Computing Probabilities - Math 3

A spinner has three sections: red (probability 0.5), blue (probability 0.3), and yellow (probability 0.2). The spinner is spun 4 times independently. What is the probability that red appears exactly twice, blue appears exactly once, and yellow appears exactly once?
Answer: $\frac{4!}{2! \cdot 1! \cdot 1!} \times (0.5)^2 \times (0.3)^1 \times (0.2)^1$ This is a multinomial probability. The number of ways to arrange 2 reds, 1 blue, and 1 yellow in 4 spins is 4!/(2!×1!×1!) = 12. Each specific arrangement has probability (0.5)²×(0.3)×(0.2). Total probability = 12×(0.5)²×(0.3)×(0.2) = 0.018. Choice B omits the multinomial coefficient. Choice C uses 4 instead of the correct multinomial coefficient 12. Choice D uses the binomial coefficient for only one color, ignoring the constraint on the other colors.
A medical test correctly identifies a disease 95% of the time when the disease is present and correctly identifies no disease 90% of the time when the disease is absent. If 2% of the population has the disease, what is the probability that a person who tests positive actually has the disease?
Answer: 0.162 Using Bayes' theorem: P(disease|positive) = P(positive|disease)×P(disease)/P(positive). P(positive) = 0.95×0.02 + 0.10×0.98 = 0.019 + 0.098 = 0.117. Therefore P(disease|positive) = (0.95×0.02)/0.117 = 0.019/0.117 ≈ 0.162. Choice B uses incorrect denominators. Choice C forgets to account for false positives. Choice D confuses this with the test sensitivity.
A restaurant serves 3 types of appetizers, 4 types of main courses, and 2 types of desserts. A customer orders one item from each category. If the customer randomly selects items and appetizer A is chosen, what is the probability that main course M₂ is also chosen?
Answer: $\frac{1}{4}$ Since the selection of appetizer A is already given, and the choices are independent, the probability of selecting main course M₂ is simply 1/4 (one choice out of 4 main courses). The appetizer choice doesn't affect the main course probability. Choice A uses the total number of meal combinations. Choice B incorrectly combines probabilities. Choice C uses the wrong denominator.
In a game, a player wins if they roll a sum of 7 or 11 with two fair dice, and loses if they roll a sum of 2, 3, or 12. For any other sum, the game continues. What is the probability that the game ends on the first roll?
Answer: $\frac{12}{36}$ Winning combinations (sum = 7 or 11): 6 ways for sum of 7, 2 ways for sum of 11 = 8 ways. Losing combinations (sum = 2, 3, or 12): 1 way for sum of 2, 2 ways for sum of 3, 1 way for sum of 12 = 4 ways. Game ends on first roll: 8 + 4 = 12 ways. Probability = 12/36. Choice A counts only winning outcomes. Choice B miscounts some combinations. Choice D includes outcomes that continue the game.
At a school carnival, the probability of winning the ring toss is 0.3, and the probability of winning the basketball shot is 0.4. These events are independent. Maria plays both games exactly once. What is the probability that she wins exactly one of the two games?
Answer: 0.58 To win exactly one game, Maria either wins ring toss and loses basketball, or loses ring toss and wins basketball. P(exactly one win) = P(ring toss win AND basketball loss) + P(ring toss loss AND basketball win) = (0.3×0.6) + (0.7×0.4) = 0.18 + 0.28 = 0.58. Choice A incorrectly calculates P(win ring toss OR win basketball) = 0.3 + 0.4 - 0.12. Choice C adds the individual probabilities without considering independence. Choice D gives the probability of winning both games.
A fair coin is flipped 5 times. What is the probability of getting at least 3 heads?
Answer: $\frac{16}{32}$ At least 3 heads means exactly 3, 4, or 5 heads. P(exactly 3) = C(5,3)/ $2^5$ = 10/32. P(exactly 4) = C(5,4)/ $2^5$ = 5/32. P(exactly 5) = C(5,5)/ $2^5$ = 1/32. Total: (10 + 5 + 1)/32 = 16/32. Choice A gives only P(exactly 3). Choice C miscalculates the combinations. Choice D includes outcomes with 2 heads.
A bag contains 8 red marbles, 6 blue marbles, and 4 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: $\frac{8}{51}$ For the first marble to be red: P(red) = 8/18. After removing one red marble, there are 17 marbles left, with 6 still blue. P(blue|red first) = 6/17. The probability of both events is (8/18) × (6/17) = 48/306. Choice A uses the wrong denominator. Choice C treats this as drawing with replacement. Choice D multiplies by 18² instead of accounting for the changing total.
Three cards are drawn without replacement from a standard 52-card deck. What is the probability that all three cards are hearts?
Answer: $\frac{13 \times 12 \times 11}{52 \times 51 \times 50}$ Without replacement, the probability changes with each draw. First heart: 13/52. Second heart: 12/51. Third heart: 11/50. Since these are dependent events, we multiply: (13×12×11)/(52×51×50). Choice B treats the draws as independent. Choice C incorrectly uses factorial. Choice D adds probabilities instead of multiplying.
In a card game, players draw cards without replacement from a standard 52-card deck. If the first player draws 2 cards and both are hearts, what is the probability that the next card drawn by the second player is also a heart?
Answer: $\frac{11}{50}$ After the first player draws 2 hearts, there are 11 hearts remaining out of 50 total cards left in the deck. The probability is 11/50. Choice B incorrectly uses the original probability before any cards were drawn. Choice C uses the correct numerator but wrong denominator, forgetting that 2 cards were already drawn. Choice D incorrectly assumes only 1 heart was drawn instead of 2.
If the test flags a batch as defective, what is the probability that the batch actually contains defective items?
Answer: 0.760 Using Bayes' theorem: P(defective|flagged) = P(flagged|defective)×P(defective)/P(flagged). P(flagged) = 0.90×0.15 + 0.05×0.85 = 0.135 + 0.0425 = 0.1775. Therefore P(defective|flagged) = (0.90×0.15)/0.1775 = 0.135/0.1775 ≈ 0.760. Choice A uses incorrect calculation of P(flagged). Choice C confuses this with the base rate. Choice D confuses this with test sensitivity.
A quality control inspector tests electronic components from two production lines. Line A has a 12% defect rate, and Line B has an 8% defect rate. If 60% of all components come from Line A and 40% come from Line B, what is the probability that a randomly selected defective component came from Line A?
Answer: 0.69 This requires Bayes' theorem. First, find P(defective) = P(defective|A)×P(A) + P(defective|B)×P(B) = 0.12×0.60 + 0.08×0.40 = 0.072 + 0.032 = 0.104. Then P(A|defective) = P(defective|A)×P(A) / P(defective) = (0.12×0.60) / 0.104 = 0.072 / 0.104 ≈ 0.69. Choice A incorrectly uses the overall proportion from Line A. Choice C uses an incorrect calculation mixing the defect rates. Choice D incorrectly uses just the defect rate of Line A.
A jar contains 8 red marbles, 6 blue marbles, and 4 green marbles. If 3 marbles are drawn simultaneously (without replacement), what is the probability that all 3 marbles are different colors? Give your answer in simplest form.
Answer: $\frac{4}{17}$ Total ways to choose 3 marbles from 18 is C(18,3) = 816. Ways to choose one of each color is 8×6×4 = 192 (choose 1 red from 8, 1 blue from 6, 1 green from 4). Probability = 192/816. Choice B incorrectly uses the total as if drawing sequentially with different denominators. Choice C uses wrong denominator, treating this as sequential draws. Choice D incorrectly treats this as replacement sampling.
In a certain high school, 65% of students play sports, 40% participate in clubs, and 25% do both. If a student is randomly selected, what is the probability that the student plays sports given that they participate in clubs?
Answer: $\frac{5}{8}$ This is a conditional probability problem. P(sports|clubs) = P(sports and clubs)/P(clubs) = 0.25/0.40 = 5/8. Choice A incorrectly uses P(both)/P(total). Choice C appears correct but doesn't simplify the fraction properly. Choice D uses P(both)/P(sports) instead of the correct conditional probability formula.
A fair six-sided die is rolled three times. What is the probability of getting exactly two even numbers?
Answer: $\frac{54}{216}$ This is a binomial probability with n=3, k=2, and p=1/2 (probability of even). The number of ways to choose 2 positions out of 3 is C(3,2) = 3. Each arrangement has probability (1/2)²(1/2)¹ = 1/8. Total probability = 3 × (1/8) = 3/8 = 54/216. Choice A uses only one arrangement. Choice B miscalculates the combinations. Choice D doubles the correct answer.
In a survey of 200 students, 120 play soccer, 80 play tennis, and 30 play both sports. If a student is selected at random and is known to play tennis, what is the probability that this student also plays soccer?
Answer: $\frac{30}{80}$ This is a conditional probability problem. P(plays soccer | plays tennis) = P(plays both) / P(plays tennis) = 30/80. We're given that the student plays tennis, so our sample space is reduced to the 80 tennis players, of whom 30 also play soccer. Choice A gives the unconditional probability of playing both sports. Choice B incorrectly uses soccer players as the denominator. Choice D incorrectly calculates tennis-only players in the numerator.
Based on the survey data, if a student is selected at random, what is the probability that the student drinks coffee or tea but not both?
Answer: $\frac{140}{200}$ Students who drink only coffee: 120 - 30 = 90. Students who drink only tea: 80 - 30 = 50. Students who drink coffee or tea but not both: 90 + 50 = 140. Probability = 140/200. Choice A includes those who drink both. Choice C miscalculates the exclusive categories. Choice D assumes all students drink at least one beverage.
Events X and Y are independent with P(X) = 0.4 and P(Y) = 0.3. What is P(X' ∩ Y'), where X' and Y' represent the complements of X and Y?
Answer: 0.42 Since X and Y are independent, X' and Y' are also independent. P(X') = 1 - 0.4 = 0.6 and P(Y') = 1 - 0.3 = 0.7. Therefore, P(X' ∩ Y') = P(X') × P(Y') = 0.6 × 0.7 = 0.42. Choice B gives P(X ∪ Y)'. Choice C gives only P(Y'). Choice D gives P(X') + P(Y') - P(X')P(Y'), which is incorrect for intersection.
In a lottery game, a player wins if they match at least 3 out of 5 numbers drawn from the set {1, 2, 3, ..., 20} without replacement. If a player's ticket has the numbers {2, 7, 13, 16, 19}, what is the probability that they win (match at least 3 numbers)?
Answer: $\frac{\binom{5}{3} \times \binom{15}{2} + \binom{5}{4} \times \binom{15}{1} + \binom{5}{5}}{\binom{20}{5}}$ When you encounter a lottery probability problem involving "at least" scenarios, think about using hypergeometric distribution. You're drawing without replacement from a finite population, where some items are "successes" (your numbers) and others are "failures" (remaining numbers). The correct approach requires calculating the probability of exactly 3, 4, or 5 matches separately, then adding them. For each case, you multiply the ways to choose matching numbers from your 5 ticket numbers by the ways to choose non-matching numbers from the remaining 15 numbers, all divided by the total ways to choose 5 numbers from 20. For exactly 3 matches: $\binom{5}{3} \times \binom{15}{2}$ (choose 3 from your 5 numbers, 2 from the other 15) For exactly 4 matches: $\binom{5}{4} \times \binom{15}{1}$ (choose 4 from your 5 numbers, 1 from the other 15) For exactly 5 matches: $\binom{5}{5} \times \binom{15}{0} = \binom{5}{5}$ (all your numbers match) Answer D correctly sums these three scenarios in the numerator, with $\binom{20}{5}$ as the denominator representing total possible draws. Answer A uses the complement approach (1 minus probability of 0, 1, or 2 matches) but has calculation errors. Answer B ignores the fact that you must also select from the non-matching numbers—it only counts ways to choose from your ticket. Answer C only calculates the probability of exactly 3 matches, missing 4 and 5 matches entirely. Remember: for hypergeometric problems, always account for both the "success" group and "failure" group when calculating each scenario, then sum all favorable outcomes.
Events A and B are mutually exclusive with P(A) = 0.3 and P(B) = 0.4. Event C has P(C) = 0.6, P(A∩C) = 0.15, and P(B∩C) = 0.2. What is P(A∪B∪C)?
Answer: 0.85 Since A and B are mutually exclusive, P(A∪B) = 0.3 + 0.4 = 0.7, and P(A∩B) = 0. Using inclusion-exclusion: P(A∪B∪C) = P(A∪B) + P(C) - P((A∪B)∩C) = 0.7 + 0.6 - P(A∩C ∪ B∩C) = 0.7 + 0.6 - (0.15 + 0.2) = 0.85. Choice B ignores the overlap with C. Choice C adds all probabilities without accounting for intersections. Choice D assumes complete coverage.
A pharmaceutical company tests a new drug on patients. The drug is effective for 85% of patients who have the target condition and ineffective for 95% of patients who do not have the target condition. If 2% of the population has the target condition, what is the probability that a patient who tests positive for drug effectiveness actually has the target condition?
Answer: 0.258 Using Bayes' theorem: P(condition | positive test) = P(positive | condition) × P(condition) / P(positive test). P(positive test) = 0.85×0.02 + 0.05×0.98 = 0.017 + 0.049 = 0.066. Therefore, P(condition | positive test) = (0.85×0.02) / 0.066 = 0.017 / 0.066 ≈ 0.258. Choice A incorrectly uses the sensitivity rate. Choice C miscalculates the denominator. Choice D incorrectly uses the base rate of the condition.

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