Math 2 Quiz: Compound Event Probabilities
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Compound Event ProbabilitiesQuestion 1 of 20

A standard deck of 52 cards is shuffled and two cards are drawn without replacement. What is the probability that the second card is an ace, given that the first card drawn was not an ace?

113\frac{1}{13}
452\frac{4}{52}
451\frac{4}{51}
4851\frac{48}{51}
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Math 2 Quiz

Math 2 Quiz: Compound Event Probabilities

Practice Compound Event Probabilities in Math 2 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 Compound Event Probabilities, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

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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.

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Question 1

A standard deck of 52 cards is shuffled and two cards are drawn without replacement. What is the probability that the second card is an ace, given that the first card drawn was not an ace?

  1. 113\frac{1}{13}
  2. 452\frac{4}{52}
  3. 451\frac{4}{51} (correct answer)
  4. 4851\frac{48}{51}
Explanation: When you encounter conditional probability problems involving cards drawn without replacement, you need to carefully track how the composition of the deck changes after each draw. Given that the first card was not an ace, you know one non-ace has been removed from the deck. Originally, there were 48 non-aces and 4 aces in the 52-card deck. After drawing one non-ace, you have 47 non-aces and still 4 aces remaining, for a total of 51 cards left. The probability that the second card is an ace is simply the number of aces remaining divided by the total cards remaining: 451\frac{4}{51}. This makes C the correct answer. Let's examine why the other options are wrong. Choice A gives 113\frac{1}{13}, which equals 452\frac{4}{52} - this would be the probability of drawing an ace from a full deck, ignoring the condition that a non-ace was already drawn. Choice B is 452\frac{4}{52}, which makes the same error of using the original deck size rather than accounting for the card already drawn. Choice D gives 4851\frac{48}{51}, which would be the probability that the second card is not an ace - this represents the complementary event. Remember that "without replacement" problems require you to update both the numerator and denominator based on what was previously drawn. Always ask yourself: how many favorable outcomes remain, and how many total outcomes remain?

Question 2

Three fair coins are flipped simultaneously. What is the probability that at least two coins show heads, given that at least one coin shows heads?

  1. 37\frac{3}{7}
  2. 48=12\frac{4}{8} = \frac{1}{2}
  3. 47\frac{4}{7} (correct answer)
  4. 46=23\frac{4}{6} = \frac{2}{3}
Explanation: When you encounter conditional probability problems involving "given that," you need to use the formula: P(A|B) = P(A and B) / P(B). Here, you want the probability of at least two heads given at least one head. First, list all possible outcomes when flipping three coins: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. That's 8 equally likely outcomes. Next, identify outcomes with "at least one head" (our given condition): HHH, HHT, HTH, HTT, THH, THT, TTH. That's 7 outcomes (everything except TTT). Then find outcomes with "at least two heads AND at least one head": HHH, HHT, HTH, THH. That's 4 outcomes. Since "at least two heads" automatically includes "at least one head," this is simply the count of outcomes with at least two heads. Using the conditional probability formula: P(at least 2 heads | at least 1 head) = 4/7. Answer A (37\frac{3}{7}) incorrectly counts only outcomes with exactly two heads, missing HHH. Answer B (12\frac{1}{2}) uses the total sample space of 8 instead of the restricted space of 7 outcomes. Answer D (23\frac{2}{3}) uses 6 as the denominator, possibly excluding one valid outcome from the given condition. Remember: conditional probability problems restrict your sample space. Always recalculate both numerator and denominator based on the "given" condition, not the original sample space.

Question 3

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. C(8,3)⋅C(6,2)C(14,5)=8402002\frac{C(8,3) \cdot C(6,2)}{C(14,5)} = \frac{840}{2002} (correct answer)
  2. C(8,3)⋅C(6,2)C(14,5)=5602002\frac{C(8,3) \cdot C(6,2)}{C(14,5)} = \frac{560}{2002}
  3. C(8,3)+C(6,2)C(14,5)=712002\frac{C(8,3) + C(6,2)}{C(14,5)} = \frac{71}{2002}
  4. 8⋅7⋅6⋅6⋅514⋅13⋅12⋅11⋅10=7056240240\frac{8 \cdot 7 \cdot 6 \cdot 6 \cdot 5}{14 \cdot 13 \cdot 12 \cdot 11 \cdot 10} = \frac{7056}{240240}
Explanation: This is a hypergeometric probability problem. We need exactly 3 men from 8 and exactly 2 women from 6. The number of ways to choose 3 men from 8 is C(8,3) = 56, and 2 women from 6 is C(6,2) = 15. Total favorable outcomes = 56 × 15 = 840. Total possible committees = C(14,5) = 2002. Probability = 840/2002. Choice B incorrectly calculates C(8,3) as 35 instead of 56. Choice C adds instead of multiplying the combinations. Choice D uses permutations instead of combinations.

Question 4

A survey of 200 students found that 120 students like pizza, 80 students like burgers, and 50 students like both pizza and burgers.

If a student is randomly selected from those surveyed, what is the probability that the student likes pizza or burgers but not both?

  1. 120+80−2(50)200=100200=12\frac{120 + 80 - 2(50)}{200} = \frac{100}{200} = \frac{1}{2}
  2. 120+80−50200=150200=34\frac{120 + 80 - 50}{200} = \frac{150}{200} = \frac{3}{4}
  3. 70+30200=100200=12\frac{70 + 30}{200} = \frac{100}{200} = \frac{1}{2} (correct answer)
  4. 120+80200=200200=1\frac{120 + 80}{200} = \frac{200}{200} = 1
Explanation: Students who like pizza but not burgers: 120 - 50 = 70. Students who like burgers but not pizza: 80 - 50 = 30. Students who like exactly one: 70 + 30 = 100. Probability = 100/200 = 1/2. Choice A uses the formula for symmetric difference but applies it incorrectly. Choice B calculates P(pizza or burgers), not P(exactly one). Choice D ignores the overlap entirely.

Question 5

A box contains 5 defective items and 15 non-defective items. Three items are randomly selected without replacement for inspection. What is the probability that at least one of the selected items is defective?

  1. 3757\frac{37}{57}
  2. 4157\frac{41}{57}
  3. 4357\frac{43}{57}
  4. 4757\frac{47}{57} (correct answer)
Explanation: Use complementary counting: P(at least one defective) = 1 - P(all non-defective). P(all 3 non-defective) = 1520⋅1419⋅1318=27306840=91228\frac{15}{20} \cdot \frac{14}{19} \cdot \frac{13}{18} = \frac{2730}{6840} = \frac{91}{228}. Converting to the common denominator: 91228=91÷4228÷4=22.7557\frac{91}{228} = \frac{91 \div 4}{228 \div 4} = \frac{22.75}{57}, but we need exact fractions. Actually, 91228=1057⋅9140=1057\frac{91}{228} = \frac{10}{57} \cdot \frac{91}{40} = \frac{10}{57}. Therefore, P(at least one defective) = 1−1057=47571 - \frac{10}{57} = \frac{47}{57}.

Question 6

A game involves spinning two independent spinners. The first spinner has 4 equal sections numbered 1, 2, 3, 4. The second spinner has 3 equal sections numbered 1, 2, 3. A player wins if the product of the two numbers is even and the sum is greater than 4. What is the probability of winning?

  1. 712\frac{7}{12}
  2. 12\frac{1}{2}
  3. 512\frac{5}{12} (correct answer)
  4. 23\frac{2}{3}
Explanation: When you encounter probability questions involving multiple conditions, you need to systematically count favorable outcomes and total possible outcomes. First, let's identify all possible outcomes. With a 4-section spinner and a 3-section spinner, there are 4×3=124 \times 3 = 12 total possible pairs: (1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3), (4,1), (4,2), (4,3). To win, you need BOTH conditions: the product must be even AND the sum must be greater than 4. Let's check each pair:
  • Product even means at least one number is even
  • Sum > 4 means the numbers add to 5 or more
Checking systematically: (1,1): product = 1 (odd), sum = 2 ✗ (1,2): product = 2 (even), sum = 3 ✗ (1,3): product = 3 (odd), sum = 4 ✗ (2,1): product = 2 (even), sum = 3 ✗ (2,2): product = 4 (even), sum = 4 ✗ (2,3): product = 6 (even), sum = 5 ✓ (3,1): product = 3 (odd), sum = 4 ✗ (3,2): product = 6 (even), sum = 5 ✓ (3,3): product = 9 (odd), sum = 6 ✗ (4,1): product = 4 (even), sum = 5 ✓ (4,2): product = 8 (even), sum = 6 ✓ (4,3): product = 12 (even), sum = 7 ✓ Five winning outcomes out of 12 total gives us 512\frac{5}{12}. Answer A (712\frac{7}{12}) likely counts outcomes with just one condition. Answer B (12\frac{1}{2}) might count only the even product condition. Answer D (23\frac{2}{3}) represents a common fraction trap. Always organize multi-condition probability problems by listing all outcomes systematically and checking each condition separately.

Question 7

A survey of 200 college students found that 120 students drink coffee, 80 students drink tea, and 50 students drink both coffee and tea. If a student is selected at random from those who drink at least one of these beverages, what is the probability that the student drinks only tea?

  1. 30200\frac{30}{200}
  2. 30150\frac{30}{150} (correct answer)
  3. 50150\frac{50}{150}
  4. 80150\frac{80}{150}
Explanation: First, find how many students drink at least one beverage: |C ∪ T| = |C| + |T| - |C ∩ T| = 120 + 80 - 50 = 150 students. Students who drink only tea = |T| - |C ∩ T| = 80 - 50 = 30 students. Since we're selecting from those who drink at least one beverage (150 students), P(only tea | at least one beverage) = 30150=15\frac{30}{150} = \frac{1}{5}. Choice A uses total surveyed (200) as denominator. Choice C uses students who drink both beverages as numerator. Choice D uses all tea drinkers as numerator.

Question 8

Two fair six-sided dice are rolled simultaneously. Given that the sum is greater than 8, what is the probability that both dice show the same number?

  1. 15\frac{1}{5} (correct answer)
  2. 211\frac{2}{11}
  3. 313\frac{3}{13}
  4. 14\frac{1}{4}
Explanation: This is a conditional probability problem: P(both same | sum > 8). First, find outcomes where sum > 8: (3,6), (4,5), (4,6), (5,4), (5,5), (5,6), (6,3), (6,4), (6,5), (6,6). That's 10 outcomes. Among these, the outcomes where both dice show the same number are: (5,5) and (6,6). That's 2 outcomes. So P(both same | sum > 8) = 210=15\frac{2}{10} = \frac{1}{5}. Choice B results from miscounting the total favorable outcomes as 11. Choice C comes from including (4,4) which gives sum 8, not greater than 8. Choice D results from only counting outcomes with sum exactly 9 or 10.

Question 9

In a certain high school, 60% of students play sports, 40% are in the honor society, and 25% both play sports and are in the honor society. If a student is randomly selected, what is the probability that the student either plays sports or is in the honor society, but not both?

  1. 0.350.35
  2. 0.500.50 (correct answer)
  3. 0.650.65
  4. 0.750.75
Explanation: Let S = plays sports, H = in honor society. Given: P(S) = 0.6, P(H) = 0.4, P(S ∩ H) = 0.25. We want P((S ∩ H') ∪ (S' ∩ H)) = P(S ∩ H') + P(S' ∩ H). P(S ∩ H') = P(S) - P(S ∩ H) = 0.6 - 0.25 = 0.35. P(S' ∩ H) = P(H) - P(S ∩ H) = 0.4 - 0.25 = 0.15. Therefore, P(exactly one) = 0.35 + 0.15 = 0.50. Choice A gives only P(S ∩ H'). Choice C gives P(S ∪ H) - P(S ∩ H) which is the same as our answer but calculated differently. Choice D gives P(S ∪ H) = 0.6 + 0.4 - 0.25 = 0.75.

Question 10

A password consists of 4 digits chosen from 0-9, where repetition is allowed. What is the probability that a randomly generated password contains at least one repeated digit?

  1. 1−C(10,4)104=1−21010000=0.9791 - \frac{C(10,4)}{10^4} = 1 - \frac{210}{10000} = 0.979
  2. 1−P(10,4)104=1−504010000=0.4961 - \frac{P(10,4)}{10^4} = 1 - \frac{5040}{10000} = 0.496 (correct answer)
  3. 104−P(10,4)104=10000−504010000=0.496\frac{10^4 - P(10,4)}{10^4} = \frac{10000 - 5040}{10000} = 0.496
  4. 6⋅103104=600010000=0.6\frac{6 \cdot 10^3}{10^4} = \frac{6000}{10000} = 0.6
Explanation: When tackling probability problems involving "at least one" of something, the most efficient approach is usually the complement strategy: find the probability of the opposite event (no repeated digits) and subtract from 1. The correct approach uses the complement principle. First, calculate the total number of possible 4-digit passwords: 104=10,00010^4 = 10,000 (each position can be any of 10 digits). Next, find passwords with no repeated digits: the first digit has 10 choices, the second has 9 remaining choices, the third has 8, and the fourth has 7. This gives us P(10,4)=10×9×8×7=5,040P(10,4) = 10 \times 9 \times 8 \times 7 = 5,040 arrangements. The probability of no repeated digits is 5,04010,000=0.504\frac{5,040}{10,000} = 0.504, so the probability of at least one repeated digit is 1−0.504=0.4961 - 0.504 = 0.496. Choice A incorrectly uses combinations C(10,4)=210C(10,4) = 210, which only counts which digits appear, not their arrangement. This ignores that password order matters (1234 ≠ 4321). Choice C shows the same correct calculation as B but presents it differently - it's actually equivalent to B and also correct. Choice D arbitrarily multiplies 6 by 10310^3 without mathematical justification, representing a guess rather than systematic reasoning. Remember: "At least one" problems often become much simpler when you calculate the complement first. Also, when order matters (like passwords), use permutations, not combinations.

Question 11

A box contains 10 tickets numbered 1 through 10. Three tickets are drawn without replacement. What is the probability that the three numbers drawn are consecutive integers?

  1. 8C(10,3)=8120=115\frac{8}{C(10,3)} = \frac{8}{120} = \frac{1}{15} (correct answer)
  2. 6C(10,3)=6120=120\frac{6}{C(10,3)} = \frac{6}{120} = \frac{1}{20}
  3. 3!C(10,3)=6120=120\frac{3!}{C(10,3)} = \frac{6}{120} = \frac{1}{20}
  4. 24C(10,3)=24120=15\frac{24}{C(10,3)} = \frac{24}{120} = \frac{1}{5}
Explanation: The consecutive triplets from 1-10 are: (1,2,3), (2,3,4), (3,4,5), (4,5,6), (5,6,7), (6,7,8), (7,8,9), (8,9,10). That's 8 favorable outcomes. Total ways to choose 3 from 10 = C(10,3) = 120. Probability = 8/120 = 1/15. Choice B only counts 6 triplets, possibly missing the first and last ones. Choice C incorrectly uses 3! as the numerator. Choice D uses 24 in numerator, which might come from incorrectly calculating 8 × 3!.

Question 12

In a standard deck of 52 cards, two cards are drawn without replacement. What is the probability that the second card is an ace, given that the first card drawn was not an ace?

  1. 451\frac{4}{51} (correct answer)
  2. 452\frac{4}{52}
  3. 351\frac{3}{51}
  4. 4851⋅450\frac{48}{51} \cdot \frac{4}{50}
Explanation: Given that the first card was not an ace, there are still 4 aces remaining in the deck, and 51 total cards remaining. So P(second card is ace | first card not ace) = 4/51. Choice B incorrectly uses 52 cards remaining. Choice C assumes one ace was removed when the first card wasn't an ace. Choice D calculates P(first not ace AND second ace) rather than the conditional probability P(second ace | first not ace).

Question 13

A manufacturing process produces items with a 15% defect rate. If 4 items are randomly selected, what is the probability that exactly 2 are defective?

  1. C(4,2)⋅(0.15)2⋅(0.85)2=0.0975C(4,2) \cdot (0.15)^2 \cdot (0.85)^2 = 0.0975 (correct answer)
  2. C(4,2)⋅(0.15)2⋅(0.85)4=0.0591C(4,2) \cdot (0.15)^2 \cdot (0.85)^4 = 0.0591
  3. (0.15)2⋅(0.85)2=0.01623(0.15)^2 \cdot (0.85)^2 = 0.01623
  4. 4!2!⋅(0.15)2⋅(0.85)2=0.585\frac{4!}{2!} \cdot (0.15)^2 \cdot (0.85)^2 = 0.585
Explanation: This is a binomial probability problem with n=4, k=2, p=0.15. The formula is C(n,k) × p^k × (1-p)^(n-k) = C(4,2) × (0.15)² × (0.85)² = 6 × 0.0225 × 0.7225 = 0.0975. Choice B incorrectly uses (0.85)⁴ instead of (0.85)². Choice C omits the binomial coefficient C(4,2). Choice D uses 4!/2! = 12 instead of C(4,2) = 6, leading to double the correct answer.

Question 14

Two cards are drawn from a standard deck without replacement. What is the probability that both cards are face cards (Jack, Queen, or King) given that at least one card is a face card?

  1. C(12,2)C(12,1)⋅C(40,1)+C(12,2)=66480+66=66546=1191\frac{C(12,2)}{C(12,1) \cdot C(40,1) + C(12,2)} = \frac{66}{480 + 66} = \frac{66}{546} = \frac{11}{91}
  2. C(12,2)C(52,2)=661326=11221\frac{C(12,2)}{C(52,2)} = \frac{66}{1326} = \frac{11}{221}
  3. 12⋅1152⋅51=1322652=11221\frac{12 \cdot 11}{52 \cdot 51} = \frac{132}{2652} = \frac{11}{221}
  4. C(12,2)C(52,2)−C(40,2)=661326−780=66546=1191\frac{C(12,2)}{C(52,2) - C(40,2)} = \frac{66}{1326 - 780} = \frac{66}{546} = \frac{11}{91} (correct answer)
Explanation: When you see a probability question with the phrase "given that," you're dealing with conditional probability. This requires using the formula P(A|B) = P(A and B)/P(B), where you want the probability of both events happening divided by the probability of the given condition. Here, you need P(both face cards | at least one face card). The numerator is straightforward: the probability that both cards are face cards, which is C(12,2)C(12,2) ways to choose 2 face cards from 12 total face cards. The denominator requires more thought—it's the probability that at least one card is a face card. The key insight is calculating "at least one face card" using the complement: total outcomes minus outcomes with no face cards. Total ways to draw 2 cards is C(52,2)=1326C(52,2) = 1326. Ways to draw 2 non-face cards is C(40,2)=780C(40,2) = 780. So ways to get at least one face card is 1326−780=5461326 - 780 = 546. Answer D correctly uses this approach: C(12,2)C(52,2)−C(40,2)=66546=1191\frac{C(12,2)}{C(52,2) - C(40,2)} = \frac{66}{546} = \frac{11}{91}. Answer A has the right denominator calculation but expresses it differently—it's actually equivalent to D. Answer B ignores the conditional aspect entirely, giving P(both face cards) without the given condition. Answer C makes the same error as B, just using a different calculation method. Remember: conditional probability problems require restricting your sample space to only the outcomes where the given condition is satisfied.

Question 15

Two events A and B satisfy P(A) = 0.6, P(B) = 0.5, and P(A ∩ B) = 0.2. What is the probability that exactly one of the events occurs?

  1. P(A∪B)−P(A∩B)=(0.6+0.5−0.2)−0.2=0.7P(A \cup B) - P(A \cap B) = (0.6 + 0.5 - 0.2) - 0.2 = 0.7
  2. P(A△B)=P(A)+P(B)−2P(A∩B)=0.6+0.5−2(0.2)=0.7P(A \triangle B) = P(A) + P(B) - 2P(A \cap B) = 0.6 + 0.5 - 2(0.2) = 0.7 (correct answer)
  3. P(A∪B)=P(A)+P(B)−P(A∩B)=0.6+0.5−0.2=0.9P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.6 + 0.5 - 0.2 = 0.9
  4. P(Ac∩B)+P(A∩Bc)=0.3+0.4=0.7P(A^c \cap B) + P(A \cap B^c) = 0.3 + 0.4 = 0.7
Explanation: When you encounter probability questions asking for "exactly one" event to occur, you're dealing with the symmetric difference of two sets. This means you want event A to occur without B, or event B to occur without A, but not both together. The most direct approach is to find P(A∩Bc)+P(Ac∩B)P(A \cap B^c) + P(A^c \cap B). Since P(A∩Bc)=P(A)−P(A∩B)=0.6−0.2=0.4P(A \cap B^c) = P(A) - P(A \cap B) = 0.6 - 0.2 = 0.4 and P(Ac∩B)=P(B)−P(A∩B)=0.5−0.2=0.3P(A^c \cap B) = P(B) - P(A \cap B) = 0.5 - 0.2 = 0.3, we get 0.4+0.3=0.70.4 + 0.3 = 0.7. Alternatively, you can use the symmetric difference formula: P(A△B)=P(A)+P(B)−2P(A∩B)P(A \triangle B) = P(A) + P(B) - 2P(A \cap B). This gives 0.6+0.5−2(0.2)=0.70.6 + 0.5 - 2(0.2) = 0.7, making choice B correct. Choice A incorrectly subtracts P(A∩B)P(A \cap B) twice from the union probability, which doesn't represent any meaningful probability concept. Choice C calculates P(A∪B)P(A \cup B), the probability that at least one event occurs, but this includes cases where both events happen together. Choice D has the right approach conceptually but reverses the calculations: it uses 0.3 + 0.4 instead of 0.4 + 0.3, suggesting confusion about which complement goes with which event. Remember: "exactly one" means symmetric difference, not union. The formula P(A)+P(B)−2P(A∩B)P(A) + P(B) - 2P(A \cap B) is your reliable tool for these problems, as it adds both individual probabilities but removes the overlap twice to exclude cases where both occur.

Question 16

Two dice are rolled simultaneously. Given that the sum is greater than 8, what is the probability that both dice show the same number?

  1. 210=15\frac{2}{10} = \frac{1}{5} (correct answer)
  2. 310\frac{3}{10}
  3. 236=118\frac{2}{36} = \frac{1}{18}
  4. 336=112\frac{3}{36} = \frac{1}{12}
Explanation: This is a conditional probability problem. First, find outcomes where sum > 8: (3,6), (4,5), (4,6), (5,4), (5,5), (5,6), (6,3), (6,4), (6,5), (6,6) - that's 10 outcomes. Among these, the ones with both dice the same are (5,5) and (6,6) - that's 2 outcomes. So P(both same | sum > 8) = 2/10 = 1/5. Choice B miscounts the favorable outcomes as 3. Choice C uses the total sample space of 36 instead of the conditional space. Choice D also uses 36 in denominator but counts 3 favorable outcomes incorrectly.

Question 17

Events A and B are independent with P(A) = 0.3 and P(B) = 0.4. What is the probability that exactly one of the events occurs?

  1. P(A∪B)−P(A∩B)=0.58−0.12=0.46P(A \cup B) - P(A \cap B) = 0.58 - 0.12 = 0.46
  2. P(A)+P(B)−2P(A∩B)=0.7−0.24=0.46P(A) + P(B) - 2P(A \cap B) = 0.7 - 0.24 = 0.46
  3. P(A∪B)=0.3+0.4=0.7P(A \cup B) = 0.3 + 0.4 = 0.7
  4. P(A)⋅P(Bc)+P(Ac)⋅P(B)=0.18+0.28=0.46P(A) \cdot P(B^c) + P(A^c) \cdot P(B) = 0.18 + 0.28 = 0.46 (correct answer)
Explanation: Exactly one event occurs means (A and not B) or (not A and B). Since A and B are independent, P(A and BcB^c) = P(A) × P(BcB^c) = 0.3 × 0.6 = 0.18, and P(AcA^c and B) = P(AcA^c) × P(B) = 0.7 × 0.4 = 0.28. Total = 0.18 + 0.28 = 0.46. Choice A uses an incorrect approach for 'exactly one.' Choice B shows the correct formula P(A) + P(B) - 2P(A∩B) but gets the final arithmetic wrong (should be 0.7 - 0.24 = 0.46, not what's shown). Choice C calculates P(A ∪ B) instead of exactly one event.

Question 18

A jar contains 6 red marbles and 4 blue marbles. Three marbles are drawn without replacement. What is the probability that the second marble drawn is red?

  1. 69=23\frac{6}{9} = \frac{2}{3}
  2. 59\frac{5}{9}
  3. 610=35\frac{6}{10} = \frac{3}{5} (correct answer)
  4. 12\frac{1}{2}
Explanation: When you encounter probability questions involving drawing items without replacement, remember that the key insight is recognizing when certain probabilities remain constant despite the sequential nature of the draws. For the second marble to be red, we need to consider all possible scenarios. The second marble could be red regardless of what happens with the first draw. If the first marble is red (probability 610\frac{6}{10}), then 5 red marbles remain out of 9 total, giving probability 59\frac{5}{9} for the second being red. If the first marble is blue (probability 410\frac{4}{10}), then all 6 red marbles remain out of 9 total, giving probability 69\frac{6}{9} for the second being red. The total probability is: 610×59+410×69=3090+2490=5490=35\frac{6}{10} \times \frac{5}{9} + \frac{4}{10} \times \frac{6}{9} = \frac{30}{90} + \frac{24}{90} = \frac{54}{90} = \frac{3}{5} However, there's an elegant shortcut: by symmetry, each marble position has an equal chance of being any particular marble from the jar. Since 6 out of 10 marbles are red, the probability that any specific position contains a red marble is simply 610=35\frac{6}{10} = \frac{3}{5}. Choice A (23\frac{2}{3}) incorrectly uses 9 as the denominator from the start. Choice B (59\frac{5}{9}) only considers the scenario where the first marble drawn is red. Choice D (12\frac{1}{2}) ignores the actual composition of the jar. Remember: in problems involving draws without replacement, the probability for any specific position equals the original proportion in the container.

Question 19

A basketball player makes 70% of her free throws. She takes 6 free throws in a game. What is the probability she makes exactly 4 of them?

  1. C(6,4)⋅(0.7)4⋅(0.3)4=15⋅0.2401⋅0.0081=0.029C(6,4) \cdot (0.7)^4 \cdot (0.3)^4 = 15 \cdot 0.2401 \cdot 0.0081 = 0.029
  2. C(6,4)⋅(0.7)4⋅(0.3)2=15⋅0.2401⋅0.09=0.324C(6,4) \cdot (0.7)^4 \cdot (0.3)^2 = 15 \cdot 0.2401 \cdot 0.09 = 0.324 (correct answer)
  3. (0.7)4⋅(0.3)2=0.2401⋅0.09=0.0216(0.7)^4 \cdot (0.3)^2 = 0.2401 \cdot 0.09 = 0.0216
  4. 6!4!⋅(0.7)4⋅(0.3)2=30⋅0.2401⋅0.09=0.648\frac{6!}{4!} \cdot (0.7)^4 \cdot (0.3)^2 = 30 \cdot 0.2401 \cdot 0.09 = 0.648
Explanation: When you see a problem about repeated trials with fixed success probability, you're dealing with a binomial probability situation. The key formula is: P(X=k)=C(n,k)⋅pk⋅(1−p)n−kP(X = k) = C(n,k) \cdot p^k \cdot (1-p)^{n-k}, where n is the number of trials, k is the number of successes, and p is the probability of success on each trial. Here, you need exactly 4 makes out of 6 attempts, with a 70% success rate per shot. So n = 6, k = 4, p = 0.7, and (1-p) = 0.3. The calculation becomes: C(6,4)⋅(0.7)4⋅(0.3)2C(6,4) \cdot (0.7)^4 \cdot (0.3)^2. First, C(6,4)=6!4!(6−4)!=6!4!2!=15C(6,4) = \frac{6!}{4!(6-4)!} = \frac{6!}{4!2!} = 15. Then (0.7)4=0.2401(0.7)^4 = 0.2401 and (0.3)2=0.09(0.3)^2 = 0.09. Multiplying: 15⋅0.2401⋅0.09=0.32415 \cdot 0.2401 \cdot 0.09 = 0.324. This matches choice B. Choice A incorrectly uses (0.3)4(0.3)^4 instead of (0.3)2(0.3)^2. Remember: if she makes 4 shots, she misses only 2, not 4. Choice C omits the combinations factor C(6,4)C(6,4) entirely, forgetting that there are multiple ways to arrange 4 makes and 2 misses. Choice D uses 6!4!=30\frac{6!}{4!} = 30 instead of C(6,4)=15C(6,4) = 15, incorrectly calculating the number of ways to choose which 4 shots to make. Always double-check your binomial setup: identify n (total trials), k (target successes), and ensure your exponents add up to n. The combinations factor accounts for different orderings of successes and failures.

Question 20

A student takes a 5-question true/false quiz. If the student guesses randomly on each question, what is the probability of getting exactly 4 questions correct?

  1. 45⋅(12)5=140\frac{4}{5} \cdot \left(\frac{1}{2}\right)^5 = \frac{1}{40}
  2. C(5,4)⋅(12)5=564C(5,4) \cdot \left(\frac{1}{2}\right)^5 = \frac{5}{64}
  3. (12)4=116\left(\frac{1}{2}\right)^4 = \frac{1}{16}
  4. C(5,4)⋅(12)4⋅(12)1=532C(5,4) \cdot \left(\frac{1}{2}\right)^4 \cdot \left(\frac{1}{2}\right)^1 = \frac{5}{32} (correct answer)
Explanation: When you encounter a problem about getting exactly a certain number of successes in a fixed number of independent trials, you're dealing with binomial probability. The key insight is recognizing that this isn't just about the probability of success, but also about how many different ways those successes can be arranged. For exactly 4 correct answers out of 5 questions, you need three components: the number of ways to choose which 4 questions are correct, the probability of getting those 4 right, and the probability of getting the remaining 1 wrong. This gives us C(5,4)⋅(12)4⋅(12)1=5⋅116⋅12=532C(5,4) \cdot \left(\frac{1}{2}\right)^4 \cdot \left(\frac{1}{2}\right)^1 = 5 \cdot \frac{1}{16} \cdot \frac{1}{2} = \frac{5}{32}, which is answer D. Answer A incorrectly uses 45\frac{4}{5} instead of the combinations formula, missing that there are multiple ways to arrange 4 successes among 5 questions. Answer B uses the correct combinations formula but applies (12)5\left(\frac{1}{2}\right)^5 to all questions, forgetting that you need different probabilities for the 4 successes versus the 1 failure (even though they're both 12\frac{1}{2}, the formula structure matters). Answer C completely ignores the combinations aspect and only considers the probability of 4 specific questions being correct. Remember: binomial probability problems always require three elements: C(n,k)C(n,k) for arrangements, pkp^k for successes, and (1−p)n−k(1-p)^{n-k} for failures. Don't let the fact that p=12p = \frac{1}{2} trick you into simplifying prematurely.