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