Compound Event Probabilities - Math 1

1. A jar contains 12 marbles: 5 red, 4 blue, and 3 green. Three marbles are drawn simultaneously. What is the probability that exactly one marble of each color is drawn?
   Answer: $\frac{3}{11}$ Total ways to choose 3 marbles from 12 = C(12,3) = 220. Ways to choose exactly 1 red, 1 blue, and 1 green = C(5,1) × C(4,1) × C(3,1) = 5 × 4 × 3 = 60. Probability = 60/220 = 3/11.
2. A committee of 4 people is randomly selected from a group of 6 men and 4 women. What is the probability that the committee contains exactly 2 men and 2 women?
   Answer: $\frac{3}{7}$ Total ways to select 4 people from 10 = C(10,4) = 210. Ways to select exactly 2 men from 6 and 2 women from 4 = C(6,2) × C(4,2) = 15 × 6 = 90. So probability = 90/210 = 3/7.
3. If two students independently make their selections, what is the probability that they choose identical meals (same main dish, side dish, and dessert)?
   Answer: $\frac{1}{24}$ When you encounter probability questions involving independent events and identical outcomes, you need to find both the total number of possible outcomes and the favorable outcomes. First, let's determine how many different meals are possible. Since students must choose one item from each category: 3 main dishes × 4 side dishes × 2 desserts = 24 possible meal combinations. For two students making independent selections, the total number of outcome pairs is 24 × 24 = 576 possible combinations of what the two students could choose. Now, for the favorable outcomes: how many ways can both students choose identical meals? The first student can choose any of the 24 possible meals, and the second student must choose that exact same meal. This gives us exactly 24 favorable outcomes. Therefore, the probability is $\frac{24}{576} = \frac{1}{24}$, which is answer D. Let's examine why the other answers are incorrect. Answer A ($\frac{9}{576}$) might result from incorrectly calculating 3 + 4 + 2 = 9 favorable outcomes, missing that all three categories must match simultaneously. Answer B ($\frac{1}{576}$) represents finding the probability of both students choosing one specific predetermined meal, rather than any identical meal. Answer C ($\frac{2}{24}$) incorrectly uses 24 as the denominator instead of recognizing this involves two independent selections. Remember: when dealing with "matching" probability problems, identify what constitutes a match, count total possible outcomes for all participants, then determine how many ways the matching condition can occur.
4. A jar contains 6 red balls and 4 white balls. Three balls are drawn simultaneously. What is the probability that exactly 2 balls are red?
   Answer: $\frac{1}{2}$ This is a combinations problem where you're selecting a subset from a larger group without regard to order. When you see "simultaneously" or "at once," think combinations, not permutations. To find the probability of exactly 2 red balls, you need to calculate favorable outcomes divided by total possible outcomes. The total ways to choose 3 balls from 10 is $C(10,3) = \frac{10!}{3!7!} = 120$. For exactly 2 red balls, you must choose 2 from the 6 red balls AND 1 from the 4 white balls. This gives you $C(6,2) \times C(4,1) = 15 \times 4 = 60$ favorable outcomes. Therefore, the probability is $\frac{60}{120} = \frac{1}{2}$, which is answer C. Let's examine why the other answers are incorrect. Answer A ($\frac{3}{10}$) might come from incorrectly thinking about the probability as simply $\frac{3}{10}$ based on drawing 3 balls from 10 total, but this ignores the specific color requirements. Answer B ($\frac{2}{5}$) could result from mistakenly calculating $\frac{6}{10} \times \frac{5}{9} \times \frac{4}{8}$ as if order mattered, but then making arithmetic errors. Answer D ($\frac{7}{15}$) might come from incorrectly calculating the combinations or confusing this with a different probability scenario. Remember: for combination problems, always identify what you're choosing from each group, calculate those combinations separately, then multiply them together. The key is recognizing that "exactly 2 red" means you must also have exactly 1 white ball.
5. Three friends each flip a fair coin. What is the probability that exactly two of them get heads?
   Answer: $\frac{3}{8}$ Total outcomes: 2³ = 8. Favorable outcomes (exactly 2 heads): HHT, HTH, THH = 3 outcomes. Probability = 3/8. Choice A gives probability for exactly 1 specific arrangement. Choice C incorrectly counts outcomes. Choice D gives probability of at least 2 heads.
6. A bag contains 3 red marbles, 4 blue marbles, and 5 green marbles. Two marbles are drawn without replacement. What is the probability that the first marble is red and the second marble is blue? Give your answer in simplest form.
   Answer: $\frac{1}{11}$ P(first red) = 3/12 = 1/4. After removing one red marble, 11 marbles remain with 4 blue. P(second blue | first red) = 4/11. P(first red and second blue) = (1/4) × (4/11) = 4/44 = 1/11. Choice B doesn't simplify the fraction. Choice C uses replacement probability. Choice D adds probabilities instead of multiplying.
7. From a deck of 52 cards, two cards are drawn without replacement. What is the probability that both cards are face cards (Jack, Queen, or King)?
   Answer: $\frac{11}{221}$ Face cards in deck: 12 (3 per suit × 4 suits). P(first face card) = 12/52. P(second face card | first face card) = 11/51. P(both face cards) = (12/52) × (11/51) = 132/2652 = 11/221. Choice B assumes replacement. Choice C uses wrong numerator calculation. Choice D doubles the correct answer.
8. Three cards are drawn without replacement from a standard 52-card deck. What is the probability that all three cards are from different suits?
   Answer: $\frac{39}{85}$ For all three cards to be from different suits, we need to select 3 suits from 4, then select one card from each chosen suit. Ways to choose 3 suits from 4 = C(4,3) = 4. For each choice of 3 suits, ways to select one card from each = 13 × 13 × 13 = 2197. Total favorable outcomes = 4 × 2197 = 8788. Wait, this is overcounting. Let me recalculate: P(all different suits) = (52/52) × (39/51) × (26/50) = 52 × 39 × 26 / (52 × 51 × 50) = 52 × 39 × 26 / (52 × 51 × 50) = 39 × 26 / (51 × 50) = 1014 / 2550 = 39/85 after simplification. Choice A: 169/425 ≠ 39/85. Choice B: 12/25 = 408/850 ≠ 39/85. Choice C: 2197/5525 ≠ 39/85. Choice D: 39/85 is correct.
9. A student council consists of 5 seniors and 4 juniors. A committee of 3 students is randomly selected. What is the probability that the committee contains exactly 2 seniors and 1 junior?
   Answer: $\frac{10}{21}$ Total ways to select 3 from 9 students: C(9,3) = 84. Ways to select exactly 2 seniors from 5 and 1 junior from 4: C(5,2) × C(4,1) = 10 × 4 = 40. Probability = 40/84 = 10/21. Choice A uses incorrect counting (5×4/84). Choice C incorrectly calculates C(5,2) as 5. Choice D uses wrong total combinations.
10. In a game, a player rolls two dice and wins if the product of the numbers is odd. What is the probability of winning this game?
    Answer: $\frac{1}{4}$ When you encounter probability questions involving products or sums of dice rolls, focus on the fundamental properties of odd and even numbers. A product is odd only when both factors are odd, since any even number multiplied by anything yields an even result. Each die shows numbers 1 through 6, where 1, 3, and 5 are odd, and 2, 4, and 6 are even. The probability of rolling an odd number on a single die is $\frac{3}{6} = \frac{1}{2}$. Since the dice are independent, you multiply the individual probabilities to find the probability that both dice show odd numbers: $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$. Choice A ($\frac{3}{4}$) represents the probability of getting an even product, which is the complement of what we want. This common trap catches students who calculate the wrong event or forget to find the complement properly. Choice B ($\frac{1}{3}$) might result from incorrectly thinking there are three odd numbers on each die out of nine total outcomes, leading to faulty reasoning about the sample space. Choice C ($\frac{1}{2}$) could come from mistakenly treating this as a simple "either/or" situation or confusing it with the probability of getting an odd number on just one die. Remember this key insight: for products involving randomness, "both conditions must be met" scenarios often involve multiplying probabilities. When dealing with odd/even properties, always consider that odd products require all factors to be odd.
11. Two standard six-sided dice are rolled simultaneously. Given that the sum of the dice is greater than 8, what is the probability that both dice show the same number? Give your answer in simplest form.
    Answer: $\frac{1}{5}$ First, find all 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 where both dice show the same number are: (5,5) and (6,6). That's 2 outcomes. So P(both same | sum > 8) = 2/10 = 1/5. Choice A is correct. Choice B is 2/10 which equals 1/5 but isn't simplified. Choice C would be if we incorrectly used all 36 possible outcomes. Choice D is 3/15 = 1/5 but uses different numbers.
12. Two cards are drawn from a standard deck without replacement. What is the probability that the first card is a heart and the second card is a spade?
    Answer: $\frac{13}{204}$ When you encounter probability questions involving multiple events without replacement, you're dealing with conditional probability where each draw affects the next. To find the probability that the first card is a heart and the second is a spade, you multiply the individual probabilities. In a standard deck, there are 13 hearts out of 52 cards, so P(first card is heart) = $\frac{13}{52} = \frac{1}{4}$. After removing one heart, 51 cards remain, with all 13 spades still in the deck, so P(second card is spade | first was heart) = $\frac{13}{51}$. The probability of both events occurring is: $\frac{1}{4} \times \frac{13}{51} = \frac{13}{204}$, which is answer D. Let's examine why the other options are wrong. Choice A ($\frac{1}{8}$) likely comes from incorrectly thinking each suit has probability $\frac{1}{4}$ and multiplying $\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$, then making an arithmetic error. Choice B ($\frac{1}{16}$) represents the flawed reasoning of treating this like sampling with replacement, where you'd multiply $\frac{1}{4} \times \frac{1}{4}$. Choice C ($\frac{13}{208}$) appears to use the wrong denominator—possibly calculating $\frac{13}{52 \times 4}$ instead of the correct $\frac{13}{52 \times 51}$. Remember: without replacement problems require you to adjust the total number of cards for subsequent draws. Always reduce your denominator by the number of cards already drawn, while carefully tracking which specific cards were removed and how that affects your numerator.
13. A committee of 4 people is formed from 7 men and 5 women. What is the probability that the committee has exactly 3 men and 1 woman?
    Answer: $\frac{35}{99}$ This is a combinations probability problem where you need to find favorable outcomes over total possible outcomes. When selecting people for committees with specific gender requirements, break the problem into parts: ways to select the required men, ways to select the required women, and total ways to form any committee. For exactly 3 men from 7: $\binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$ For exactly 1 woman from 5: $\binom{5}{1} = 5$ Favorable outcomes = $35 \times 5 = 175$ Total ways to select 4 people from 12: $\binom{12}{4} = \frac{12!}{4!8!} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495$ Therefore: $P = \frac{175}{495} = \frac{35}{99}$ Answer A ($\frac{7}{22}$) likely comes from incorrectly using just the ratio of men to total people without proper combinations. Answer B ($\frac{5}{22}$) similarly uses the ratio of women to total people. Answer D ($\frac{21}{66}$) appears to result from calculation errors in the combinations or incorrect simplification of fractions. Study tip: For committee probability problems, always use the multiplication principle: (ways to select required men) × (ways to select required women) ÷ (total ways to select committee). Double-check your combinations calculations, and remember that $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.
14. In a class of 30 students, 18 play soccer, 12 play basketball, and 8 play both sports. If a student is selected at random, what is the probability that the student plays soccer or basketball but not both?
    Answer: $\frac{7}{15}$ Students playing only soccer: 18 - 8 = 10. Students playing only basketball: 12 - 8 = 4. Students playing exactly one sport: 10 + 4 = 14. Probability = 14/30 = 7/15. Choice A includes students who play both sports. Choice B doesn't simplify the fraction. Choice D incorrectly calculates total students playing either sport.
15. If a student is selected at random from this group, what is the probability that the student likes pizza or burgers, but not both?
    Answer: $\frac{11}{20}$ This is a probability question involving set theory and the principle of inclusion-exclusion. When you see problems about "or" versus "and" with overlapping groups, you need to carefully track what's being counted. To find students who like pizza or burgers but not both, you need to find those who like exactly one of these foods. Start by identifying the distinct groups: students who like only pizza, students who like only burgers, and students who like both. Students who like only pizza = Total pizza lovers - Those who like both = 60 - 25 = 35 students Students who like only burgers = Total burger lovers - Those who like both = 45 - 25 = 20 students Students who like pizza or burgers but not both = 35 + 20 = 55 students The probability is $\frac{55}{100} = \frac{11}{20}$, which is answer C. Looking at the wrong answers: A) $\frac{3}{5} = \frac{60}{100}$ represents the probability of liking pizza, ignoring the "but not both" condition. B) $\frac{13}{20} = \frac{65}{100}$ might result from incorrectly adding only pizza (35) plus those who like both (25) plus something else. D) $\frac{4}{5} = \frac{80}{100}$ represents students who like pizza or burgers (including both), calculated as 60 + 45 - 25 = 80, but this ignores the "not both" requirement. Remember: "Or but not both" means exactly one condition, not both. Always subtract the overlap when the question excludes students who satisfy both conditions.
16. A spinner has 8 equal sections numbered 1 through 8. The spinner is spun three times. What is the probability that all three spins result in even numbers?
    Answer: $\frac{1}{8}$ Even numbers on spinner: 2, 4, 6, 8 (4 out of 8). P(even on one spin) = 4/8 = 1/2. P(even on all three spins) = (1/2)³ = 1/8. Choice B calculates probability for exactly one even spin incorrectly. Choice C uses (1/8)³. Choice D uses incorrect calculation with 3/8 probability.
17. Two fair dice are rolled simultaneously. Given that the sum is greater than 8, what is the probability that both dice show the same number?
    Answer: $\frac{1}{5}$ Outcomes with 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, only (5,5) and (6,6) have matching numbers, so 2 outcomes. Probability = 2/10 = 1/5. Choice A miscounts favorable outcomes as 1. Choice C uses total outcomes of 9 instead of 10. Choice D uses 3 favorable outcomes incorrectly.

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