 12$.
3. Apply the formula $|A \cup B| = |A| + |B| - |A \cap B|$.
4. Calculate $40 + 30 - 12 = 58$.

Answer: 58

1. Sum the individual sets: $50 + 40 + 30 = 120$.
2. Subtract the pairwise intersections: $120 - (15 + 10 + 12) = 120 - 37 = 83$.
3. Add back the triple intersection: $83 + 5 = 88$.

Answer: 88

1. Use the formula $|A \cup B| = |A| + |B| - |A \cap B|$.
2. Substitute the known values: $100 = 70 + 50 - |A \cap B|$.
3. Simplify the right side: $100 = 120 - |A \cap B|$.
4. Solve for $|A \cap B|$: $|A \cap B| = 120 - 100 = 20$.

Answer: 20

1. Use the formula $D(6) = 6! \sum_{k=0}^{6} \frac{(-1)^k}{k!}$.
2. Expand the sum: $720 \times (1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} + \frac{1}{720})$.
3. Simplify the terms: $720 \times (0 + 0.5 - 0.1666... + 0.04166... - 0.00833... + 0.00138...)$.
4. Calculate the sum inside: $720 \times (\frac{360 - 120 + 30 - 6 + 1}{720}) = 720 \times \frac{265}{720}$.
5. The result is 265.

Answer: 265

1. The total number of permutations is $5! = 120$.
2. The number of derangements $D(5)$ is $120 \times (1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}) = 44$.
3. The probability is $\frac{D(5)}{5!} = \frac{44}{120}$.
4. Simplify the fraction: $\frac{11}{30}$.

Answer: \frac{11}{30}

1. First, choose which 2 elements are fixed: $\binom{7}{2} = \frac{7 \times 6}{2} = 21$.
2. The remaining $7 - 2 = 5$ elements must be deranged so that no more elements are in their original positions.
3. The number of derangements for 5 elements is $D(5) = 44$.
4. The total number of such permutations is $\binom{7}{2} \times D(5) = 21 \times 44$.
5. Calculate $21 \times 44 = 924$.

Answer: 924

1. The order of people in a committee does not matter, so we use combinations.
2. We need to calculate 10 choose 4.
3. $\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$.

Answer: 210

1. Since the apples are identical, we use stars and bars.
2. Each child must get at least one, so we pre-allocate 3 apples.
3. Remaining apples $n = 7 - 3 = 4$. Number of children $k = 3$.
4. The number of ways is $\binom{4+3-1}{3-1} = \binom{6}{2} = 15$.

Answer: 15

1. A derangement means no letter is in its original position.
2. We use the formula $D(n)$ for $n=4$.
3. $D(4) = 4! (1/0! - 1/1! + 1/2! - 1/3! + 1/4!)$.
4. $D(4) = 24 (1 - 1 + 1/2 - 1/6 + 1/24) = 12 - 4 + 1 = 9$.

Answer: 9

1. The order of people in a committee does not matter, so we use combinations.
2. We apply the formula $\binom{10}{4}$.
3. Calculation: $\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$.

Answer: 210

1. The apples are identical and the children are distinct, so we use stars and bars.
2. Here $n=7$ (apples) and $k=4$ (children).
3. The formula is $\binom{n+k-1}{k-1} = \binom{7+4-1}{4-1} = \binom{10}{3}$.
4. Calculation: $\frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.

Answer: 120

1. First, we choose which 2 people get their own umbrellas: $\binom{5}{2} = 10$ ways.
2. The remaining 3 people must all get the wrong umbrella, which is a derangement of 3 items: $D(3)$.
3. Calculate $D(3) = 3!(1 - 1 + 1/2 - 1/6) = 6(1/3) = 2$.
4. Total ways = $\binom{5}{2} \times D(3) = 10 \times 2 = 20$.

Answer: 20

1. The total number of ways to choose 2 cards from 10 is $\binom{10}{2} = \frac{10 \times 9}{2} = 45$.
2. The number of ways to choose 2 Aces from the 4 available is $\binom{4}{2} = \frac{4 \times 3}{2} = 6$.
3. The probability is the ratio of favorable outcomes to total outcomes: $P = \frac{6}{45}$.

Answer: 2/15

1. The sample space size for two rolls is $6 \times 6 = 36$.
2. The outcomes that sum to 7 are $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$.
3. There are 6 such outcomes.
4. The probability is $P = \frac{6}{36}$.

Answer: 1/6

1. The total number of ways to choose 3 people from 9 is $\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
2. The event 'at least 2 women' includes two cases: exactly 2 women or exactly 3 women.
3. Case 1 (2 women, 1 man): $\binom{4}{2} \times \binom{5}{1} = 6 \times 5 = 30$.
4. Case 2 (3 women, 0 men): $\binom{4}{3} \times \binom{5}{0} = 4 \times 1 = 4$.
5. Total favorable outcomes = $30 + 4 = 34$.
6. The probability is $P = \frac{34}{84}$.

Answer: 17/42

1. Check Axiom 1: All values $0.4, 0.3, 0.2, 0.2$ are $\ge 0$.
2. Check Axiom 2: Sum the probabilities: $0.4 + 0.3 + 0.2 + 0.2 = 1.1$.
3. Since the sum is $1.1$, which is not equal to $1$, the normalization axiom is violated.

Answer: No, it is invalid because the sum of probabilities is $1.1$, violating $P(S)=1$.

1. Let $W$ be winning, $D$ be drawing, and $L$ be losing.
2. The sample space is $S = \{W, D, L\}$.
3. By Axiom 2, $P(W) + P(D) + P(L) = 1$.
4. Substitute the known values: $0.15 + 0.30 + P(L) = 1$.
5. Solve for $P(L)$: $P(L) = 1 - 0.45 = 0.55$.

Answer: 0.55

1. Let $P(G) = x$. Then $P(B) = 2x$.
2. We know $P(R) = 1/3$.
3. By Axiom 2, $P(R) + P(B) + P(G) = 1$.
4. Substitute: $1/3 + 2x + x = 1$.
5. Combine terms: $3x = 1 - 1/3 = 2/3$.
6. Divide by 3: $x = 2/9$.

Answer: 2/9

1. The total number of ways to choose 5 cards from 52 is $\binom{52}{5} = 2598960$.
2. There are 4 suits. For each suit, there are 13 cards, and we choose 5 of them: $\binom{13}{5} = 1287$.
3. The total number of favorable outcomes is $4 \times 1287 = 5148$.
4. The probability is $\frac{5148}{2598960} \approx 0.00198$.

Answer: \frac{5148}{2598960} \approx 0.00198

1. The total number of ways to choose 3 marbles from 12 is $\binom{12}{3} = 220$.
2. The number of ways to choose 2 red marbles from 7 is $\binom{7}{2} = 21$.
3. The number of ways to choose 1 blue marble from 5 is $\binom{5}{1} = 5$.
4. The number of favorable outcomes is $21 \times 5 = 105$.
5. The probability is $\frac{105}{220} = \frac{21}{44}$.

Answer: \frac{21}{44}

1. The total number of arrangements is $4! = 24$.
2. First, find the probability that they ARE sitting next to each other. Treat Alice and Bob as one block. There are $3!$ ways to arrange the block and the other two people, and $2!$ ways to arrange Alice and Bob within the block: $3! \times 2! = 6 \times 2 = 12$.
3. The probability they are adjacent is $\frac{12}{24} = \frac{1}{2}$.
4. The probability they are NOT adjacent is $1 - \frac{1}{2} = \frac{1}{2}$.

Answer: \frac{1}{2}

1. Find the probability that all $3$ people have different birthdays: $P(\text{no match}) = \frac{365}{365} \times \frac{364}{365} \times \frac{363}{365}$.
2. Calculate the product: $P(\text{no match}) = \frac{132132}{133225} \approx 0.9918$.
3. Subtract from $1$: $P(\text{match}) = 1 - 0.9918 = 0.0082$.

Answer: 0.0082

1. Calculate $P(\text{no match}) = \frac{P(365, 30)}{365^{30}}$.
2. Using the formula, $P(\text{no match}) \approx 0.2937$.
3. Find $P(\text{match}) = 1 - 0.2937 = 0.7063$.
4. Since $0.7063 > 0.70$, the probability is greater than $70\%$.

Answer: Greater than 70%

1. We need $P(\text{no match}) < 0.5$, where $P(\text{no match}) = \frac{P(100, n)}{100^n}$.
2. For $n=12$: $P(\text{no match}) = \frac{100 \times 99 \times \dots \times 89}{100^{12}} \approx 0.501$.
3. For $n=13$: $P(\text{no match}) = \frac{100 \times 99 \times \dots \times 88}{100^{13}} \approx 0.441$.
4. The probability first exceeds $0.5$ when $n=13$.

Answer: 13

1. The probability of success (heads) is $p = 0.5$.
2. The probability of failure (tails) is $1 - p = 0.5$.
3. The probability of zero heads in 5 flips is $(0.5)^5 = 0.03125$.
4. The probability of at least one head is $1 - 0.03125 = 0.96875$.

Answer: 0.96875

1. The probability of success is $p = 0.7$.
2. The probability of failure (missing) is $1 - p = 0.3$.
3. The probability of missing all 3 shots is $(0.3)^3 = 0.027$.
4. The probability of making at least one is $1 - 0.027 = 0.973$.

Answer: 0.973

1. Find the probability that each alarm fails: $P(F_1) = 0.1$, $P(F_2) = 0.2$, $P(F_3) = 0.3$.
2. The probability that all three fail is $0.1 \times 0.2 \times 0.3 = 0.006$.
3. The probability that at least one succeeds is $1 - 0.006 = 0.994$.

Answer: 0.994

1. Identify the restricted sample space $B$: {Monday, Tuesday, Wednesday, Thursday, Friday}. The size $|B| = 5$.
2. Identify the intersection $A \cap B$: Days that are weekdays AND start with 'T'. These are {Tuesday, Thursday}. The size $|A \cap B| = 2$.
3. Apply the formula $P(A|B) = \frac{|A \cap B|}{|B|} = \frac{2}{5}$.

Answer: 2/5

1. Let $M$ be liking Math and $S$ be liking Science.
2. The condition is liking Science, so $P(S) = \frac{10}{30}$.
3. The intersection is liking both, so $P(M \cap S) = \frac{7}{30}$.
4. Calculate $P(M|S) = \frac{7/30}{10/30} = \frac{7}{10}$.

Answer: 7/10

1. The sample space is $S = \{HH, HT, TH, TT\}$.
2. The condition $B$ is 'at least one head': $B = \{HH, HT, TH\}$. Thus $|B| = 3$.
3. The event $A$ is 'both heads': $A = \{HH\}$.
4. The intersection $A \cap B$ is just $\{HH\}$, so $|A \cap B| = 1$.
5. The conditional probability is $P(A|B) = \frac{1}{3}$.

Answer: 1/3

1. The probability the first card is a heart is $13/52$.
2. Given the first is a heart, there are 12 hearts left out of 51 cards, so the second is $12/51$.
3. Given the first two are hearts, there are 11 hearts left out of 50 cards, so the third is $11/50$.
4. Multiply these: $(13/52) \cdot (12/51) \cdot (11/50) = (1/4) \cdot (4/17) \cdot (11/50) = 11/850$.

Answer: 11/850

1. The probability the first ball is white is $7/10$.
2. Given the first was white, there are still 3 black balls, but only 9 balls total. The probability the second is black is $3/9$.
3. Multiply the probabilities: $(7/10) \cdot (3/9) = 21/90 = 7/30$.

Answer: 7/30

1. There are $3! = 6$ different orders to draw one of each color (RGB, RBG, GRB, GBR, BRG, BGR).
2. The probability of any one specific order, say Red then Green then Blue, is $(5/12) \cdot (4/11) \cdot (3/10) = 60/1320 = 1/22$.
3. Since all 6 orders have the same probability, the total probability is $6 \cdot (1/22) = 6/22 = 3/11$.

Answer: 3/11

1. Calculate the product of the individual probabilities: $0.5 \cdot 0.4 = 0.2$.
2. Compare this product to the given joint probability $P(A \cap B) = 0.2$.
3. Since $0.2 = 0.2$, the condition for independence is met.

Answer: Yes, they are independent because $P(A \cap B) = P(A)P(B)$.

1. Since they are disjoint, $P(X \cap Y) = 0$.
2. Since $P(X) > 0$ and $P(Y) > 0$, their product $P(X)P(Y)$ must be greater than zero.
3. Therefore, $P(X \cap Y) \neq P(X)P(Y)$.

Answer: No, they are dependent because the occurrence of one makes the probability of the other zero.

1. Identify the events: $A$ is tails, $B$ is die $> 4$.
2. The events are independent because the coin does not affect the die.
3. Compute $P(A) = 1/2$.
4. Compute $P(B) = 2/6 = 1/3$ (outcomes 5 and 6).
5. Multiply the probabilities: $1/2 \cdot 1/3 = 1/6$.

Answer: $1/6$

1. Identify the scenarios: $P(A) = 0.40$, $P(B) = 0.35$, $P(C) = 0.25$.
2. Identify conditional probabilities: $P(Def|A) = 0.02$, $P(Def|B) = 0.03$, $P(Def|C) = 0.05$.
3. Apply the Law of Total Probability: $P(Def) = (0.40)(0.02) + (0.35)(0.03) + (0.25)(0.05)$.
4. Calculate: $P(Def) = 0.008 + 0.0105 + 0.0125 = 0.031$.

Answer: 0.031

1. Identify scenarios: $P(Rain) = 0.30$, $P(No Rain) = 0.70$.
2. Identify conditional probabilities: $P(Late|Rain) = 0.60$, $P(Late|No Rain) = 0.10$.
3. Apply the Law of Total Probability: $P(Late) = (0.30)(0.60) + (0.70)(0.10)$.
4. Calculate: $P(Late) = 0.18 + 0.07 = 0.25$.

Answer: 0.25

1. Identify scenarios: $P(Box 1) = 0.5$, $P(Box 2) = 0.5$.
2. Identify conditional probabilities: $P(Silver|Box 1) = 7/10 = 0.7$, $P(Silver|Box 2) = 2/10 = 0.2$.
3. Apply the Law of Total Probability: $P(Silver) = (0.5)(0.7) + (0.5)(0.2)$.
4. Calculate: $P(Silver) = 0.35 + 0.10 = 0.45$.

Answer: 0.45

1. Identify priors: $P(L1) = 0.7$, $P(L2) = 0.3$.
2. Identify likelihoods: $P(D|L1) = 0.03$, $P(D|L2) = 0.08$.
3. Calculate total probability of defect: $P(D) = (0.03 \times 0.7) + (0.08 \times 0.3) = 0.021 + 0.024 = 0.045$.
4. Apply Bayes' theorem: $P(L1|D) = \frac{0.021}{0.045} = \frac{21}{45} = \frac{7}{15}$.

Answer: 7/15 \approx 0.4667

1. Priors: $P(Rain) = 0.2$, $P(No Rain) = 0.8$.
2. Likelihoods: $P(Pred|Rain) = 0.8$, $P(Pred|No Rain) = 1 - 0.9 = 0.1$.
3. Total probability of prediction: $P(Pred) = (0.8 \times 0.2) + (0.1 \times 0.8) = 0.16 + 0.08 = 0.24$.
4. Posterior: $P(Rain|Pred) = \frac{0.16}{0.24} = \frac{2}{3}$.

Answer: 2/3 \approx 0.6667

1. Priors: $P(A) = P(B) = P(C) = 1/3$.
2. Likelihoods: $P(R|A) = 2/5$, $P(R|B) = 4/5$, $P(R|C) = 3/5$.
3. Total probability of red: $P(R) = (1/3 \times 2/5) + (1/3 \times 4/5) + (1/3 \times 3/5) = (2+4+3)/15 = 9/15 = 3/5$.
4. Posterior: $P(B|R) = \frac{(1/3 \times 4/5)}{3/5} = \frac{4/15}{9/15} = 4/9$.

Answer: 4/9 \approx 0.4444

1. Identify priors: $P(D) = 0.002$ and $P(D^c) = 0.998$.
2. Identify likelihoods: $P(+|D) = 0.95$ and $P(+|D^c) = 0.03$.
3. Calculate total probability of positive test: $P(+) = (0.95 \times 0.002) + (0.03 \times 0.998) = 0.0019 + 0.02994 = 0.03184$.
4. Apply Bayes' theorem: $P(D|+) = \frac{0.0019}{0.03184} \approx 0.0597$.

Answer: 0.0597

1. Priors: $P(S) = 0.01$, $P(S^c) = 0.99$.
2. Likelihoods: $P(+|S) = 0.90$. Since specificity is $90\%$, the false positive rate $P(+|S^c) = 1 - 0.90 = 0.10$.
3. Total probability of positive: $P(+) = (0.90 \times 0.01) + (0.10 \times 0.99) = 0.009 + 0.099 = 0.108$.
4. Posterior: $P(S|+) = \frac{0.009}{0.108} = \frac{1}{12} \approx 0.0833$.

Answer: 0.0833

1. Priors: $P(B) = 0.0001$, $P(B^c) = 0.9999$.
2. Likelihoods: $P(+|B) = 0.999$, $P(+|B^c) = 0.001$.
3. Total probability of alarm: $P(+) = (0.999 \times 0.0001) + (0.001 \times 0.9999) = 0.0000999 + 0.0009999 = 0.0010998$.
4. Posterior: $P(B|+) = \frac{0.0000999}{0.0010998} \approx 0.0908$.

Answer: 0.0908

1. The initial probability of picking the car is $1/5$.
2. The probability that the car is among the other $4$ doors is $4/5$.
3. The host eliminates $2$ of those $4$ doors, leaving $2$ closed doors in that group.
4. The $4/5$ probability is now split equally between the $2$ remaining doors.
5. The probability for each remaining door is $(4/5) / 2 = 2/5$.

Answer: $2/5$

1. Let $C_i$ be the event the car is behind door $i$, and $G_3$ be the event the host opens door $3$ and finds a goat.
2. The prior probabilities are $P(C_1) = P(C_2) = P(C_3) = 1/3$.
3. The likelihoods are $P(G_3 | C_1) = 1/2$ (he could pick door $2$ or $3$), $P(G_3 | C_2) = 1/2$ (he picks from doors $2$ and $3$, and $3$ is a goat), $P(G_3 | C_3) = 0$.
4. Using Bayes: $P(C_2 | G_3) = \frac{(1/2)(1/3)}{(1/2)(1/3) + (1/2)(1/3)} = 1/2$.

Answer: $1/2$

1. The probability your first choice was correct is $1/100$.
2. The probability the car is in the other $99$ doors is $99/100$.
3. The host eliminates $98$ of those $99$ doors.
4. The entire $99/100$ probability is now concentrated in the single remaining door.
5. Switching wins if the car was in the group of $99$ doors.

Answer: $99/100$

1. The complement of 'at least 2 heads' is '0 heads' or '1 head'.
2. The probability of 0 heads is $\binom{10}{0}(0.5)^{10} = 1/1024$.
3. The probability of 1 head is $\binom{10}{1}(0.5)^{10} = 10/1024$.
4. The probability of the complement is $11/1024$.
5. The final probability is $1 - 11/1024 = 1013/1024$.

Answer: $1013/1024 \approx 0.9893$

1. The total number of ways to choose 3 balls from 10 is $\binom{10}{3} = 120$.
2. The number of ways to choose 2 white balls from 4 is $\binom{4}{2} = 6$.
3. The number of ways to choose 1 black ball from 6 is $\binom{6}{1} = 6$.
4. The number of successful outcomes is $6 \times 6 = 36$.
5. The probability is $36/120 = 3/10$.

Answer: $0.3$

1. Let $F$ be fair, $H$ be double-headed, $T$ be double-tailed. $P(F)=P(H)=P(T)=1/3$.
2. The probability of heads given fair is $P(\text{heads}|F) = 1/2$.
3. The probability of heads given double-headed is $P(\text{heads}|H) = 1$.
4. The probability of heads given double-tailed is $P(\text{heads}|T) = 0$.
5. Total probability of heads: $P(\text{heads}) = (1/3)(1/2) + (1/3)(1) + (1/3)(0) = 1/6 + 1/3 = 1/2$.
6. Using Bayes' theorem: $P(F|\text{heads}) = \frac{P(\text{heads}|F)P(F)}{P(\text{heads})} = \frac{(1/2)(1/3)}{1/2} = 1/3$.

Answer: $1/3$

1. Calculate total ways to choose 4 from 14: $\binom{14}{4} = 1001$.
2. Calculate ways to choose 2 men from 6: $\binom{6}{2} = 15$.
3. Calculate ways to choose 2 women from 8: $\binom{8}{2} = 28$.
4. Multiply successful outcomes: $15 \times 28 = 420$.
5. Divide by total: $420 / 1001 \approx 0.4196$.

Answer: 0.4196

1. Use the complement rule: $P(\text{at least one red}) = 1 - P(\text{no red})$.
2. No red means all 3 are blue.
3. Ways to choose 3 blue from 5: $\binom{5}{3} = 10$.
4. Total ways to choose 3 from 10: $\binom{10}{3} = 120$.
5. Probability of no red: $10 / 120 = 1/12$.
6. Subtract from one: $1 - 1/12 = 11/12 \approx 0.9167$.

Answer: 0.9167

1. Prior $P(D) = 0.01$, so $P(D^c) = 0.99$.
2. Likelihoods: $P(+|D) = 0.99$ and $P(+|D^c) = 1 - 0.95 = 0.05$.
3. Total probability of positive $P(+) = (0.99)(0.01) + (0.05)(0.99) = 0.0099 + 0.0495 = 0.0594$.
4. Apply Bayes: $P(D|+) = \frac{P(+|D)P(D)}{P(+)} = \frac{0.0099}{0.0594}$.
5. Calculate result: $0.0099 / 0.0594 = 1/6 \approx 0.1667$.

Answer: 0.1667

1. The possible values for $X$ are 0, 1, or 2.
2. For $X=0$, both balls must be blue: $P(X=0) = \frac{2}{5} \times \frac{1}{4} = \frac{2}{20} = 0.1$.
3. For $X=2$, both balls must be red: $P(X=2) = \frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = 0.3$.
4. For $X=1$, one red and one blue in any order: $P(X=1) = (\frac{3}{5} \times \frac{2}{4}) + (\frac{2}{5} \times \frac{3}{4}) = \frac{6}{20} + \frac{6}{20} = \frac{12}{20} = 0.6$.
5. Check sum: $0.1 + 0.6 + 0.3 = 1.0$.

Answer: P(X=0)=0.1, P(X=1)=0.6, P(X=2)=0.3

1. The total number of outcomes in the sample space is $2^4 = 16$.
2. The number of ways to get exactly 2 heads is given by the combination formula $\binom{4}{2}$.
3. $\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$.
4. The probability is the number of successful outcomes divided by the total: $P(X=2) = \frac{6}{16} = \frac{3}{8} = 0.375$.

Answer: 0.375

1. The sum of all probabilities must be 1: $0.2 + 0.5 + k = 1$.
2. Solving for $k$: $0.7 + k = 1 \implies k = 0.3$.
3. The event $X < 3$ includes the values $X=1$ and $X=2$.
4. Summing these probabilities: $P(X < 3) = P(X=1) + P(X=2) = 0.2 + 0.5 = 0.7$.

Answer: k=0.3, P(X < 3)=0.7

1. Step 1: $F(0) = p(0) = 0.7$.
2. Step 2: $F(1) = p(0) + p(1) = 0.7 + 0.2 = 0.9$.
3. Step 3: $F(2) = p(0) + p(1) + p(2) = 0.7 + 0.2 + 0.1 = 1.0$.

Answer: F(0)=0.7, F(1)=0.9, F(2)=1.0

1. Step 1: Use the formula $P(a < X \le b) = F(b) - F(a)$.
2. Step 2: Substitute $a = 1$ and $b = 3$.
3. Step 3: $P(1 < X \le 3) = F(3) - F(1) = 1.0 - 0.3 = 0.7$.

Answer: 0.7

1. Step 1: $p(0) = F(0) - 0 = 0.2 - 0 = 0.2$.
2. Step 2: $p(1) = F(1) - F(0) = 0.5 - 0.2 = 0.3$.
3. Step 3: $p(2) = F(2) - F(1) = 1.0 - 0.5 = 0.5$.

Answer: p(0)=0.2, p(1)=0.3, p(2)=0.5

1. Identify the values $x_i$ and their probabilities $p_i$: $(10, 0.2), (20, 0.5), (30, 0.3)$.
2. Apply the expectation formula: $E[X] = 10(0.2) + 20(0.5) + 30(0.3)$.
3. Compute each term: $2 + 10 + 9$.
4. Sum the terms: $21$.

Answer: 21

1. The possible outcomes are $x_1 = 5$ and $x_2 = -2$.
2. Since the coin is fair, $P(X=5) = 0.5$ and $P(X=-2) = 0.5$.
3. Calculate the expectation: $E[X] = 5(0.5) + (-2)(0.5)$.
4. Simplify: $2.5 - 1.0 = 1.5$.

Answer: 1.5

1. First, find $E[X]$: $E[X] = 1(0.1) + 2(0.3) + 3(0.4) + 4(0.2)$.
2. Compute $E[X]$: $0.1 + 0.6 + 1.2 + 0.8 = 2.7$.
3. Determine the values of $Y$ for each $X$: $Y$ takes values $2(1)+3=5, 2(2)+3=7, 2(3)+3=9, 2(4)+3=11$.
4. Calculate $E[Y]$ using the same probabilities: $E[Y] = 5(0.1) + 7(0.3) + 9(0.4) + 11(0.2)$.
5. Compute $E[Y]$: $0.5 + 2.1 + 3.6 + 2.2 = 8.4$.

Answer: 8.4

1. Define $I_i$ as the indicator variable that the $i$-th card is an ace for $i=1, \dots, 5$.
2. The probability that any single card is an ace is $P(I_i = 1) = \frac{4}{52} = \frac{1}{13}$.
3. The expected value of each indicator is $E[I_i] = \frac{1}{13}$.
4. By linearity of expectation, the total expected number of aces is $E[X] = \sum_{i=1}^5 E[I_i] = 5 \cdot \frac{1}{13} = \frac{5}{13}$.

Answer: $\frac{5}{13}$

1. Let $X$ be the number of 6s and $Y$ be the number of 1s.
2. The total winnings are $W = 2X - 1Y$.
3. The expected number of 6s in 30 rolls is $E[X] = 30 \cdot \frac{1}{6} = 5$.
4. The expected number of 1s in 30 rolls is $E[Y] = 30 \cdot \frac{1}{6} = 5$.
5. By linearity, $E[W] = 2E[X] - 1E[Y] = 2(5) - 1(5) = 10 - 5 = 5$.

Answer: $5$

1. There are $\binom{10}{2} = \frac{10 \cdot 9}{2} = 45$ possible pairs of people.
2. For any specific pair, the probability they share a birth-month is $\frac{12}{12 \cdot 12} = \frac{1}{12}$.
3. Let $I_{ij}$ be the indicator that pair $(i, j)$ shares a month. $E[I_{ij}] = \frac{1}{12}$.
4. The total number of pairs $X$ is the sum of all 45 indicators.
5. By linearity, $E[X] = 45 \cdot \frac{1}{12} = \frac{45}{12} = 3.75$.

Answer: $3.75$

1. First, find the expected value: $E[X] = 1(0.2) + 2(0.5) + 3(0.3) = 0.2 + 1.0 + 0.9 = 2.1$.
2. Next, find the squared deviations: $(1-2.1)^2 = 1.21$, $(2-2.1)^2 = 0.01$, $(3-2.1)^2 = 0.81$.
3. Weight these by probabilities: $1.21(0.2) + 0.01(0.5) + 0.81(0.3) = 0.242 + 0.005 + 0.243 = 0.49$.
4. The variance is $0.49$.
5. The standard deviation is $\sqrt{0.49} = 0.7$.

Answer: \text{Var}(X) = 0.49, \sigma = 0.7

1. Calculate $E[X] = 0(0.4) + 10(0.4) + 20(0.2) = 0 + 4 + 4 = 8$.
2. Calculate $E[X^2] = 0^2(0.4) + 10^2(0.4) + 20^2(0.2) = 0 + 40 + 80 = 120$.
3. Apply the shortcut: $\text{Var}(X) = 120 - 8^2 = 120 - 64 = 56$.

Answer: \text{Var}(X) = 56

1. The variance is defined as $\sum (x - E[X])^2 P(X=x)$.
2. Since $(x - E[X])^2$ is always non-negative and $P(X=x)$ is non-negative, the sum can only be zero if every term is zero.
3. This means for every $x$ where $P(X=x) > 0$, we must have $x = E[X]$.
4. Therefore, $X$ must be a constant random variable that takes only one value with probability $1$.

Answer: X is a constant; $P(X=E[X]) = 1$.

1. Identify parameters: $n = 6$, $p = 1/6$, $k = 2$.
2. Use the binomial formula: $P(X=2) = \binom{6}{2} (1/6)^2 (5/6)^{6-2}$.
3. Calculate the combination: $\binom{6}{2} = 15$.
4. Calculate the powers: $(1/6)^2 = 1/36$ and $(5/6)^4 = 625/1296$.
5. Multiply together: $15 \cdot (1/36) \cdot (625/1296) = 9375 / 46656 \approx 0.2009$.

Answer: 0.2009

1. Identify parameters: $n = 20$, $p = 0.03$.
2. Calculate mean: $E[X] = np = 20 \cdot 0.03 = 0.6$.
3. Calculate variance: $\text{Var}(X) = np(1-p) = 20 \cdot 0.03 \cdot 0.97$.
4. Compute the product: $0.6 \cdot 0.97 = 0.582$.

Answer: E[X] = 0.6, \text{Var}(X) = 0.582

1. Identify parameters: $n = 10$, $p = 1/4 = 0.25$, $k = 3$.
2. Use the binomial formula: $P(X=3) = \binom{10}{3} (0.25)^3 (0.75)^7$.
3. Calculate the combination: $\binom{10}{3} = 120$.
4. Calculate the powers: $(0.25)^3 = 0.015625$ and $(0.75)^7 \approx 0.13348$.
5. Multiply together: $120 \cdot 0.015625 \cdot 0.13348 \approx 0.2503$.

Answer: 0.2503

1. Identify parameters: $n=15$, $p=0.1$, $k=2$.
2. Apply the binomial formula: $P(X=2) = \binom{15}{2} (0.1)^2 (0.9)^{13}$.
3. Compute the combination: $\binom{15}{2} = \frac{15 \times 14}{2} = 105$.
4. Compute the powers: $(0.1)^2 = 0.01$ and $(0.9)^{13} \approx 0.2542$.
5. Multiply the results: $105 \times 0.01 \times 0.2542 \approx 0.2669$.

Answer: 0.2669

1. Identify parameters: $n=6$, $p=0.7$.
2. We need $P(X \ge 5) = P(X=5) + P(X=6)$.
3. Calculate $P(X=5) = \binom{6}{5} (0.7)^5 (0.3)^1 = 6 \times 0.16807 \times 0.3 = 0.3025$.
4. Calculate $P(X=6) = \binom{6}{6} (0.7)^6 (0.3)^0 = 1 \times 0.1176 \times 1 = 0.1176$.
5. Sum the probabilities: $0.3025 + 0.1176 = 0.4201$.

Answer: 0.4201

1. Identify parameters: $n=100$, $p=0.05$.
2. Calculate expectation: $E[X] = np = 100 \times 0.05 = 5$.
3. Calculate variance: $\text{Var}(X) = np(1-p) = 100 \times 0.05 \times 0.95 = 4.75$.
4. Calculate standard deviation: $\sigma = \sqrt{4.75} \approx 2.1794$.

Answer: E[X]=5, \sigma \approx 2.1794

1. Identify the probability of success $p = 0.5$ and the trial number $k = 4$.
2. Use the geometric PMF: $P(X=4) = (1-0.5)^{4-1} \cdot 0.5$.
3. Calculate $(0.5)^3 \cdot 0.5 = 0.125 \cdot 0.5 = 0.0625$.

Answer: 0.0625

1. Identify $p = 0.05$ and $k = 10$.
2. Use the cumulative formula: $P(X \le 10) = 1 - (1-0.05)^{10}$.
3. Calculate $1 - (0.95)^{10} \approx 1 - 0.5987 = 0.4013$.

Answer: 0.4013

1. The expected value is $E[X] = 1/p = 1/0.3 \approx 3.33$.
2. The probability of taking more than 5 shots is the probability of failing the first 5 shots: $P(X > 5) = (1-0.3)^5$.
3. Calculate $(0.7)^5 = 0.16807$.

Answer: E[X] \approx 3.33, P(X > 5) \approx 0.1681

1. Identify parameters: $r = 4$, $k = 7$, and $p = 0.3$.
2. Use the negative binomial formula: $P(X=7) = \binom{7-1}{4-1} (0.3)^4 (1-0.3)^{7-4}$.
3. Calculate the combination: $\binom{6}{3} = 20$.
4. Calculate the powers: $(0.3)^4 = 0.0081$ and $(0.7)^3 = 0.343$.
5. Multiply together: $20 \cdot 0.0081 \cdot 0.343 = 0.055566$.

Answer: 0.0556

1. Identify parameters: $r = 6$ and $p = 0.2$.
2. Use the expected value formula for the negative binomial distribution: $E[X] = \frac{r}{p}$.
3. Substitute the values: $E[X] = \frac{6}{0.2}$.
4. Calculate the result: $6 \div 0.2 = 30$.

Answer: 30

1. Identify parameters: $r = 3$, $k = 6$, and $p = 0.5$.
2. Use the formula: $P(X=6) = \binom{6-1}{3-1} (0.5)^3 (0.5)^{6-3}$.
3. Calculate the combination: $\binom{5}{2} = 10$.
4. Calculate the powers: $(0.5)^3 \cdot (0.5)^3 = (0.5)^6 = 0.015625$.
5. Multiply together: $10 \cdot 0.015625 = 0.15625$.

Answer: 0.1563

1. Identify the rate $\lambda = 3.5$ and the number of events $k = 2$.
2. Apply the Poisson formula: $P(X=2) = \frac{3.5^2 e^{-3.5}}{2!}$.
3. Calculate $3.5^2 = 12.25$ and $2! = 2$.
4. Compute $P(X=2) = \frac{12.25 \times 0.0302}{2} \approx 0.1850$.

Answer: 0.1850

1. First, scale the rate to the $5$-page interval: $\lambda = 0.2 \times 5 = 1$.
2. Identify $k = 1$.
3. Apply the formula: $P(X=1) = \frac{1^1 e^{-1}}{1!}$.
4. Simplify to $e^{-1} \approx 0.3679$.

Answer: 0.3679

1. Identify $\lambda = 4$. We want $P(X > 1)$.
2. Use the complement: $P(X > 1) = 1 - [P(X=0) + P(X=1)]$.
3. Calculate $P(X=0) = \frac{4^0 e^{-4}}{0!} = e^{-4} \approx 0.0183$.
4. Calculate $P(X=1) = \frac{4^1 e^{-4}}{1!} = 4 e^{-4} \approx 0.0733$.
5. Sum the probabilities: $0.0183 + 0.0733 = 0.0916$.
6. Subtract from one: $1 - 0.0916 = 0.9084$.

Answer: 0.9084

1. Identify $n = 500$ and $p = 0.004$.
2. Calculate $\lambda = np = 500 \times 0.004 = 2$.
3. Use the Poisson PMF for $k = 2$: $P(X=2) = \frac{2^2 e^{-2}}{2!}$.
4. Compute $P(X=2) = \frac{4 \times 0.1353}{2} = 0.2707$.

Answer: 0.2707

1. Identify $n = 2000$ and $p = 0.001$.
2. Calculate $\lambda = np = 2000 \times 0.001 = 2$.
3. Find $P(X \le 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)$.
4. Calculate each: $P(0) = e^{-2} \approx 0.1353$, $P(1) = 2e^{-2} \approx 0.2707$, $P(2) = \frac{4e^{-2}}{2} \approx 0.2707$, $P(3) = \frac{8e^{-2}}{6} \approx 0.1804$.
5. Sum them: $0.1353 + 0.2707 + 0.2707 + 0.1804 = 0.8571$.

Answer: 0.8571

1. For Poisson: $n = 10,000, p = 0.0001 \implies \lambda = 1$.
2. Calculate $P(X=1) = \frac{1^1 e^{-1}}{1!} = e^{-1} \approx 0.3679$.
3. For Binomial: $P(X=1) = \binom{10000}{1} (0.0001)^1 (0.9999)^{9999}$.
4. Calculate $10000 \times 0.0001 \times (0.9999)^{9999} = 1 \times 0.36788 \approx 0.3679$.
5. The results are nearly identical.

Answer: 0.3679

1. Identify parameters: $N=15$, $K=3$, $n=4$, $k=1$.
2. Calculate ways to pick 1 defective: $\binom{3}{1} = 3$.
3. Calculate ways to pick 3 non-defective: $\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$.
4. Calculate total ways to pick 4 bulbs: $\binom{15}{4} = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1365$.
5. Compute probability: $P(X=1) = \frac{3 \times 220}{1365} = \frac{660}{1365} \approx 0.4835$.

Answer: 0.4835

1. Identify parameters: $N=12$, $K=5$, $n=3$. We need $P(X \ge 2) = P(X=2) + P(X=3)$.
2. For $k=2$: $P(X=2) = \frac{\binom{5}{2} \binom{7}{1}}{\binom{12}{3}} = \frac{10 \times 7}{220} = \frac{70}{220}$.
3. For $k=3$: $P(X=3) = \frac{\binom{5}{3} \binom{7}{0}}{\binom{12}{3}} = \frac{10 \times 1}{220} = \frac{10}{220}$.
4. Sum the probabilities: $\frac{70}{220} + \frac{10}{220} = \frac{80}{220} \approx 0.3636$.

Answer: 0.3636

1. Identify parameters: $N=20$, $K=6$, $n=5$, $k=3$.
2. Calculate ways to pick 3 red: $\binom{6}{3} = 20$.
3. Calculate ways to pick 2 black: $\binom{14}{2} = \frac{14 \times 13}{2} = 91$.
4. Calculate total ways to pick 5 cards: $\binom{20}{5} = 15504$.
5. Compute probability: $P(X=3) = \frac{20 \times 91}{15504} = \frac{1820}{15504} \approx 0.1174$.

Answer: 0.1174

1. Identify $n=3$.
2. Calculate the 3rd harmonic number: $H_3 = 1 + 1/2 + 1/3 = 11/6$.
3. Multiply by $n$: $E[T] = 3 \times (11/6) = 11/2 = 5.5$.

Answer: 5.5

1. You need the 4th and 5th characters.
2. The probability of getting the 4th is $p_4 = (5-3)/5 = 2/5$. Expected time $E[T_4] = 5/2 = 2.5$.
3. The probability of getting the 5th is $p_5 = (5-4)/5 = 1/5$. Expected time $E[T_5] = 5/1 = 5$.
4. Total additional time is $2.5 + 5 = 7.5$.

Answer: 7.5

1. Substitute $n=100$ into the approximation: $H_{100} \approx \ln(100) + 0.577$.
2. Calculate $\ln(100) \approx 4.605$.
3. Sum them: $H_{100} \approx 4.605 + 0.577 = 5.182$.
4. Multiply by $n$: $E[T] \approx 100 \times 5.182 = 518.2$.

Answer: 518.2

1. Identify parameters: $n=12$, $p=1/6$, $k=3$.
2. Use Binomial PMF: $P(X=3) = \binom{12}{3} (1/6)^3 (5/6)^9$.
3. Compute $\binom{12}{3} = 220$.
4. Compute $(1/6)^3 \approx 0.00463$ and $(5/6)^9 \approx 0.1938$.
5. Multiply: $220 \times 0.00463 \times 0.1938 \approx 0.1974$.

Answer: 0.1974

1. Identify as a Poisson process with $\lambda = 3$.
2. We want $P(X=5)$.
3. Use Poisson PMF: $P(X=5) = \frac{3^5 e^{-3}}{5!}$.
4. Compute $3^5 = 243$ and $5! = 120$.
5. Compute $e^{-3} \approx 0.0498$.
6. Calculate: $(243 \times 0.0498) / 120 \approx 0.1008$.

Answer: 0.1008

1. Identify as Hypergeometric: $N=18$, $K=8$ (women), $n=5$, $k=3$.
2. Use formula: $P(X=3) = \frac{\binom{8}{3} \binom{10}{2}}{\binom{18}{5}}$.
3. Compute $\binom{8}{3} = 56$ and $\binom{10}{2} = 45$.
4. Compute $\binom{18}{5} = 8568$.
5. Calculate: $(56 \times 45) / 8568 = 2520 / 8568 \approx 0.2941$.

Answer: 0.2941

1. Check non-negativity: $x^2$ is always non-negative, so $f(x) \ge 0$.
2. Check total area: $\int_{0}^{2} \frac{3}{8}x^2 \, dx = [\frac{1}{8}x^3]_{0}^{2} = \frac{8}{8} - 0 = 1$.
3. Compute probability: $P(1 \le X \le 2) = \int_{1}^{2} \frac{3}{8}x^2 \, dx = [\frac{1}{8}x^3]_{1}^{2} = \frac{8}{8} - \frac{1}{8} = \frac{7}{8}$.

Answer: 0.875

1. Set the total integral to 1: $\int_{0}^{2} c(4x - 2x^2) \, dx = 1$.
2. Integrate: $c [2x^2 - \frac{2}{3}x^3]_{0}^{2} = 1$.
3. Evaluate: $c (2(4) - \frac{2}{3}(8)) = c (8 - \frac{16}{3}) = c (\frac{24-16}{3}) = c \frac{8}{3}$.
4. Solve for $c$: $c \frac{8}{3} = 1 \implies c = \frac{3}{8}$.

Answer: 3/8

1. Set up the integral: $P(0 \le X \le 1) = \int_{0}^{1} \frac{1}{\pi(1+x^2)} \, dx$.
2. Recognize the antiderivative: $\int \frac{1}{1+x^2} \, dx = \arctan(x)$.
3. Evaluate: $\frac{1}{\pi} [\arctan(x)]_{0}^{1} = \frac{1}{\pi} (\arctan(1) - \arctan(0))$.
4. Substitute values: $\frac{1}{\pi} (\frac{\pi}{4} - 0) = \frac{1}{4}$.

Answer: 0.25

1. For $x < 0$, $F(x) = 0$.
2. For $0 \le x \le 1$, $F(x) = \int_{0}^{x} 3t^2 \, dt$.
3. The antiderivative of $3t^2$ is $t^3$.
4. Evaluating from $0$ to $x$ gives $x^3 - 0^3 = x^3$.
5. For $x > 1$, $F(x) = 1$.

Answer: F(x) = x^3 \text{ for } 0 \le x \le 1

1. The PDF is the derivative of the CDF: $f(x) = \frac{d}{dx}(1 - e^{-x}) = e^{-x}$.
2. The probability $P(1 \le X \le 2) = F(2) - F(1)$.
3. Substitute values: $(1 - e^{-2}) - (1 - e^{-1})$.
4. Simplify: $e^{-1} - e^{-2} \approx 0.3679 - 0.1353 = 0.2326$.

Answer: f(x) = e^{-x}, P(1 \le X \le 2) \approx 0.2326

1. First, find the CDF $F(x) = \int_{0}^{x} \frac{1}{2}t \, dt = [\frac{1}{4}t^2]_{0}^{x} = \frac{1}{4}x^2$.
2. We want to find $m$ such that $F(m) = 0.5$.
3. Set up the equation: $\frac{1}{4}m^2 = 0.5$.
4. Multiply by 4: $m^2 = 2$.
5. Solve for $m$: $m = \sqrt{2} \approx 1.414$.

Answer: m = \sqrt{2} \approx 1.414

1. Compute the mean: $E[X] = \int_{0}^{1} x(\frac{3}{2}x^2) \, dx = \int_{0}^{1} \frac{3}{2}x^3 \, dx = [\frac{3}{8}x^4]_{0}^{1} = \frac{3}{8}$.
2. Compute the second moment: $E[X^2] = \int_{0}^{1} x^2(\frac{3}{2}x^2) \, dx = \int_{0}^{1} \frac{3}{2}x^4 \, dx = [\frac{3}{10}x^5]_{0}^{1} = \frac{3}{10}$.
3. Compute variance: $Var(X) = E[X^2] - (E[X])^2 = \frac{3}{10} - (\frac{3}{8})^2 = \frac{3}{10} - \frac{9}{64}$.
4. Find common denominator: $\frac{96}{320} - \frac{45}{320} = \frac{51}{320}$.

Answer: E[X] = \frac{3}{8}, Var(X) = \frac{51}{320}

1. Set up the integral: $E[X] = \int_{0}^{1} x(2 - 2x) \, dx$.
2. Expand the integrand: $\int_{0}^{1} (2x - 2x^2) \, dx$.
3. Integrate: $[x^2 - \frac{2}{3}x^3]_{0}^{1}$.
4. Evaluate: $1 - \frac{2}{3} = \frac{1}{3}$.

Answer: E[X] = \frac{1}{3}

1. Find $E[X]$ using integration by parts: $\int_{0}^{\infty} x \frac{1}{2}e^{-x/2} \, dx = 2$.
2. Find $E[X^2]$ using integration by parts twice: $\int_{0}^{\infty} x^2 \frac{1}{2}e^{-x/2} \, dx = 8$.
3. Apply variance formula: $Var(X) = E[X^2] - (E[X])^2 = 8 - 2^2 = 8 - 4 = 4$.

Answer: Var(X) = 4

1. Identify the parameters: $a = 5$ and $b = 15$.
2. Calculate the total width: $b - a = 15 - 5 = 10$.
3. Calculate the target width: $12 - 8 = 4$.
4. The probability is the ratio of the target width to the total width: $4 / 10 = 0.4$.

Answer: 0.4

1. The mean is the average of the endpoints: $E[X] = (10 + 20) / 2 = 30 / 2 = 15$.
2. The variance is the square of the width divided by 12: $Var(X) = (20 - 10)^2 / 12$.
3. Compute the square: $10^2 = 100$.
4. Divide by 12: $100 / 12 = 25 / 3 \approx 8.33$.

Answer: E[X] = 15, Var(X) = 8.33

1. From the mean: $(a + b) / 2 = 10 \implies a + b = 20$.
2. From the variance: $(b - a)^2 / 12 = 3 \implies (b - a)^2 = 36$.
3. Taking the square root: $b - a = 6$.
4. Solve the system: $a + b = 20$ and $b - a = 6$. Adding the equations gives $2b = 26$, so $b = 13$.
5. Subtracting gives $2a = 14$, so $a = 7$.

Answer: a = 7, b = 13

1. Identify the mean $E[X] = 100$.
2. Calculate the rate $\lambda = 1/100 = 0.01$.
3. Use the survival function $P(X > x) = e^{-\lambda x}$.
4. Substitute $x = 150$ and $\lambda = 0.01$: $P(X > 150) = e^{-0.01 \cdot 150} = e^{-1.5}$.

Answer: $e^{-1.5} \approx 0.2231$

1. Convert the time to minutes to match the rate: $30$ seconds $= 0.5$ minutes.
2. Use the CDF $F(x) = 1 - e^{-\lambda x}$.
3. Substitute $\lambda = 2$ and $x = 0.5$: $F(0.5) = 1 - e^{-2 \cdot 0.5} = 1 - e^{-1}$.

Answer: $1 - e^{-1} \approx 0.6321$

1. Identify the mean $E[X] = 500$, so $\lambda = 1/500$.
2. Recognize the memoryless property: $P(X > 200 + 500 \mid X > 200) = P(X > 500)$.
3. Calculate $P(X > 500) = e^{-\lambda \cdot 500} = e^{-(1/500) \cdot 500} = e^{-1}$.

Answer: $e^{-1} \approx 0.3679$

1. Identify the rate $\lambda = 2$ crashes per month.
2. We want the probability that the time until the first crash $T$ is greater than $3$.
3. Use the exponential survival function $P(T > t) = e^{-\lambda t}$.
4. Substitute $\lambda = 2$ and $t = 3$ to get $e^{-(2)(3)} = e^{-6}$.

Answer: $e^{-6} \approx 0.0025$

1. Convert the rate to the given time unit: $\lambda = 12$ per hour is $3$ per $15$ minutes.
2. Alternatively, $\mu = \lambda t = 12 \times (15/60) = 3$.
3. Use the Poisson PMF for $k = 4$ and $\mu = 3$.
4. Calculate $P(X=4) = \frac{e^{-3} 3^4}{4!} = \frac{81 e^{-3}}{24} = \frac{27}{8} e^{-3}$.

Answer: \frac{27}{8} e^{-3} \approx 0.1680

1. The second failure occurs after $20$ years if and only if there are $0$ or $1$ failures in the first $20$ years.
2. Calculate the expected number of failures: $\mu = \lambda t = 0.1 \times 20 = 2$.
3. Use the Poisson distribution to find $P(X=0) + P(X=1)$.
4. Compute $e^{-2} \frac{2^0}{0!} + e^{-2} \frac{2^1}{1!} = e^{-2} + 2e^{-2} = 3e^{-2}$.

Answer: $3e^{-2} \approx 0.4060$

1. The empirical rule states that the middle $95\%$ of a normal distribution lies within 2 standard deviations of the mean.
2. Calculate the lower bound: $50 - 2(4) = 50 - 8 = 42$.
3. Calculate the upper bound: $50 + 2(4) = 50 + 8 = 58$.

Answer: [42, 58]

1. First, find the z-score for $174$ grams: $z = (174 - 150) / 12 = 24 / 12 = 2$.
2. The value $174$ is exactly 2 standard deviations above the mean.
3. The empirical rule says $95\%$ of data is within 2 standard deviations, meaning $5\%$ is outside this range.
4. Because the curve is symmetric, half of that $5\%$ is in the upper tail: $5\% / 2 = 2.5\%$.

Answer: 2.5%

1. First, find the standard deviation $\sigma$ from the variance: $\sigma = \sqrt{0.0016} = 0.04$ mm.
2. The acceptable range is $\mu \pm 3\sigma$.
3. Lower bound: $10.00 - 3(0.04) = 10.00 - 0.12 = 9.88$ mm.
4. Upper bound: $10.00 + 3(0.04) = 10.00 + 0.12 = 10.12$ mm.

Answer: [9.88, 10.12]

1. The normalization constant is given by the formula $\frac{1}{\sigma\sqrt{2\pi}}$
2. Substitute $\sigma = \sqrt{2}$ into the formula: $\frac{1}{\sqrt{2}\sqrt{2\pi}}$
3. Simplify the denominator: $\sqrt{2} \cdot \sqrt{2} \cdot \sqrt{\pi} = 2\sqrt{\pi}$
4. The constant is $\frac{1}{2\sqrt{\pi}}$

Answer: $\frac{1}{2\sqrt{\pi}}$

1. The maximum height of the PDF occurs at $x = \mu$ and is equal to the constant $\frac{1}{\sigma\sqrt{2\pi}}$
2. Let the original height be $H_1 = \frac{1}{\sigma\sqrt{2\pi}}$
3. Let the new standard deviation be $\sigma' = 2\sigma$
4. The new height is $H_2 = \frac{1}{2\sigma\sqrt{2\pi}} = \frac{1}{2} H_1$

Answer: The maximum height is halved.

1. Start with the integral $I = \int_{-\infty}^{\infty} \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} \, dx$
2. Apply substitution $u = \frac{x-\mu}{\sigma\sqrt{2}}$, which means $dx = \sigma\sqrt{2} \, du$
3. Substitute into the integral: $I = \int_{-\infty}^{\infty} \frac{1}{\sigma\sqrt{2\pi}} e^{-u^2} (\sigma\sqrt{2} \, du)$
4. Cancel $\sigma$ and simplify the constants: $I = \frac{\sqrt{2}}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-u^2} \, du = \frac{1}{\sqrt{\pi}} \int_{-\infty}^{\infty} e^{-u^2} \, du$
5. Using the Gaussian integral result $\int_{-\infty}^{\infty} e^{-u^2} \, du = \sqrt{\pi}$, we get $I = \frac{1}{\sqrt{\pi}} \cdot \sqrt{\pi} = 1$

Answer: The integral equals 1.

1. Calculate the z-score: $z = \frac{62 - 50}{5} = \frac{12}{5} = 2.4$.
2. Find the cumulative probability from the z-table: $P(Z < 2.4) \approx 0.9918$.
3. Since we want the probability of being greater than 62, subtract from 1: $P(X > 62) = 1 - 0.9918 = 0.0082$.

Answer: 0.0082

1. Determine how many standard deviations 70 and 130 are from the mean.
2. For 70: $z = \frac{70 - 100}{15} = -2$.
3. For 130: $z = \frac{130 - 100}{15} = 2$.
4. The empirical rule states that approximately 95.45% of data falls within 2 standard deviations.

Answer: 0.9545

1. Find the z-score that corresponds to a cumulative area of 0.025 in the z-table.
2. The z-score for 0.025 is $z = -1.96$.
3. Use the formula $x = \mu + z\sigma$.
4. Calculate $x = 200 + (-1.96)(20) = 200 - 39.2 = 160.8$.

Answer: 160.8

1. Calculate the z-score: $z = (165 - 150) / 10 = 1.5$.
2. Find the cumulative probability for $z = 1.5$ from the table: $P(Z < 1.5) = 0.9332$.
3. Since we want the probability of weighing more, use the complement: $1 - 0.9332 = 0.0668$.

Answer: 0.0668

1. Calculate z-scores for both bounds: $z_1 = (42 - 50) / 5 = -1.6$ and $z_2 = (58 - 50) / 5 = 1.6$.
2. Find cumulative probabilities: $P(Z < 1.6) = 0.9452$ and $P(Z < -1.6) = 0.0548$.
3. Subtract the lower from the upper: $0.9452 - 0.0548 = 0.8904$.

Answer: 0.8904

1. The top $2.5\%$ means the area to the left is $1 - 0.025 = 0.975$.
2. Find the z-score for $0.975$ in the table: $z = 1.96$.
3. Use the inverse formula: $x = 100 + (1.96)(15)$.
4. Calculate: $100 + 29.4 = 129.4$.

Answer: 129.4

1. The PDF of $X$ is $f_X(x) = 1$ for $0 \le x \le 1$.
2. The range of $Y$ is $[0^3, 1^3] = [0, 1]$.
3. The CDF of $Y$ is $F_Y(y) = P(X^3 \le y) = P(X \le y^{1/3}) = y^{1/3}$.
4. The PDF is the derivative: $f_Y(y) = \frac{d}{dy} y^{1/3} = \frac{1}{3} y^{-2/3}$.

Answer: f_Y(y) = \frac{1}{3y^{2/3}} \text{ for } 0 < y \le 1

1. The PDF of $X$ is $f_X(x) = 1/2$ for $0 \le x \le 2$.
2. The range of $Y$ is $[e^0, e^2] = [1, e^2]$.
3. The CDF of $Y$ is $F_Y(y) = P(e^X \le y) = P(X \le \ln y)$.
4. Since $X$ is uniform on $[0, 2]$, $F_X(x) = x/2$. Thus $F_Y(y) = \frac{\ln y}{2}$.
5. The PDF is $f_Y(y) = \frac{d}{dy} \frac{\ln y}{2} = \frac{1}{2y}$.

Answer: f_Y(y) = \frac{1}{2y} \text{ for } 1 \le y \le e^2

1. The PDF of $X$ is $f_X(x) = 1$ for $1 \le x \le 2$.
2. The range of $Y$ is $[1/2, 1/1] = [1/2, 1]$.
3. The CDF of $Y$ is $F_Y(y) = P(1/X \le y) = P(X \ge 1/y)$.
4. This is $1 - P(X < 1/y) = 1 - F_X(1/y)$.
5. Since $F_X(x) = x - 1$ for $x \in [1, 2]$, we have $F_Y(y) = 1 - (1/y - 1) = 2 - 1/y$.
6. The PDF is $f_Y(y) = \frac{d}{dy} (2 - 1/y) = 1/y^2$.

Answer: f_Y(y) = \frac{1}{y^2} \text{ for } 1/2 \le y \le 1

1. The PDF is $f(x) = \frac{0.5^2}{\Gamma(2)} x^{2-1} e^{-0.5x} = 0.25 x e^{-0.5x}$.
2. Integrate $\int_0^1 0.25 x e^{-0.5x} dx$ using integration by parts.
3. Let $u = 0.25x$ and $dv = e^{-0.5x} dx$, so $du = 0.25 dx$ and $v = -2e^{-0.5x}$.
4. The integral is $[-0.5x e^{-0.5x}]_0^1 - \int_0^1 -0.5 e^{-0.5x} dx$.
5. This evaluates to $-0.5 e^{-0.5} + [-e^{-0.5x}]_0^1 = -0.5 e^{-0.5} - e^{-0.5} + 1 = 1 - 1.5 e^{-0.5}$.

Answer: 1 - 1.5 e^{-0.5} \approx 0.0902

1. First find $B(3, 2) = \frac{\Gamma(3)\Gamma(2)}{\Gamma(5)} = \frac{2 \cdot 1}{24} = \frac{1}{12}$.
2. The PDF is $f(x) = 12 x^{3-1} (1-x)^{2-1} = 12 x^2 (1-x) = 12x^2 - 12x^3$.
3. Integrate from $0.5$ to $1$: $\int_{0.5}^1 (12x^2 - 12x^3) dx$.
4. The antiderivative is $4x^3 - 3x^4$.
5. Evaluate at boundaries: $(4(1)^3 - 3(1)^4) - (4(0.5)^3 - 3(0.5)^4) = 1 - (0.5 - 0.1875) = 1 - 0.3125 = 0.6875$.

Answer: 0.6875

1. The sum of $k$ independent exponential variables with rate $\beta$ follows a Gamma distribution with shape $\alpha = k$ and rate $\beta$.
2. Here $k = 3$ and $\beta = 4$.
3. The PDF is $f(y) = \frac{4^3}{\Gamma(3)} y^{3-1} e^{-4y}$.
4. Since $\Gamma(3) = 2! = 2$, the constant is $64 / 2 = 32$.
5. The final PDF is $f(y) = 32 y^2 e^{-4y}$ for $y > 0$.

Answer: f(y) = 32 y^2 e^{-4y}

1. Set the CDF equal to 0.75: $1 - e^{-2x} = 0.75$.
2. Rearrange to isolate the exponential: $e^{-2x} = 0.25$.
3. Take the natural logarithm of both sides: $-2x = \ln(0.25)$.
4. Solve for $x$: $x = \frac{\ln(0.25)}{-2} = \frac{-1.3863}{-2} \approx 0.6931$.

Answer: 0.6931

1. The CDF for a uniform distribution is $F(x) = \frac{x - a}{b - a}$.
2. Substitute $a = 20$ and $b = 100$: $F(x) = \frac{x - 20}{80}$.
3. Set $F(x) = 0.5$ for the median: $\frac{x - 20}{80} = 0.5$.
4. Multiply by 80: $x - 20 = 40$.
5. Solve for $x$: $x = 60$.

Answer: 60

1. The condition $P(X > x) = 0.10$ is equivalent to $P(X \le x) = 0.90$.
2. Set the CDF equal to 0.90: $1 - e^{-x} = 0.90$.
3. Isolate the exponential: $e^{-x} = 0.10$.
4. Take the natural logarithm: $-x = \ln(0.10)$.
5. Solve for $x$: $x = -(-2.3026) = 2.3026$.

Answer: 2.3026

1. Set the total integral equal to 1: $\int_{1}^{2} k x^3 \, dx = 1$.
2. Compute the integral: $k [\frac{1}{4} x^4]_{1}^{2} = 1$.
3. Evaluate the bounds: $k (\frac{16}{4} - \frac{1}{4}) = 1$.
4. Solve for $k$: $k (\frac{15}{4}) = 1 \implies k = \frac{4}{15}$.

Answer: k = 4/15

1. The CDF of an exponential distribution is $F(x) = 1 - e^{-\lambda x}$.
2. Set $F(m) = 0.5$: $1 - e^{-0.5 m} = 0.5$.
3. Rearrange: $e^{-0.5 m} = 0.5$.
4. Take the natural log: $-0.5 m = \ln(0.5)$.
5. Solve for $m$: $m = \frac{\ln(0.5)}{-0.5} = -2 \ln(0.5) = 2 \ln(2) \approx 1.386$.

Answer: m \approx 1.386

1. Find the CDF of $Y$: $F_Y(y) = P(X^2 \le y) = P(X \le \sqrt{y})$.
2. Since $X$ is uniform on $[0, 1]$, $P(X \le \sqrt{y}) = \sqrt{y}$ for $0 \le y \le 1$.
3. Differentiate the CDF to find the PDF: $f_Y(y) = \frac{d}{dy} \sqrt{y}$.
4. The result is $f_Y(y) = \frac{1}{2\sqrt{y}}$ for $0 < y \le 1$.

Answer: f_Y(y) = \frac{1}{2\sqrt{y}}

1. For $x=1$, sum $p(1, y)$ for $y=1, 2, 3$: $\frac{1+1}{21} + \frac{1+2}{21} + \frac{1+3}{21} = \frac{2+3+4}{21} = \frac{9}{21} = 3/7$.
2. For $x=2$, sum $p(2, y)$ for $y=1, 2, 3$: $\frac{2+1}{21} + \frac{2+2}{21} + \frac{2+3}{21} = \frac{3+4+5}{21} = \frac{12}{21} = 4/7$.

Answer: $p_X(1) = 3/7, p_X(2) = 4/7$

1. The region is a triangle with vertices $(0,0), (1,0), (1,1)$.
2. To find $f_Y(y)$, we integrate $f(x, y)$ with respect to $x$.
3. For a fixed $y$, $x$ ranges from $y$ to $1$.
4. $\int_{y}^{1} 2 \, dx = [2x]_{y}^{1} = 2(1 - y)$.
5. The marginal PDF is $f_Y(y) = 2(1-y)$ for $0 \le y \le 1$.

Answer: $f_Y(y) = 2(1-y)$ for $0 \le y \le 1$

1. We integrate the joint PDF over the region where $0 \le x \le 0.5$ and $0 \le y \le x$.
2. Inner integral: $\int_{0}^{x} 8xy \, dy = [4xy^2]_{0}^{x} = 4x^3$.
3. Outer integral: $\int_{0}^{0.5} 4x^3 \, dx = [x^4]_{0}^{0.5} = (0.5)^4 = 0.0625$.

Answer: 0.0625

1. Sum the rows to find $p_X(x)$: $p_X(0) = 0.1+0.1+0.1 = 0.3$, $p_X(1) = 0.1+0.2+0.1 = 0.4$, $p_X(2) = 0.1+0.1+0.1 = 0.3$.
2. Check sum: $0.3+0.4+0.3 = 1.0$.
3. For $p(X=1 | Y=1)$, find the marginal $p_Y(1) = p(0,1)+p(1,1)+p(2,1) = 0.1+0.2+0.1 = 0.4$.
4. Then $p(X=1 | Y=1) = p(1,1) / p_Y(1) = 0.2 / 0.4 = 0.5$.

Answer: p_X(0)=0.3, p_X(1)=0.4, p_X(2)=0.3; p(X=1|Y=1)=0.5

1. Integrate $f(x, y)$ over $y$: $f_X(x) = \int_{0}^{1} 4xy \, dy = [2xy^2]_{0}^{1} = 2x$.
2. Check: $\int_{0}^{1} 2x \, dx = [x^2]_{0}^{1} = 1$. Correct.
3. Find conditional: $f_{Y|X}(y | x) = f(x, y) / f_X(x) = (4xy) / (2x) = 2y$.

Answer: f_X(x)=2x, f_{Y|X}(y|x)=2y

1. Since $X$ and $Y$ are independent, the conditional distribution of $X$ given $Y$ is simply the marginal distribution of $X$.
2. Therefore, $f_{X|Y}(x | y) = f_X(x) = e^{-x}$ for $x > 0$.

Answer: f_{X|Y}(x|y)=e^{-x}

1. Compute marginal $p_X(0) = 0.2 + 0.2 = 0.4$ and $p_X(1) = 0.1 + 0.5 = 0.6$.
2. Compute marginal $p_Y(0) = 0.2 + 0.1 = 0.3$ and $p_Y(1) = 0.2 + 0.5 = 0.7$.
3. Check $p(0,0) = 0.2$. The product $p_X(0) p_Y(0) = 0.4 \times 0.3 = 0.12$.
4. Since $0.2 \neq 0.12$, the variables are dependent.

Answer: Dependent

1. Find marginal $f_X(x) = \int_0^1 6x^2 y \, dy = [3x^2 y^2]_0^1 = 3x^2$.
2. Find marginal $f_Y(y) = \int_0^1 6x^2 y \, dx = [2x^3 y]_0^1 = 2y$.
3. Check product: $f_X(x) f_Y(y) = 3x^2 \times 2y = 6x^2 y$.
4. The product equals the joint PDF and the support is a rectangle, so they are independent.

Answer: Independent

1. Identify the support of the distribution: $S = \{ (x, y) : 0 < x < 1, x < y < 1 \}$.
2. Observe that the range of $y$ depends on the value of $x$ (specifically, $y$ must be greater than $x$).
3. Because the support is a triangle and not a rectangle aligned with the axes, the variables are dependent.

Answer: Dependent

1. Find marginals: $p_X(0)=0.5, p_X(1)=0.5$ and $p_Y(0)=0.3, p_Y(1)=0.7$.
2. Compute means: $E[X]=0.5$ and $E[Y]=0.7$.
3. Compute $E[XY]$: only $p(1,1)$ contributes, so $E[XY]=1 \cdot 1 \cdot 0.4 = 0.4$.
4. Apply formula: $\text{Cov}(X, Y) = 0.4 - (0.5)(0.7) = 0.4 - 0.35 = 0.05$.

Answer: 0.05

1. Recall that for independent variables, $E[XY] = E[X]E[Y]$.
2. Substitute this into the covariance formula: $\text{Cov}(X, Y) = E[X]E[Y] - E[X]E[Y] = 0$.

Answer: 0

1. Find marginal $f_X(x) = \int_0^1 4xy \, dy = [2xy^2]_0^1 = 2x$.
2. Compute $E[X] = \int_0^1 x(2x) \, dx = [2x^3/3]_0^1 = 2/3$. By symmetry, $E[Y] = 2/3$.
3. Compute $E[XY] = \int_0^1 \int_0^1 xy(4xy) \, dx \, dy = \int_0^1 \int_0^1 4x^2y^2 \, dx \, dy = \int_0^1 [4x^3y^2/3]_0^1 \, dy = \int_0^1 4y^2/3 \, dy = [4y^3/9]_0^1 = 4/9$.
4. Apply formula: $\text{Cov}(X, Y) = 4/9 - (2/3)(2/3) = 4/9 - 4/9 = 0$.

Answer: 0

1. Use the formula $\rho = \frac{\text{Cov}(X, Y)}{\sigma_X \sigma_Y}$.
2. Substitute the values: $\rho = \frac{12}{4 \times 3}$.
3. Calculate the denominator: $4 \times 3 = 12$.
4. Divide: $12 / 12 = 1$.

Answer: 1

1. The correlation formula is $\rho = \frac{\text{Cov}(X, Y)}{\sigma_X \sigma_Y}$.
2. If $\sigma_X$ is doubled, the covariance $\text{Cov}(X, Y) = E[(X-\mu_X)(Y-\mu_Y)]$ also doubles because the deviations $(X-\mu_X)$ are doubled.
3. The new correlation is $\rho_{new} = \frac{2 \times \text{Cov}(X, Y)}{(2 \sigma_X) \sigma_Y}$.
4. The factor of 2 in the numerator and denominator cancels out.

Answer: The correlation remains -0.7.

1. The correlation is $\rho = \frac{\text{Cov}(X, Y)}{\sigma_X \sigma_Y}$.
2. $\text{Cov}(X, Y) = E[XY] - E[X]E[Y]$.
3. Since $E[X] = 0$, $\text{Cov}(X, Y) = E[X \cdot X^2] = E[X^3]$.
4. For a symmetric distribution about 0, the expected value of any odd power of $X$ is 0.
5. Therefore, $\text{Cov}(X, Y) = 0$, which implies $\rho = 0$.

Answer: \rho = 0

1. Identify possible values of $Z$: $0+1=1$, $0+2=2$, $1+1=2$, $1+2=3$. So $Z \in \{1, 2, 3\}$.
2. Compute $P(Z=1)$: Only $X=0$ and $Y=1$ works. $0.4 \times 0.7 = 0.28$.
3. Compute $P(Z=2)$: $X=0, Y=2$ or $X=1, Y=1$. $(0.4 \times 0.3) + (0.6 \times 0.7) = 0.12 + 0.42 = 0.54$.
4. Compute $P(Z=3)$: Only $X=1$ and $Y=2$ works. $0.6 \times 0.3 = 0.18$.
5. Check sum: $0.28 + 0.54 + 0.18 = 1.0$.

Answer: P(Z=1)=0.28, P(Z=2)=0.54, P(Z=3)=0.18

1. The total number of outcomes is $4 \times 4 = 16$.
2. List pairs $(x, y)$ that sum to $5$: $(1, 4), (2, 3), (3, 2), (4, 1)$.
3. There are $4$ such pairs.
4. The probability is $4/16 = 1/4$.

Answer: 0.25

1. Find PMF of $Z$: $P(Z=2)=0.1$, $P(Z=3)=0.4+0.3=0.7$, $P(Z=4)=0.2$.
2. Compute $E[Z] = (2 \times 0.1) + (3 \times 0.7) + (4 \times 0.2) = 0.2 + 2.1 + 0.8 = 3.1$.
3. Compute $E[Z^2] = (2^2 \times 0.1) + (3^2 \times 0.7) + (4^2 \times 0.2) = 0.4 + 6.3 + 3.2 = 9.9$.
4. Compute $\text{Var}(Z) = E[Z^2] - (E[Z])^2 = 9.9 - (3.1)^2 = 9.9 - 9.61 = 0.29$.

Answer: 0.29

1. Identify the formula for the variance of a sum: $\text{Var}(X+Y) = \text{Var}(X) + \text{Var}(Y) + 2\text{Cov}(X, Y)$.
2. Substitute the given values: $5 + 8 + 2(3)$.
3. Calculate the sum: $13 + 6 = 19$.

Answer: 19

1. Since $X$ and $Y$ are independent, their covariance $\text{Cov}(X, Y) = 0$.
2. The formula simplifies to $\text{Var}(X+Y) = \text{Var}(X) + \text{Var}(Y)$.
3. Substitute the values: $12 + 7 = 19$.

Answer: 19

1. First, find the covariance using $\text{Cov}(X, Y) = \rho \sigma_X \sigma_Y$.
2. Calculate standard deviations: $\sigma_X = \sqrt{9} = 3$ and $\sigma_Y = \sqrt{16} = 4$.
3. Compute covariance: $-0.5 \times 3 \times 4 = -6$.
4. Apply the variance of sums formula: $9 + 16 + 2(-6)$.
5. Calculate the final result: $25 - 12 = 13$.

Answer: 13

1. First, calculate the z-score of $X$: $z_X = (12 - 10) / 2 = 1$.
2. Use the conditional mean formula: $E[Y \mid X=x] = \mu_Y + \rho \sigma_Y z_X$.
3. Substitute the values: $50 + (-0.4) \cdot 10 \cdot 1$.
4. Calculate the result: $50 - 4 = 46$.

Answer: 46

1. Use the conditional variance formula: $\text{Var}(Y \mid X) = \sigma_Y^2 (1 - \rho^2)$.
2. Substitute the values: $10^2 \cdot (1 - (-0.4)^2)$.
3. Calculate the square: $100 \cdot (1 - 0.16)$.
4. Final result: $100 \cdot 0.84 = 84$.

Answer: 84

1. If $\rho = 1$, the conditional mean formula gives $E[z_Y \mid z_X] = 1 \cdot z_X$, so $z_Y = z_X$.
2. The conditional variance becomes $\sigma_Y^2 (1 - 1^2) = 0$.
3. This means for any given $X$, there is only one possible value for $Y$.
4. The 3D surface collapses from a bell into a 2D plane (a ridge) slicing through the 3D space.

Answer: $z_X = z_Y$; the surface collapses into a 2D ridge.

1. Find the marginal $P(X=1) = 0.2 + 0.3 = 0.5$.
2. Find conditional probabilities: $P(Y=1 \mid X=1) = 0.2/0.5 = 0.4$ and $P(Y=2 \mid X=1) = 0.3/0.5 = 0.6$.
3. Compute expectation: $E[Y \mid X=1] = 1(0.4) + 2(0.6) = 0.4 + 1.2 = 1.6$.

Answer: 1.6

1. Find marginal $f_X(x) = \int_0^x 2 \, dy = 2x$.
2. Find conditional density $f_{Y \mid X}(y \mid x) = 2 / (2x) = 1/x$ for $0 < y < x$.
3. Compute expectation: $E[Y \mid X=x] = \int_0^x y(1/x) \, dy = [y^2/(2x)]_0^x = x^2/(2x) = x/2$.

Answer: x/2

1. By independence, the conditional density $f_{Y \mid X}(y \mid x) = f_Y(y)$.
2. Substitute this into the conditional expectation formula: $E[Y \mid X=x] = \int y f_Y(y) \, dy$.
3. The resulting integral is the definition of the marginal expectation $E[Y]$.

Answer: E[Y]

1. First, find the conditional expectation $E[Y \mid X]$. For a binomial distribution, the mean is $np$, so $E[Y \mid X] = X \times 0.4$.
2. Next, apply the Law of Iterated Expectation: $E[Y] = E[E[Y \mid X]] = E[0.4 X]$.
3. Use linearity of expectation: $E[Y] = 0.4 E[X]$.
4. Compute $E[X] = (1 \times 0.6) + (2 \times 0.4) = 0.6 + 0.8 = 1.4$.
5. Finally, $E[Y] = 0.4 \times 1.4 = 0.56$.

Answer: 0.56

1. Find the conditional expectation $E[Y \mid X=x]$. Since $Y$ is uniform on $[0, x]$, the mean is $x/2$.
2. Thus, $E[Y \mid X] = X/2$.
3. Apply the Law of Iterated Expectation: $E[Y] = E[X/2] = \frac{1}{2} E[X]$.
4. Since $X$ is uniform on $[0, 10]$, $E[X] = (0+10)/2 = 5$.
5. Therefore, $E[Y] = \frac{1}{2} \times 5 = 2.5$.

Answer: 2.5

1. Find the conditional expectation $E[Y \mid X]$. Since $Y$ given $X$ is Poisson with parameter $X$, $E[Y \mid X] = X$.
2. Apply the Law of Iterated Expectation: $E[Y] = E[E[Y \mid X]] = E[X]$.
3. Since $X$ is Poisson with $\lambda = 3$, $E[X] = 3$.
4. Therefore, $E[Y] = 3$.

Answer: 3

1. Identify values: $w=0.6, \sigma_X=0.15, \sigma_Y=0.25, \rho=0.3$.
2. Compute first term: $0.6^2 \times 0.15^2 = 0.36 \times 0.0225 = 0.0081$.
3. Compute second term: $0.4^2 \times 0.25^2 = 0.16 \times 0.0625 = 0.01$.
4. Compute interaction term: $2 \times 0.6 \times 0.4 \times 0.3 \times 0.15 \times 0.25 = 0.48 \times 0.3 \times 0.0375 = 0.0054$.
5. Sum the terms: $0.0081 + 0.01 + 0.0054 = 0.0235$.

Answer: 0.0235

1. Identify values: $w=0.5, \sigma_X=0.2, \sigma_Y=0.2, \rho=-1$.
2. Compute variance: $\text{Var}(R) = 0.5^2(0.2^2) + 0.5^2(0.2^2) + 2(0.5)(0.5)(-1)(0.2)(0.2)$.
3. Simplify: $0.25(0.04) + 0.25(0.04) - 0.5(0.04) = 0.01 + 0.01 - 0.02 = 0$.
4. The standard deviation is $\sqrt{0} = 0$.

Answer: 0

1. Identify values: $w=0.3, \sigma_X=0.4, \sigma_Y=0, \rho=0$ (since $Y$ is constant).
2. Apply formula: $\text{Var}(R) = 0.3^2(0.4^2) + 0.7^2(0^2) + 2(0.3)(0.7)(0)(0.4)(0)$.
3. Simplify: $\text{Var}(R) = 0.09 \times 0.16 = 0.0144$.
4. Compute standard deviation: $\sigma_P = \sqrt{0.0144} = 0.12$.

Answer: 0.12

1. To find $f_X(x)$, integrate $8xy$ with respect to $y$ from $0$ to $x$.
2. $\int_0^x 8xy dy = [4xy^2]_0^x = 4x^3$.
3. To find $f_Y(y)$, integrate $8xy$ with respect to $x$ from $y$ to $1$.
4. $\int_y^1 8xy dx = [4x^2y]_y^1 = 4y(1 - y^2) = 4y - 4y^3$.

Answer: $f_X(x) = 4x^3, f_Y(y) = 4y - 4y^3$

1. Compute $E[X] = \int_0^1 x(4x^3) dx = [4x^5/5]_0^1 = 0.8$.
2. Compute $E[Y] = \int_0^1 y(4y - 4y^3) dy = [4y^3/3 - 4y^5/5]_0^1 = 4/3 - 4/5 = 8/15 \approx 0.533$.
3. Compute $E[XY] = \int_0^1 \int_0^x xy(8xy) dy dx = \int_0^1 [8x^2 y^3/3]_0^x dx = \int_0^1 8x^5/3 dx = [8x^6/18]_0^1 = 4/9 \approx 0.444$.
4. $\text{Cov}(X,Y) = E[XY] - E[X]E[Y] = 4/9 - (4/5)(8/15) = 4/9 - 32/75 = (100 - 96)/225 = 4/225$.

Answer: $\text{Cov}(X,Y) = 4/225$

1. We are given the conditional expectation $E[Y|X=x] = x$.
2. Therefore, the conditional expectation as a random variable is $E[Y|X] = X$.
3. By the Law of Iterated Expectations, $E[Y] = E[ E[Y|X] ]$.
4. Substituting the conditional expectation, we get $E[Y] = E[X]$.
5. Since $X$ has a mean of $0$, $E[Y] = 0$.

Answer: $E[Y] = 0$

1. Identify the given values: $\sigma = 12$ and $n = 36$.
2. Use the formula for standard error: $\text{SE}_{\bar{X}} = \frac{\sigma}{\sqrt{n}}$.
3. Substitute the values: $\text{SE}_{\bar{X}} = \frac{12}{\sqrt{36}}$.
4. Calculate the square root: $\sqrt{36} = 6$.
5. Divide: $12 / 6 = 2$.

Answer: 2

1. Use the formula for the variance of the sample mean: $\text{Var}(\bar{X}) = \frac{\sigma^2}{n}$.
2. Substitute the known values: $4 = \frac{400}{n}$.
3. Multiply both sides by $n$: $4n = 400$.
4. Divide by $4$: $n = 100$.

Answer: 100

1. Set up the inequality for standard error: $\frac{\sigma}{\sqrt{n}} \le 0.5$.
2. Substitute $\sigma = 5$: $\frac{5}{\sqrt{n}} \le 0.5$.
3. Rearrange to solve for $\sqrt{n}$: $\sqrt{n} \ge \frac{5}{0.5}$.
4. Simplify the fraction: $\sqrt{n} \ge 10$.
5. Square both sides: $n \ge 100$.

Answer: 100

1. Identify the knowns: $E[X] = 12$ and $a = 36$.
2. Apply Markov's Inequality: $P(X \ge a) \le \frac{E[X]}{a}$.
3. Substitute the values: $P(X \ge 36) \le \frac{12}{36}$.
4. Simplify the fraction: $\frac{12}{36} = \frac{1}{3}$.

Answer: 1/3

1. Find the standard deviation: $\sigma = \sqrt{16} = 4$.
2. Determine the distance from the mean to the boundary: $62 - 50 = 12$ (or $50 - 38 = 12$).
3. Calculate $k$, the number of standard deviations: $k = \frac{12}{4} = 3$.
4. Apply Chebyshev's Inequality: $P(|Y - 50| \ge 12) \le \frac{1}{3^2}$.
5. Compute the result: $\frac{1}{9} \approx 0.1111$.

Answer: 1/9

1. Recall the standard error of the mean: $\text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{\sigma}{\sqrt{400}} = \frac{\sigma}{20} = 0.05\sigma$.
2. The distance we are interested in is $\epsilon = 0.05\sigma$.
3. Calculate $k$ by dividing the distance by the standard error: $k = \frac{0.05\sigma}{0.05\sigma} = 1$.
4. Apply Chebyshev's Inequality: $P(|\bar{X} - \mu| \ge 0.05\sigma) \le \frac{1}{1^2}$.
5. The result is $1$, which means the probability is at most $100\%$. (Note: This shows that for this specific sample size and distance, the universal bound is not restrictive).

Answer: 1

1. We use the bound $P(|\bar{X}_n - \mu| \ge \epsilon) \le \frac{\sigma^2}{n\epsilon^2}$.
2. Substitute the given values: $\epsilon = 0.1$ and $\sigma = 0.5$.
3. Set the inequality: $\frac{0.5^2}{n \cdot 0.1^2} < 0.05$.
4. Simplify the constants: $\frac{0.25}{n \cdot 0.01} < 0.05$, which becomes $\frac{25}{n} < 0.05$.
5. Solve for $n$: $n > \frac{25}{0.05} = 500$.

Answer: n > 500

1. The deviation from the mean $\mu = -2$ to the threshold $0$ is $\epsilon = 2$.
2. The variance of the sample mean is $\text{Var}(\bar{X}_n) = \frac{\sigma^2}{n} = \frac{10^2}{10000} = \frac{100}{10000} = 0.01$.
3. Apply Chebyshev's inequality: $P(|\bar{X}_n - (-2)| \ge 2) \le \frac{0.01}{2^2}$.
4. Calculate the result: $\frac{0.01}{4} = 0.0025$.

Answer: 0.0025

1. Start with the complement: $P(|\bar{X}_n - \mu| < \epsilon) = 1 - P(|\bar{X}_n - \mu| \ge \epsilon)$.
2. Use the Law of Large Numbers bound: $P(|\bar{X}_n - \mu| \ge \epsilon) \le \frac{\sigma^2}{n\epsilon^2}$.
3. Therefore, $P(|\bar{X}_n - \mu| < \epsilon) \ge 1 - \frac{\sigma^2}{n\epsilon^2}$.
4. Take the limit as $n \to \infty$: $\lim_{n \to \infty} (1 - \frac{\sigma^2}{n\epsilon^2}) = 1 - 0 = 1$.
5. Since the probability is bounded above by $1$ and below by a value approaching $1$, it must converge to $1$.

Answer: The probability converges to 1.

1. Calculate the standard error: $\text{SE} = \sqrt{2.9167 / 40} \approx 0.2700$.
2. Find the z-score for $3.3$: $z_1 = (3.3 - 3.5) / 0.2700 \approx -0.74$.
3. Find the z-score for $3.7$: $z_2 = (3.7 - 3.5) / 0.2700 \approx 0.74$.
4. Find the area between $z = -0.74$ and $z = 0.74$ using the normal table: $\Phi(0.74) - \Phi(-0.74) \approx 0.7704 - 0.2296 = 0.5408$.

Answer: 0.5408

1. Calculate the standard error: $\text{SE} = 20 / \sqrt{64} = 20 / 8 = 2.5$.
2. Calculate the z-score: $z = (105 - 100) / 2.5 = 5 / 2.5 = 2.0$.
3. Find the probability $P(Z > 2.0) = 1 - \Phi(2.0) \approx 1 - 0.9772 = 0.0228$.

Answer: 0.0228

1. For an exponential distribution, the mean $\mu = 1/\lambda$ and the variance $\sigma^2 = 1/\lambda^2$. Since $\mu = 2$, then $\sigma^2 = 2^2 = 4$.
2. Calculate the standard error: $\text{SE} = \sqrt{4 / 100} = 2 / 10 = 0.2$.
3. Calculate the z-score: $z = (1.8 - 2.0) / 0.2 = -0.2 / 0.2 = -1.0$.
4. Find the probability $P(Z < -1.0) = \Phi(-1.0) \approx 0.1587$.

Answer: 0.1587

1. Calculate the mean of the sum: $E[S] = 20 \times 5 = 100$.
2. Calculate the variance of the sum: $\text{Var}(S) = 20 \times 4 = 80$.
3. Calculate the standard deviation of the sum: $\sigma_S = \sqrt{80} \approx 8.9443$.
4. Find the z-score for 120: $z = (120 - 100) / 8.9443 \approx 2.236$.
5. Find the probability $P(Z > 2.236)$ using the standard normal table: $1 - 0.9873 = 0.0127$.

Answer: 0.0127

1. The Central Limit Theorem states that the sampling distribution approaches normality as $n$ increases.
2. For symmetric populations, this happens quickly (small $n$).
3. For highly skewed populations, a larger $n$ (typically $n \ge 30$) is required to overcome the skewness.
4. Since $n=5$ is very small and the population is highly skewed, the approximation would be poor.

Answer: No, because high skewness requires a larger sample size (typically $n \ge 30$) for the sampling distribution to become approximately normal.

1. The given PDF is that of a Cauchy distribution.
2. The integral for the variance $\int x^2 f(x) dx$ diverges, meaning the variance is infinite.
3. The Central Limit Theorem requires the population to have a finite variance $\sigma^2 < \infty$.
4. Since the variance is infinite, the sum will not converge to a normal distribution regardless of the sample size.

Answer: No, because the distribution has infinite variance, which violates the primary requirement of the Central Limit Theorem.

1. Check conditions: $np = 80 \times 0.6 = 48$ and $nq = 80 \times 0.4 = 32$. Both are $\ge 5$.
2. Parameters: $\mu = 48$, $\sigma = \sqrt{80 \times 0.6 \times 0.4} = \sqrt{19.2} \approx 4.38$.
3. Continuity correction: $P(X \le 45) \approx P(Y < 45.5)$.
4. Z-score: $Z = \frac{45.5 - 48}{4.38} = \frac{-2.5}{4.38} \approx -0.57$.
5. Probability: $P(Z < -0.57) \approx 0.2843$.

Answer: 0.2843

1. Check conditions: $np = 200 \times 0.05 = 10$ and $nq = 200 \times 0.95 = 190$. Both are $\ge 5$.
2. Parameters: $\mu = 10$, $\sigma = \sqrt{200 \times 0.05 \times 0.95} = \sqrt{9.5} \approx 3.08$.
3. Continuity correction: $P(X > 12) \approx P(Y > 12.5)$.
4. Z-score: $Z = \frac{12.5 - 10}{3.08} = \frac{2.5}{3.08} \approx 0.81$.
5. Probability: $P(Z > 0.81) = 1 - 0.7910 = 0.2090$.

Answer: 0.2090

1. Check conditions: $np = 50 \times 0.25 = 12.5$ and $nq = 50 \times 0.75 = 37.5$. Both are $\ge 5$.
2. Parameters: $\mu = 12.5$, $\sigma = \sqrt{50 \times 0.25 \times 0.75} = \sqrt{9.375} \approx 3.06$.
3. Continuity correction: $P(10 \le X \le 15) \approx P(9.5 < Y < 15.5)$.
4. Z-scores: $Z_1 = \frac{9.5 - 12.5}{3.06} \approx -0.98$ and $Z_2 = \frac{15.5 - 12.5}{3.06} \approx 0.98$.
5. Probability: $P(-0.98 < Z < 0.98) = 0.8365 - 0.1635 = 0.6730$.

Answer: 0.6730

1. The theoretical probability is $P = \frac{2L}{\pi d}$. Substituting $L = d/2$, we get $P = \frac{2(d/2)}{\pi d} = \frac{1}{\pi}$.
2. The observed fraction is $\frac{112}{500} = 0.224$.
3. Setting the theoretical probability equal to the observed fraction: $\frac{1}{\pi} \approx 0.224$.
4. Solving for $\pi$, we get $\pi \approx \frac{1}{0.224} \approx 4.464$.

Answer: 4.464

1. The area of the bounding box (the unit square) is $1 \times 1 = 1$.
2. The fraction of points falling inside the region is $\frac{330}{1000} = 0.33$.
3. The estimated area is $\text{Area}_{\text{box}} \times \text{fraction} = 1 \times 0.33 = 0.33$.

Answer: 0.33

1. The area of the box is $2 \times 1 = 2$.
2. The estimated area of the region is $2 \times \frac{1580}{2000} = 2 \times 0.79 = 1.58$.
3. The region is a semi-circle of radius $1$, so its exact area is $\frac{1}{2}\pi(1)^2 = \frac{\pi}{2}$.
4. Setting $\frac{\pi}{2} \approx 1.58$, we find $\pi \approx 1.58 \times 2 = 3.16$.

Answer: 3.16

1. Let $r$ be right steps and $l$ be left steps. We need $r - l = 4$ and $r + l = 12$.
2. Adding the equations: $2r = 16$, so $r = 8$. Then $l = 4$.
3. The probability is $\binom{12}{8} (0.5)^{12}$.
4. $\binom{12}{8} = 495$.
5. $495 / 4096 \approx 0.1208$.

Answer: 0.1208

1. Total paths are $4^2 = 16$.
2. To return, the walker must move in opposite directions: (Right, Left), (Left, Right), (Up, Down), or (Down, Up).
3. There are $4$ such paths.
4. Probability is $4 / 16 = 0.25$.

Answer: 0.25

1. The mean is $E[S_{100}] = 0$.
2. The standard deviation is $\sigma = \sqrt{100} = 10$.
3. We calculate the z-score: $Z = (20 - 0) / 10 = 2$.
4. We need $P(Z > 2)$.
5. Using the standard normal table, $P(Z \le 2) \approx 0.9772$.
6. The probability is $1 - 0.9772 = 0.0228$.

Answer: 0.0228

1. The mean of a single die is $\mu = 3.5$ and the variance is $\sigma^2 = 35/12 \approx 2.0833$.
2. The standard deviation of the mean is $\sigma_{\bar{X}} = \sqrt{2.0833/100} \approx 0.1443$.
3. Calculate z-scores: $z_1 = (3.2 - 3.5)/0.1443 \approx -2.08$ and $z_2 = (3.8 - 3.5)/0.1443 \approx 2.08$.
4. Find the area between these z-scores: $P(-2.08 < Z < 2.08) \approx 0.9812 - 0.0188 = 0.9624$.

Answer: 0.9624

1. The interval $[20, 80]$ is centered at the mean $50$.
2. The distance from the mean to the boundary is $80 - 50 = 30$.
3. Express this distance in terms of standard deviations: $k = 30/10 = 3$.
4. Apply Chebyshev's inequality: $P(|Y - 50| \ge 30) \le 1/3^2 = 1/9$.

Answer: 1/9

1. The sample size $n=3$ is very small, which violates the general rule of thumb ($n \ge 30$) for the CLT to be reliable unless the components are already normally distributed.
2. A Monte Carlo simulation would allow us to generate thousands of triplets of lifetimes based on the actual distribution of the components.
3. By summing these triplets and counting the proportion that fall below $250$, we get an empirical probability that doesn't rely on the normal approximation.

Answer: n=3 is too small for CLT; simulation provides empirical evidence without distribution assumptions.

1. Identify the parameters: $\mu=10$, $\sigma=2$, $n=50$.
2. Calculate the standard error: $\sigma_{\bar{X}} = \frac{2}{\sqrt{50}} \approx 0.2828$.
3. Calculate the z-score: $z = \frac{10.5 - 10}{0.2828} \approx 1.768$.
4. Find the probability $P(Z > 1.768)$ using the normal table, which is approximately $0.0385$.

Answer: 0.0385

1. The range is $100 \pm 45$.
2. Calculate $k$ as the number of standard deviations: $k = \frac{45}{15} = 3$.
3. Apply Chebyshev's inequality: $P(|X - 100| \ge 45) \le \frac{1}{3^2}$.
4. Compute the result: $\frac{1}{9} \approx 0.1111$.

Answer: 0.1111

1. Expected value of one step: $E[X] = 0.7 - 0.3 = 0.4$.
2. Variance of one step: $\text{Var}(X) = 1 - (0.4)^2 = 0.84$.
3. For $n=100$: $\mu_{S} = 100(0.4) = 40$ and $\sigma_{S} = \sqrt{100(0.84)} = \sqrt{84} \approx 9.165$.
4. Apply continuity correction: $P(S_{100} \ge 50) \approx P(S_{norm} \ge 49.5)$.
5. Calculate z-score: $z = \frac{49.5 - 40}{9.165} \approx 1.036$.
6. Find $P(Z > 1.036) \approx 0.1501$.

Answer: 0.1501

1. Identify the knowns: The success rate $p = 0.2$ is given.
2. Identify the goal: Find the probability of a specific data outcome (at least $3$ recoveries).
3. Since we move from a known parameter to a predicted outcome, it is a forward problem.

Answer: Forward probability problem.

1. Identify the knowns: The data is fixed ($2$ defects out of $50$).
2. Identify the goal: Estimate the population parameter $p$.
3. Since we move from observed data to an unknown parameter, it is an inverse inference problem.

Answer: Inverse inference problem.

1. Task 1: Parameters $\mu=100, \sigma=15$ are known. We predict the data (percentage above $130$). This is the forward direction.
2. Task 2: Data (sample mean $105$) is known. We estimate the parameter $\mu$. This is the inverse direction.
3. The shift from Task 1 to Task 2 represents the U-turn from probability to statistics.

Answer: Task 1 is forward (probability); Task 2 is inverse (inference).

1. Identify the population standard deviation $\sigma = 24$ and sample size $n = 64$.
2. Use the formula $SE = \frac{\sigma}{\sqrt{n}}$.
3. Calculate the square root of 64, which is 8.
4. Divide 24 by 8 to get 3.

Answer: 3

1. Calculate the standard error: $SE = \frac{10}{\sqrt{25}} = \frac{10}{5} = 2$.
2. Calculate the z-score for $\bar{X} = 48$: $z = \frac{48 - 50}{2} = \frac{-2}{2} = -1$.
3. Find the probability $P(Z < -1)$ using the standard normal distribution.
4. The area to the left of $z = -1$ is approximately $0.1587$.

Answer: 0.1587

1. Set up the inequality $SE = \frac{12}{\sqrt{n}} \le 1.5$.
2. Rearrange to solve for $\sqrt{n}$: $\sqrt{n} \ge \frac{12}{1.5}$.
3. Calculate $12 / 1.5 = 8$.
4. Square both sides to find $n$: $n \ge 8^2 = 64$.

Answer: 64

1. Find the expected value: $E[\hat{\theta}] = E[\frac{X_1 + 2X_2}{3}]$
2. Use linearity of expectation: $E[\hat{\theta}] = \frac{1}{3}(E[X_1] + 2E[X_2])$
3. Substitute $E[X_1] = \mu$ and $E[X_2] = \mu$: $E[\hat{\theta}] = \frac{1}{3}(\mu + 2\mu) = \frac{3\mu}{3} = \mu$
4. Since $E[\hat{\theta}] = \mu$, the bias is zero.

Answer: Yes, it is unbiased.

1. Identify the given values: $\text{Var}(\hat{\theta}) = 0.25$ and $\text{Bias}(\hat{\theta}) = 0.5$
2. Use the MSE formula: $\text{MSE} = \text{Var}(\hat{\theta}) + [\text{Bias}(\hat{\theta})]^2$
3. Substitute the values: $\text{MSE} = 0.25 + (0.5)^2$
4. Compute the square: $0.5^2 = 0.25$
5. Sum the results: $0.25 + 0.25 = 0.5$

Answer: 0.5

1. Find the expectation: $E[\hat{\theta}] = E[\bar{X} + \frac{1}{n}] = E[\bar{X}] + \frac{1}{n}$
2. Since $E[\bar{X}] = \mu$, we have $E[\hat{\theta}] = \mu + \frac{1}{n}$
3. Calculate bias: $\text{Bias} = E[\hat{\theta}] - \mu = (\mu + \frac{1}{n}) - \mu = \frac{1}{n}$
4. Take the limit as $n \to \infty$: $\lim_{n \to \infty} \frac{1}{n} = 0$

Answer: The bias is $1/n$; it becomes unbiased in the limit as $n$ approaches infinity.

1. Calculate the sample mean: $\bar{X} = (2+4+6+8)/4 = 5$.
2. Calculate squared deviations: $(2-5)^2=9$, $(4-5)^2=1$, $(6-5)^2=1$, $(8-5)^2=9$.
3. Sum the squared deviations: $9+1+1+9 = 20$.
4. Divide by $n-1 = 3$: $s^2 = 20/3 \approx 6.67$.

Answer: 6.67

1. The biased variance is $s^2_{\text{biased}} = 18$.
2. The relationship is $s^2 = \frac{n}{n-1} s^2_{\text{biased}}$.
3. Substitute the values: $s^2 = \frac{10}{9} \times 18$.
4. Calculate: $s^2 = 10 \times 2 = 20$.

Answer: 20

1. The bias exists because the sample mean is used instead of the population mean, which underestimates the spread.
2. The correction factor is $\frac{n}{n-1}$.
3. As $n \to \infty$, the term $\frac{n}{n-1} \to 1$, meaning the correction becomes smaller.
4. For $n=100$, the ratio is $100/99 \approx 1.0101$.

Answer: 1.0101

1. The likelihood is $\mathcal{L}(\theta) = \theta^3(1-\theta)^1$.
2. The log-likelihood is $l(\theta) = 3\ln(\theta) + \ln(1-\theta)$.
3. The derivative is $l'(\theta) = \frac{3}{\theta} - \frac{1}{1-\theta}$.
4. Setting $l'(\theta) = 0$ gives $3(1-\theta) = \theta$, so $3 - 3\theta = \theta$.
5. Solving for $\theta$ gives $4\theta = 3$, so $\theta = 0.75$.

Answer: $\hat{\theta}_{MLE} = 0.75$

1. The likelihood is $\mathcal{L}(\lambda) = \lambda e^{-5\lambda}$.
2. The log-likelihood is $l(\lambda) = \ln(\lambda) - 5\lambda$.
3. The derivative is $l'(\lambda) = \frac{1}{\lambda} - 5$.
4. Setting $l'(\lambda) = 0$ gives $\frac{1}{\lambda} = 5$.
5. Solving for $\lambda$ gives $\lambda = 0.2$.

Answer: $\hat{\lambda}_{MLE} = 0.2$

1. The likelihood is $\mathcal{L}(\lambda) = \frac{e^{-\lambda}\lambda^2}{2!} \cdot \frac{e^{-\lambda}\lambda^4}{4!}$.
2. The log-likelihood is $l(\lambda) = -2\lambda + 6\ln(\lambda) - \ln(2! \cdot 4!)$.
3. The derivative is $l'(\lambda) = -2 + \frac{6}{\lambda}$.
4. Setting $l'(\lambda) = 0$ gives $\frac{6}{\lambda} = 2$.
5. Solving for $\lambda$ gives $\lambda = 3$.

Answer: $\hat{\lambda}_{MLE} = 3$

1. Identify the number of trials $n = 15$ and the number of successes $\sum x_i = 12$.
2. Use the derived Bernoulli MLE formula $\hat{p} = \frac{\sum x_i}{n}$.
3. Calculate $\hat{p} = \frac{12}{15}$.
4. Simplify the fraction to $0.8$.

Answer: 0.8

1. Identify the sample data: $5, 8, 2$ with $n = 3$.
2. Use the Poisson MLE formula $\hat{\lambda} = \frac{\sum x_i}{n}$.
3. Sum the observations: $5 + 8 + 2 = 15$.
4. Divide by the sample size: $15 / 3 = 5$.

Answer: 5

1. Write the log-likelihood: $l(\mu) = -\frac{n}{2}\ln(2\pi\sigma^2) - \frac{1}{2\sigma^2} \sum (x_i - \mu)^2$.
2. Differentiate with respect to $\mu$: $\frac{d}{d\mu} l(\mu) = \frac{1}{\sigma^2} \sum (\mu - x_i)$.
3. Set the derivative to zero: $\sum (\mu - x_i) = 0$.
4. Expand the sum: $n\mu - \sum x_i = 0$.
5. Solve for $\mu$: $\mu = \frac{\sum x_i}{n} = \bar{X}$.

Answer: \hat{\mu} = \bar{X}

1. Option A is incorrect because the true mean is a fixed constant, not a random variable; it is either in the interval or it is not.
2. Option B is incorrect because the confidence interval estimates the population mean, not the distribution of individual values.
3. Option C correctly describes the confidence level as the long-run success rate of the interval-generating procedure.

Answer: C

1. To be more confident that the interval captures the true mean, we must account for more of the sampling distribution.
2. This requires a larger multiplier for the margin of error.
3. A larger multiplier increases the distance from the point estimate to the boundaries.

Answer: The interval becomes wider.

1. The width of the interval is proportional to the standard error, which is $\frac{\sigma}{\sqrt{n}}$.
2. The original width is proportional to $1/\sqrt{25} = 1/5$.
3. The new width is proportional to $1/\sqrt{100} = 1/10$.
4. The ratio of the new width to the old width is $(1/10) / (1/5) = 1/2$.
5. Therefore, the new width is $10 \times 0.5 = 5$.

Answer: 5 units

1. Identify the knowns: $\bar{x} = 72$, $\sigma = 6$, $n = 36$.
2. For $99\%$ confidence, the critical value $z$ is $2.58$.
3. Calculate the standard error: $\text{SE} = 6 / \sqrt{36} = 6 / 6 = 1$.
4. Calculate the margin of error: $\text{ME} = 2.58 \times 1 = 2.58$.
5. Construct the interval: $72 \pm 2.58$.

Answer: [69.42, 74.58]

1. Identify the knowns: $\bar{x} = 16.2$, $\sigma = 0.5$, $n = 64$.
2. For $95\%$ confidence, $z = 1.96$.
3. Calculate the standard error: $\text{SE} = 0.5 / \sqrt{64} = 0.5 / 8 = 0.0625$.
4. Calculate the margin of error: $\text{ME} = 1.96 \times 0.0625 = 0.1225$.
5. Construct the interval: $16.2 \pm 0.1225$.

Answer: [16.0775, 16.3225]

1. Set up the margin of error formula: $\text{ME} = z \frac{\sigma}{\sqrt{n}}$.
2. Substitute the knowns: $0.5 = 1.96 \times \frac{3}{\sqrt{n}}$.
3. Solve for $\sqrt{n}$: $\sqrt{n} = (1.96 \times 3) / 0.5 = 5.88 / 0.5 = 11.76$.
4. Square both sides: $n = 11.76^2 = 138.2976$.
5. Since $n$ must be an integer, round up to the nearest whole number.

Answer: 139

1. Identify the given values: $\bar{x}=82$, $s=6$, $n=15$.
2. Calculate the standard error: $\text{SE} = 6 / \sqrt{15} \approx 1.549$.
3. Calculate the margin of error: $\text{ME} = 2.145 \times 1.549 \approx 3.323$.
4. Construct the interval: $82 \pm 3.323$.

Answer: [78.677, 85.323]

1. Identify the given values: $\bar{x}=98.5$, $s=0.8$, $n=20$.
2. Calculate the standard error: $\text{SE} = 0.8 / \sqrt{20} \approx 0.1789$.
3. Calculate the margin of error: $\text{ME} = 2.861 \times 0.1789 \approx 0.512$.
4. Construct the interval: $98.5 \pm 0.512$.

Answer: [97.988, 99.012]

1. Calculate the standard error: $\text{SE} = 2 / \sqrt{5} \approx 0.8944$.
2. Calculate the margin of error for $90\%$ confidence: $\text{ME} = 2.132 \times 0.8944 \approx 1.907$.
3. Compare the calculated margin of error $1.907$ to the required $1.5$.
4. Since $1.907 > 1.5$, the interval is too wide.

Answer: No, the margin of error is $1.907$, which exceeds $1.5$.

1. Calculate the sample proportion: $\hat{p} = 340 / 500 = 0.68$.
2. Find the standard error: $\text{SE} = \sqrt{(0.68 \times 0.32) / 500} \approx 0.0209$.
3. Use $z^* = 1.96$ for $95\%$ confidence: $\text{MOE} = 1.96 \times 0.0209 \approx 0.041$.
4. Construct the interval: $0.68 \pm 0.041$.

Answer: [0.639, 0.721]

1. Identify values: $z^* = 1.645$ for $90\%$ confidence, $\text{MOE} = 0.03$, and $\hat{p} = 0.5$.
2. Apply the formula: $n = (1.645 / 0.03)^2 \times 0.25$.
3. Calculate: $n = (54.833)^2 \times 0.25 \approx 3006.7$.
4. Round up to the nearest whole number.

Answer: 3007

1. The sample proportion is $\hat{p} = 0.45$.
2. The margin of error is $\text{MOE} = 0.028$.
3. The confidence interval is $0.45 \pm 0.028$, which is $[0.422, 0.478]$.
4. Check if $0.50$ is within the range $[0.422, 0.478]$.

Answer: No

1. Identify the critical value for $95\%$ confidence: $z^* = 1.96$.
2. Use the conservative estimate $\hat{p} = 0.5$, so $\hat{p}(1-\hat{p}) = 0.25$.
3. Plug values into the formula: $n = \frac{0.25(1.96)^2}{(0.04)^2}$.
4. Calculate: $n = \frac{0.9604}{0.0016} = 600.25$.
5. Round up to the nearest integer: $n = 601$.

Answer: 601

1. Identify the critical value for $90\%$ confidence: $z^* = 1.645$.
2. Use the prior estimate $\hat{p} = 0.3$, so $\hat{p}(1-\hat{p}) = 0.3 \times 0.7 = 0.21$.
3. Set the margin of error $\text{MOE} = 0.03$.
4. Plug into the formula: $n = \frac{(1.645)^2 (0.21)}{(0.03)^2}$.
5. Calculate: $n = \frac{2.706025 \times 0.21}{0.0009} = \frac{0.56826525}{0.0009} \approx 631.41$.
6. Round up to the nearest integer: $n = 632$.

Answer: 632

1. The formula for $n$ is $n = \frac{C}{\text{MOE}^2}$ where $C$ is a constant for a fixed $z^*$ and $\hat{p}$.
2. Let $n_1$ be the size for $\text{MOE}_1 = 0.05$ and $n_2$ be the size for $\text{MOE}_2 = 0.025$.
3. The ratio is $\frac{n_2}{n_1} = \frac{C / (0.025)^2}{C / (0.05)^2} = \frac{(0.05)^2}{(0.025)^2}$.
4. Simplify the ratio: $\left( \frac{0.05}{0.025} \right)^2 = 2^2 = 4$.
5. The sample size must increase by a factor of 4.

Answer: 4

1. Identify that $\sigma$ is unknown, so use the t-distribution with $\text{df} = 12 - 1 = 11$.
2. Find the critical value $t^*$ for $99\%$ confidence and $11$ degrees of freedom, which is approximately $3.106$.
3. Calculate the standard error: $\text{SE} = 6 / \sqrt{12} \approx 1.732$.
4. Calculate the margin of error: $\text{MOE} = 3.106 \times 1.732 \approx 5.38$.
5. The interval is $84 \pm 5.38$.

Answer: [78.62, 89.38]

1. Calculate the sample proportion: $\hat{p} = 210 / 500 = 0.42$.
2. Identify the critical value $z^*$ for $95\%$ confidence, which is $1.96$.
3. Calculate the standard error: $\text{SE} = \sqrt{(0.42 \times 0.58) / 500} \approx 0.0220$.
4. Calculate the margin of error: $\text{MOE} = 1.96 \times 0.0220 \approx 0.043$.
5. The interval is $0.42 \pm 0.043$.

Answer: [0.377, 0.463]

1. Use the conservative estimate $\hat{p} = 0.5$.
2. Identify the critical value $z^*$ for $90\%$ confidence, which is $1.645$.
3. Set the margin of error $\text{MOE} = 0.02$.
4. Apply the formula: $n = (1.645 / 0.02)^2 \times 0.5 \times 0.5$.
5. Calculate: $n = (82.25)^2 \times 0.25 = 6765.0625 \times 0.25 = 1691.26$.
6. Round up to the nearest whole number.

Answer: 1692