Discrete Math Quiz: Probabilities In Finite Spaces
17 questions · exam conditions
0:00
Probabilities In Finite SpacesQuestion 1 of 17

A bag contains 8 balls: 3 red, 3 blue, and 2 yellow. Two balls are drawn without replacement. What is the probability that the second ball is red, given that the first ball drawn was not red?

27\frac{2}{7}
38\frac{3}{8}
37\frac{3}{7}
514\frac{5}{14}
← Back to quizzes

Discrete Math Quiz

Discrete Math Quiz: Probabilities In Finite Spaces

Practice Probabilities In Finite Spaces in Discrete Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Probabilities In Finite Spaces, giving you a quick way to practice the rules, question types, and explanations that matter most for Discrete Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A bag contains 8 balls: 3 red, 3 blue, and 2 yellow. Two balls are drawn without replacement. What is the probability that the second ball is red, given that the first ball drawn was not red?

  1. 27\frac{2}{7}
  2. 38\frac{3}{8}
  3. 37\frac{3}{7} (correct answer)
  4. 514\frac{5}{14}
Explanation: This is a conditional probability problem where you need to update your sample space based on given information. When you see "given that" in a probability question, you're working with a restricted scenario that changes the total number of possible outcomes. Since the first ball drawn was not red, it must have been either blue or yellow. This removes one non-red ball from the bag, leaving us with 7 balls total: still 3 red balls, but only 4 non-red balls remaining (either 2 blue + 2 yellow, or 3 blue + 1 yellow, depending on what was drawn first). The key insight is that the condition doesn't change the number of red balls available for the second draw—there are still 3 red balls among the remaining 7 balls. Therefore, the probability that the second ball is red is 37\frac{3}{7}, which is answer C. Looking at the wrong answers: A) 27\frac{2}{7} incorrectly assumes only 2 red balls remain, perhaps confusing this with the number of yellow balls. B) 38\frac{3}{8} uses the original total of 8 balls, ignoring that one ball was already removed. D) 514\frac{5}{14} appears to come from incorrectly calculating 3×58×7\frac{3 \times 5}{8 \times 7}, possibly attempting to use a more complex conditional probability formula unnecessarily. Remember: in conditional probability problems, always identify your new sample space first. The condition "given that the first ball was not red" eliminates one non-red ball but leaves all red balls available, making the calculation straightforward.

Question 2

A password consists of 4 distinct digits chosen from {1, 2, 3, 4, 5, 6, 7, 8}. What is the probability that the password contains both the digits 3 and 7?

  1. 28\frac{2}{8}
  2. 628\frac{6}{28}
  3. 3601680\frac{360}{1680}
  4. 314\frac{3}{14} (correct answer)
Explanation: This is a conditional probability problem involving combinations. When you see "probability that something contains specific elements," you need to count favorable outcomes where those elements are present, then divide by total possible outcomes. First, find the total number of 4-digit passwords using distinct digits from 8 available digits. This is a permutation: P(8,4)=8×7×6×5=1680P(8,4) = 8 \times 7 \times 6 \times 5 = 1680 possible passwords. Next, count passwords containing both 3 and 7. Since these two digits must be included, you need to choose 2 more digits from the remaining 6 digits {1, 2, 4, 5, 6, 8}, then arrange all 4 digits. Choose 2 from 6: (62)=15\binom{6}{2} = 15 ways. Then arrange these 4 digits: 4!=244! = 24 ways. Total favorable outcomes: 15×24=36015 \times 24 = 360. Therefore, the probability is 3601680=314\frac{360}{1680} = \frac{3}{14}. Choice A (28\frac{2}{8}) incorrectly treats this as simply selecting 2 specific digits from 8, ignoring the constraint of 4 distinct digits. Choice B (628\frac{6}{28}) appears to use combinations incorrectly, possibly confusing (62)\binom{6}{2} and (82)\binom{8}{2}. Choice C (3601680\frac{360}{1680}) shows the correct calculation before simplification—this fraction reduces to 314\frac{3}{14}. Remember: for "contains both" problems, fix the required elements first, then count ways to fill remaining positions. Always check if your final fraction can be simplified, as exam answers typically appear in lowest terms.

Question 3

A computer system has 4 independent components, each with reliability 0.9 (probability of functioning correctly). The system functions if at least 3 of the 4 components work.

What is the probability that the system functions?

  1. (0.9)3+4×(0.9)3×(0.1)(0.9)^3 + 4 \times (0.9)^3 \times (0.1)
  2. (0.9)4+(43)×(0.9)3×(0.1)(0.9)^4 + \binom{4}{3} \times (0.9)^3 \times (0.1) (correct answer)
  3. (43)×(0.9)3×(0.1)+(44)×(0.9)4\binom{4}{3} \times (0.9)^3 \times (0.1) + \binom{4}{4} \times (0.9)^4
  4. 1−(42)×(0.9)2×(0.1)21 - \binom{4}{2} \times (0.9)^2 \times (0.1)^2
Explanation: The system works if exactly 3 or exactly 4 components work. P(exactly 4 work) = (0.9)4(0.9)^4. P(exactly 3 work) = (43)×(0.9)3×(0.1)1\binom{4}{3} \times (0.9)^3 \times (0.1)^1. Total probability = (0.9)4+(43)×(0.9)3×(0.1)(0.9)^4 + \binom{4}{3} \times (0.9)^3 \times (0.1). Choice A uses 4 instead of (43)\binom{4}{3} and omits the case where all 4 work. Choice C correctly shows both terms with proper binomial coefficients but in reverse order from choice B. Choice D attempts complement but only subtracts P(exactly 2 work), missing P(exactly 0 work) and P(exactly 1 work) terms.

Question 4

A password consists of 4 distinct digits selected from {0,1,2,3,4,5,6,7,8,9}\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}. If a password is chosen at random, what is the probability that it contains both the digit 0 and the digit 9?

  1. (82)(104)\frac{\binom{8}{2}}{\binom{10}{4}} (correct answer)
  2. 2×(82)(104)\frac{2 \times \binom{8}{2}}{\binom{10}{4}}
  3. (82)×4!(104)×4!\frac{\binom{8}{2} \times 4!}{\binom{10}{4} \times 4!}
  4. P(8,2)P(10,4)\frac{P(8,2)}{P(10,4)}
Explanation: We need passwords containing both 0 and 9. After including 0 and 9, we need to choose 2 more digits from the remaining 8 digits {1,2,3,4,5,6,7,8}\{1,2,3,4,5,6,7,8\}. Ways to do this: (82)\binom{8}{2}. Total ways to choose 4 digits from 10: (104)\binom{10}{4}. Since we're only concerned with which digits appear (not their arrangement), the probability is (82)(104)\frac{\binom{8}{2}}{\binom{10}{4}}. Choice B incorrectly multiplies by 2. Choice C unnecessarily includes 4!4! terms that cancel out. Choice D uses permutations instead of combinations, suggesting order matters when the problem asks about containment.

Question 5

A committee of 5 people is randomly selected from a group of 8 men and 6 women. What is the probability that the committee contains exactly 3 men and 2 women?

  1. 2801001\frac{280}{1001}
  2. 4201001\frac{420}{1001} (correct answer)
  3. 3361001\frac{336}{1001}
  4. 5601001\frac{560}{1001}
Explanation: The total number of ways to select 5 people from 14 is (145)=2002\binom{14}{5} = 2002. The number of ways to select exactly 3 men from 8 and 2 women from 6 is (83)×(62)=56×15=840\binom{8}{3} \times \binom{6}{2} = 56 \times 15 = 840. Therefore, the probability is 8402002=4201001\frac{840}{2002} = \frac{420}{1001}. Choice A uses (145)=1001\binom{14}{5} = 1001 instead of 2002. Choice C incorrectly calculates (83)=42\binom{8}{3} = 42 instead of 56. Choice D doubles the numerator incorrectly.

Question 6

An urn contains 3 red balls, 4 white balls, and 5 blue balls. Three balls are drawn without replacement. What is the probability that no two balls have the same color?

  1. 60220\frac{60}{220}
  2. 1255\frac{12}{55}
  3. 14\frac{1}{4}
  4. 311\frac{3}{11} (correct answer)
Explanation: This problem tests combinations with conditional probability — specifically, calculating favorable outcomes when drawing items without replacement under specific constraints. To find the probability that all three balls have different colors, you need exactly one red, one white, and one blue ball. The total number of ways to draw 3 balls from 12 is (123)=12!3!⋅9!=220\binom{12}{3} = \frac{12!}{3! \cdot 9!} = 220. For the favorable outcomes, you must choose 1 ball from each color: (31)×(41)×(51)=3×4×5=60\binom{3}{1} \times \binom{4}{1} \times \binom{5}{1} = 3 \times 4 \times 5 = 60. Therefore, the probability is 60220=311\frac{60}{220} = \frac{3}{11}, which is answer D. A (60220\frac{60}{220}) represents the unreduced fraction — mathematically correct but not simplified. Always reduce fractions to lowest terms on multiple-choice exams. B (1255\frac{12}{55}) likely comes from incorrectly calculating the total combinations as (113)=165\binom{11}{3} = 165 instead of (123)=220\binom{12}{3} = 220, then computing 60165=1233\frac{60}{165} = \frac{12}{33}. This suggests miscounting the total balls. C (14\frac{1}{4}) might result from oversimplifying the problem by assuming each outcome is equally likely without properly accounting for the different quantities of each color. Key strategy: In "without replacement" problems, always use combinations to count total possible outcomes, then count favorable outcomes using the multiplication principle. Remember to reduce your final fraction — test makers often include the unreduced version as a distractor.

Question 7

In a class of 20 students, 12 like mathematics, 8 like physics, and 5 like both subjects. If a student is selected at random, what is the probability that the student likes mathematics or physics but not both?

  1. 1020\frac{10}{20} (correct answer)
  2. 1520\frac{15}{20}
  3. 520\frac{5}{20}
  4. 1220\frac{12}{20}
Explanation: Students who like only mathematics: 12 - 5 = 7. Students who like only physics: 8 - 5 = 3. Students who like exactly one subject: 7 + 3 = 10. The probability is 1020=12\frac{10}{20} = \frac{1}{2}. Choice B gives the probability of liking at least one subject (12 + 8 - 5 = 15). Choice C gives the probability of liking both subjects. Choice D gives the probability of liking mathematics.

Question 8

Two fair dice are rolled simultaneously. Given that the sum is at least 9, what is the probability that both dice show the same number?

  1. 15\frac{1}{5} (correct answer)
  2. 210\frac{2}{10}
  3. 310\frac{3}{10}
  4. 13\frac{1}{3}
Explanation: The outcomes with sum ≥ 9 are: (3,6), (4,5), (4,6), (5,4), (5,5), (5,6), (6,3), (6,4), (6,5), (6,6) - total of 10 outcomes. Among these, only (5,5) and (6,6) have both dice showing the same number - 2 outcomes. The probability is 210=15\frac{2}{10} = \frac{1}{5}. Choice B fails to reduce the fraction. Choice C incorrectly counts 3 favorable outcomes. Choice D uses 6 total outcomes instead of 10.

Question 9

A box contains 6 identical balls numbered 1 through 6. Two balls are drawn simultaneously. What is the probability that the sum of the numbers is prime?

  1. 12\frac{1}{2}
  2. 815\frac{8}{15}
  3. 715\frac{7}{15} (correct answer)
  4. 25\frac{2}{5}
Explanation: This is a combination probability problem where you need to find favorable outcomes over total possible outcomes. When drawing two balls simultaneously from six numbered balls, you're looking at combinations, not permutations, since order doesn't matter. First, calculate the total possible outcomes. Choosing 2 balls from 6 gives us (62)=6!2!(6−2)!=15\binom{6}{2} = \frac{6!}{2!(6-2)!} = 15 possible pairs. Next, list all possible pairs and their sums: (1,2)→3, (1,3)→4, (1,4)→5, (1,5)→6, (1,6)→7, (2,3)→5, (2,4)→6, (2,5)→7, (2,6)→8, (3,4)→7, (3,5)→8, (3,6)→9, (4,5)→9, (4,6)→10, (5,6)→11. Now identify which sums are prime. A prime number has exactly two factors: 1 and itself. From our sums {3, 4, 5, 6, 7, 8, 9, 10, 11}, the primes are: 3, 5, 7, and 11. Count the pairs that give prime sums: sum=3 (1 pair), sum=5 (2 pairs), sum=7 (3 pairs), sum=11 (1 pair). That's 7 favorable outcomes. The probability is 715\frac{7}{15}, which is answer C. Answer A (12\frac{1}{2}) would mean 7.5 favorable outcomes, which is impossible. Answer B (815\frac{8}{15}) overcounts by including a non-prime sum like 4 or 6. Answer D (25=615\frac{2}{5} = \frac{6}{15}) undercounts the favorable outcomes by one. Remember: always verify which numbers are actually prime, and use combinations (not permutations) when order doesn't matter in selection problems.

Question 10

In a lottery, 6 numbers are drawn without replacement from the integers 1 through 45. What is the probability that exactly 3 of the drawn numbers are less than or equal to 15?

  1. (153)×(303)(456)\frac{\binom{15}{3} \times \binom{30}{3}}{\binom{45}{6}} (correct answer)
  2. (153)×(302)(455)\frac{\binom{15}{3} \times \binom{30}{2}}{\binom{45}{5}}
  3. (156)×(300)(456)\frac{\binom{15}{6} \times \binom{30}{0}}{\binom{45}{6}}
  4. 153×303456\frac{15^3 \times 30^3}{45^6}
Explanation: We need exactly 3 numbers from {1,2,…,15}\{1,2,\ldots,15\} and exactly 3 numbers from {16,17,…,45}\{16,17,\ldots,45\}. There are 15 numbers ≤15 and 30 numbers >15. Ways to choose 3 from first group: (153)\binom{15}{3}. Ways to choose 3 from second group: (303)\binom{30}{3}. Total ways to choose 6 from 45: (456)\binom{45}{6}. Probability is (153)×(303)(456)\frac{\binom{15}{3} \times \binom{30}{3}}{\binom{45}{6}}. Choice B uses (302)\binom{30}{2} and (455)\binom{45}{5}, suggesting only 5 total numbers drawn. Choice C requires all 6 numbers ≤15, which is impossible given the constraint. Choice D uses powers instead of combinations, treating draws as independent with replacement.

Question 11

In a group of 20 people, each person chooses a number from 1 to 5 independently and uniformly at random. What is the probability that exactly 3 people choose the number 1?

  1. (15)3×(45)17\left(\frac{1}{5}\right)^3 \times \left(\frac{4}{5}\right)^{17}
  2. (203)×(15)20\binom{20}{3} \times \left(\frac{1}{5}\right)^{20}
  3. 320×15\frac{3}{20} \times \frac{1}{5}
  4. (203)×(15)3×(45)17\binom{20}{3} \times \left(\frac{1}{5}\right)^3 \times \left(\frac{4}{5}\right)^{17} (correct answer)
Explanation: When you encounter a problem asking for the probability that exactly a certain number of people do something independently, you're dealing with a binomial probability distribution. This framework applies when you have a fixed number of trials (20 people), each with the same probability of success (choosing 1), and independence between trials. To find the probability that exactly 3 people choose the number 1, you need three components: the number of ways to choose which 3 people select 1, the probability those 3 people choose 1, and the probability the remaining 17 people don't choose 1. First, there are (203)\binom{20}{3} ways to select which 3 of the 20 people choose the number 1. Next, each person has a 15\frac{1}{5} probability of choosing 1, so the probability that your chosen 3 people all select 1 is (15)3\left(\frac{1}{5}\right)^3. Finally, each of the remaining 17 people must choose something other than 1, which happens with probability 45\frac{4}{5}, giving us (45)17\left(\frac{4}{5}\right)^{17}. Multiplying these together gives answer D: (203)×(15)3×(45)17\binom{20}{3} \times \left(\frac{1}{5}\right)^3 \times \left(\frac{4}{5}\right)^{17}. Answer A omits the combinatorial factor (203)\binom{20}{3}, failing to account for the different ways to select which people choose 1. Answer B incorrectly uses (15)20\left(\frac{1}{5}\right)^{20}, treating all 20 people as choosing 1. Answer C uses a completely incorrect approach, treating this as a simple fraction rather than a binomial probability. Remember the binomial formula: (nk)×pk×(1−p)n−k\binom{n}{k} \times p^k \times (1-p)^{n-k} for exactly k successes in n trials with probability p.

Question 12

From a standard deck of 52 cards, 5 cards are dealt. What is the probability that the hand contains exactly one ace and exactly two kings?

  1. (83)×(442)(525)\frac{\binom{8}{3} \times \binom{44}{2}}{\binom{52}{5}}
  2. (41)×(42)×(482)(525)\frac{\binom{4}{1} \times \binom{4}{2} \times \binom{48}{2}}{\binom{52}{5}}
  3. 4×6×(442)(525)\frac{4 \times 6 \times \binom{44}{2}}{\binom{52}{5}}
  4. (41)×(42)×(442)(525)\frac{\binom{4}{1} \times \binom{4}{2} \times \binom{44}{2}}{\binom{52}{5}} (correct answer)
Explanation: When you encounter probability problems involving specific combinations from different groups, you're dealing with hypergeometric probability. The key is to identify what you're selecting from each distinct group and multiply those possibilities together. For this problem, you need exactly 1 ace AND exactly 2 kings from your 5-card hand. A standard deck has 4 aces, 4 kings, and 44 other cards (52 - 4 - 4 = 44). The correct approach uses the multiplication principle:
  • Ways to choose 1 ace from 4: (41)=4\binom{4}{1} = 4
  • Ways to choose 2 kings from 4: (42)=6\binom{4}{2} = 6
  • Ways to choose the remaining 2 cards from the other 44: (442)\binom{44}{2}
  • Total ways to choose any 5 cards: (525)\binom{52}{5}
This gives us (41)×(42)×(442)(525)\frac{\binom{4}{1} \times \binom{4}{2} \times \binom{44}{2}}{\binom{52}{5}}, which is answer D. Answer A incorrectly treats aces and kings as one group of 8 cards, then tries to pick 3 from them. This doesn't guarantee you get exactly 1 ace and 2 kings. Answer B makes a critical error by using (482)\binom{48}{2} for the remaining cards. Since we need 5 total cards and already selected 3 (1 ace + 2 kings), we only need 2 more cards from the remaining 44, not 48. Answer C uses the correct logic but writes 4×64 \times 6 instead of the proper combination notation (41)×(42)\binom{4}{1} \times \binom{4}{2}, though numerically equivalent. Remember: always account for all cards in the deck and ensure your selections add up to the total hand size.

Question 13

A fair six-sided die is rolled repeatedly until a 6 appears. What is the probability that exactly 4 rolls are needed?

  1. (56)3×16\left(\frac{5}{6}\right)^3 \times \frac{1}{6} (correct answer)
  2. (56)4×16\left(\frac{5}{6}\right)^4 \times \frac{1}{6}
  3. (41)×(56)3×16\binom{4}{1} \times \left(\frac{5}{6}\right)^3 \times \frac{1}{6}
  4. 164\frac{1}{6^4}
Explanation: For exactly 4 rolls to be needed, the first 3 rolls must not be 6 (probability 56\frac{5}{6} each), and the 4th roll must be 6 (probability 16\frac{1}{6}). This gives (56)3×16\left(\frac{5}{6}\right)^3 \times \frac{1}{6}. Choice B incorrectly uses 4 non-6 rolls instead of 3. Choice C incorrectly applies binomial coefficient, treating this as a binomial distribution rather than geometric. Choice D treats all outcomes as equally likely, ignoring the constraint structure.

Question 14

Three fair six-sided dice are rolled. What is the probability that the sum of the three dice is 10, given that no die shows a 1?

  1. 18125\frac{18}{125}
  2. 27216\frac{27}{216}
  3. 1853\frac{18}{5^3} (correct answer)
  4. 2763\frac{27}{6^3}
Explanation: Given no die shows 1, each die shows values from {2,3,4,5,6}\{2,3,4,5,6\}, giving 53=1255^3 = 125 total outcomes. For sum = 10 with values ≥ 2, we need combinations of (a,b,c)(a,b,c) where a+b+c=10a+b+c=10 and 2≤a,b,c≤62 \leq a,b,c \leq 6. Substituting x=a−2,y=b−2,z=c−2x=a-2, y=b-2, z=c-2 gives x+y+z=4x+y+z=4 with 0≤x,y,z≤40 \leq x,y,z \leq 4. Valid solutions: (0,0,4),(0,1,3),(0,2,2),(0,3,1),(0,4,0),(1,0,3),(1,1,2),(1,2,1),(1,3,0),(2,0,2),(2,1,1),(2,2,0),(3,0,1),(3,1,0),(4,0,0)(0,0,4), (0,1,3), (0,2,2), (0,3,1), (0,4,0), (1,0,3), (1,1,2), (1,2,1), (1,3,0), (2,0,2), (2,1,1), (2,2,0), (3,0,1), (3,1,0), (4,0,0) and their permutations. This gives 18 total arrangements. Probability is 18125=1853\frac{18}{125} = \frac{18}{5^3}. Choice A shows the numerical form. Choice B uses 216 = 636^3 (unconditional sample space). Choice D uses 27, which might count some invalid cases.

Question 15

A student takes a 10-question true/false exam by guessing randomly on each question. What is the probability that the student gets exactly 7 questions correct?

  1. (107)×(12)7×(12)3\binom{10}{7} \times \left(\frac{1}{2}\right)^7 \times \left(\frac{1}{2}\right)^3
  2. (107)×(12)10\binom{10}{7} \times \left(\frac{1}{2}\right)^{10} (correct answer)
  3. 710\frac{7}{10}
  4. (107)×(12)7\binom{10}{7} \times \left(\frac{1}{2}\right)^7
Explanation: This is a binomial distribution with n=10n=10, p=12p=\frac{1}{2}, and we want P(X=7)P(X=7). The formula is (107)×(12)7×(12)10−7=(107)×(12)10\binom{10}{7} \times \left(\frac{1}{2}\right)^7 \times \left(\frac{1}{2}\right)^{10-7} = \binom{10}{7} \times \left(\frac{1}{2}\right)^{10}. Choice A separates the probability terms but this equals choice B since (12)7×(12)3=(12)10\left(\frac{1}{2}\right)^7 \times \left(\frac{1}{2}\right)^3 = \left(\frac{1}{2}\right)^{10}. Choice B is the more standard form. Choice C treats this as a simple proportion. Choice D omits the probability of the 3 incorrect answers.

Question 16

A bag contains 4 red balls, 3 blue balls, and 5 green balls. Three balls are drawn without replacement. What is the probability that all three balls are different colors?

  1. 60220\frac{60}{220}
  2. 311\frac{3}{11} (correct answer)
  3. 1255\frac{12}{55}
  4. 180220\frac{180}{220}
Explanation: Total ways to draw 3 balls from 12 is (123)=220\binom{12}{3} = 220. To get one of each color, we select 1 red from 4, 1 blue from 3, and 1 green from 5: 4×3×5=604 \times 3 \times 5 = 60. The probability is 60220=311\frac{60}{220} = \frac{3}{11}. Choice A doesn't reduce the fraction. Choice C uses (41)×(31)×(51)=60\binom{4}{1} \times \binom{3}{1} \times \binom{5}{1} = 60 but incorrectly calculates the total as 12×11×10/6=22012 \times 11 \times 10 / 6 = 220, then makes an arithmetic error. Choice D incorrectly multiplies by 3! = 6.

Question 17

A fair coin is flipped 5 times. What is the probability of getting exactly 3 heads, given that at least 2 heads occur?

  1. 516\frac{5}{16}
  2. 1031\frac{10}{31}
  3. 513\frac{5}{13} (correct answer)
  4. 1032\frac{10}{32}
Explanation: When you encounter conditional probability problems involving "given that" language, you're dealing with a restricted sample space. The key insight is that you only consider outcomes where the given condition is true. This problem asks for P(exactly 3 heads | at least 2 heads). Using the conditional probability formula: P(A|B) = P(A and B)/P(B). First, find P(at least 2 heads). With 5 coin flips, it's easier to calculate 1 - P(fewer than 2 heads). P(0 heads) = (50)(12)5=132\binom{5}{0}(\frac{1}{2})^5 = \frac{1}{32} and P(1 head) = (51)(12)5=532\binom{5}{1}(\frac{1}{2})^5 = \frac{5}{32}. So P(at least 2 heads) = 1−132−532=26321 - \frac{1}{32} - \frac{5}{32} = \frac{26}{32}. Next, find P(exactly 3 heads and at least 2 heads). Since having exactly 3 heads automatically satisfies "at least 2 heads," this equals P(exactly 3 heads) = (53)(12)5=1032\binom{5}{3}(\frac{1}{2})^5 = \frac{10}{32}. Therefore: P(exactly 3 heads | at least 2 heads) = 10/3226/32=1026=513\frac{10/32}{26/32} = \frac{10}{26} = \frac{5}{13}, which is answer C. Answer A (516\frac{5}{16}) incorrectly uses the total sample space instead of the restricted one. Answer B (1031\frac{10}{31}) makes an arithmetic error in the denominator calculation. Answer D (1032\frac{10}{32}) gives the unconditional probability of exactly 3 heads, ignoring the "given that" condition entirely. Remember: conditional probability problems require you to recalculate everything within the restricted sample space defined by the given condition.