ISEE Upper Level Quiz: Multi Event Probability
16 questions · exam conditions
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Multi Event ProbabilityQuestion 1 of 16

In a game, a player rolls two dice. The player wins if the sum is 7 or 11, loses if the sum is 2, 3, or 12, and continues playing if any other sum occurs. What is the probability that the player wins on the first roll?

29\frac{2}{9}
14\frac{1}{4}
518\frac{5}{18}
13\frac{1}{3}
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ISEE Upper Level Quiz

ISEE Upper Level Quiz: Multi Event Probability

Practice Multi Event Probability in ISEE Upper Level 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 Multi Event Probability, giving you a quick way to practice the rules, question types, and explanations that matter most for ISEE Upper Level.

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

In a game, a player rolls two dice. The player wins if the sum is 7 or 11, loses if the sum is 2, 3, or 12, and continues playing if any other sum occurs. What is the probability that the player wins on the first roll?

  1. 29\frac{2}{9} (correct answer)
  2. 14\frac{1}{4}
  3. 518\frac{5}{18}
  4. 13\frac{1}{3}
Explanation: When you see a probability question involving dice, start by identifying all possible outcomes and then count the favorable ones. With two dice, there are 6×6=366 \times 6 = 36 total possible outcomes. To find the probability of winning on the first roll, you need to count outcomes where the sum is 7 or 11. For a sum of 7, the winning combinations are: (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1) — that's 6 ways. For a sum of 11, the combinations are: (5,6) and (6,5) — that's 2 ways. In total, there are 8 favorable outcomes out of 36 possible outcomes, giving us 836=29\frac{8}{36} = \frac{2}{9}. Looking at the wrong answers: Choice B (14\frac{1}{4}) would equal 936\frac{9}{36}, suggesting someone miscounted and found 9 favorable outcomes instead of 8. Choice C (518\frac{5}{18}) equals 1036\frac{10}{36}, which might result from incorrectly including some losing combinations or double-counting. Choice D (13\frac{1}{3}) equals 1236\frac{12}{36}, which could happen if someone mistakenly included all the losing outcomes (sums of 2, 3, 12 have 4 total ways) plus the winning ones. The correct answer is A: 29\frac{2}{9}. Study tip: In dice probability problems, always systematically list the combinations for each target sum rather than trying to count in your head. This prevents miscounting, which is the most common error on these questions.

Question 2

A factory produces widgets, and quality control tests show that 5% are defective. If widgets are tested independently, what is the probability that among the first 4 widgets tested, exactly 2 are found to be defective?

  1. (42)(0.05)2(0.95)2\binom{4}{2}(0.05)^2(0.95)^2 (correct answer)
  2. (42)(0.05)2(0.95)4\binom{4}{2}(0.05)^2(0.95)^4
  3. (42)(0.05)4(0.95)2\binom{4}{2}(0.05)^4(0.95)^2
  4. (24)(0.05)2(0.95)2\binom{2}{4}(0.05)^2(0.95)^2
Explanation: When you encounter a problem about finding the probability of a specific number of successes in a fixed number of independent trials, you're dealing with a binomial probability situation. The key formula is: P(X=k)=(nk)pk(1−p)n−kP(X = k) = \binom{n}{k}p^k(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. In this problem, you have 4 widgets tested (n = 4), you want exactly 2 defective ones (k = 2), and the probability of any widget being defective is 5% or 0.05 (p = 0.05). The probability of a widget NOT being defective is 0.95. Applying the formula: P(X=2)=(42)(0.05)2(0.95)2P(X = 2) = \binom{4}{2}(0.05)^2(0.95)^2. The binomial coefficient (42)\binom{4}{2} counts the ways to choose which 2 of the 4 widgets are defective. The term (0.05)2(0.05)^2 represents the probability that those 2 specific widgets are defective, and (0.95)2(0.95)^2 represents the probability that the remaining 2 widgets are not defective. Choice A matches this exactly. Choice B incorrectly uses (0.95)4(0.95)^4 instead of (0.95)2(0.95)^2, suggesting all 4 widgets are non-defective rather than just 2. Choice C reverses the exponents, treating defective as more likely than non-defective. Choice D uses the impossible notation (24)\binom{2}{4}, which violates the rule that you can't choose more items than you have. Remember: in binomial problems, the exponents must add up to the total number of trials, and the binomial coefficient follows the pattern (nk)\binom{n}{k}, never (kn)\binom{k}{n} when k > n.

Question 3

A survey found that 60% of people like chocolate, 70% like vanilla, and 40% like both chocolate and vanilla. If a person is selected at random, what is the probability that they like chocolate but not vanilla?

  1. 0.20 (correct answer)
  2. 0.30
  3. 0.40
  4. 0.50
Explanation: When you encounter probability questions involving overlapping groups, you're dealing with set theory and the principle of inclusion-exclusion. The key is to carefully identify what each percentage represents and avoid double-counting. Let's organize the given information: 60% like chocolate, 70% like vanilla, and 40% like both flavors. To find the probability someone likes chocolate but not vanilla, you need to subtract the overlap from the chocolate group. Think of it this way: of all the chocolate lovers (60%), some also like vanilla (40%), so the chocolate-only group is 60%−40%=20%60\% - 40\% = 20\% or 0.20. Looking at the wrong answers: B) 0.30 represents vanilla-only lovers (70% - 40% = 30%), which answers a different question. C) 0.40 is the percentage who like both flavors, not chocolate exclusively. D) 0.50 doesn't correspond to any meaningful calculation with these numbers and likely results from incorrect arithmetic. The correct answer is A) 0.20. You can verify this using a Venn diagram approach: if 40% like both, 20% like only chocolate, and 30% like only vanilla, then 10% like neither flavor. These percentages total 100%, confirming our calculation. Strategy tip: For overlapping probability questions, always draw a quick Venn diagram or use the formula: P(A only) = P(A) - P(A and B). Watch out for answer choices that represent related but different probabilities—test makers often include these as distractors.

Question 4

In a class of 30 students, 18 play basketball, 15 play soccer, and 8 play both sports. If a student is selected at random, what is the probability that the student plays basketball or soccer but not both?

  1. 1730\frac{17}{30} (correct answer)
  2. 2530\frac{25}{30}
  3. 3330\frac{33}{30}
  4. 1030\frac{10}{30}
Explanation: When you encounter problems about students playing multiple sports or activities, you're dealing with set theory and overlapping groups. The key is to carefully track what each number represents and avoid double-counting. Let's organize the given information: 30 total students, 18 play basketball, 15 play soccer, and 8 play both sports. The question asks for the probability that a randomly selected student plays basketball or soccer but not both. "But not both" means we want students who play exactly one sport. To find this, calculate students who play only basketball plus students who play only soccer. Students who play only basketball = Total basketball players - Those who play both = 18 - 8 = 10 students Students who play only soccer = Total soccer players - Those who play both = 15 - 8 = 7 students Students who play exactly one sport = 10 + 7 = 17 students Therefore, the probability is 1730\frac{17}{30}, which is answer A. Looking at the wrong answers: Answer B (2530\frac{25}{30}) represents students who play at least one sport (basketball or soccer, including both), calculated as 18 + 15 - 8 = 25. Answer C (3330\frac{33}{30}) incorrectly adds all players without accounting for overlap (18 + 15 = 33), creating an impossible probability greater than 1. Answer D (1030\frac{10}{30}) only counts students who play basketball but not soccer, missing those who play only soccer. Remember: when dealing with overlapping groups, always subtract the overlap to avoid double-counting, and pay close attention to whether the question asks for "or" (at least one) versus "but not both" (exactly one).

Question 5

A game show has three doors. Behind one door is a prize, behind the other two are empty. A contestant chooses a door. The host, who knows what's behind each door, opens a different door that is empty. The contestant can then switch to the remaining unopened door or stay with their original choice.

If the contestant uses the switching strategy, what is the probability of winning the prize?

  1. 13\frac{1}{3}
  2. 12\frac{1}{2}
  3. 23\frac{2}{3} (correct answer)
  4. 34\frac{3}{4}
Explanation: This is the famous Monty Hall problem, which tests your understanding of conditional probability. When you encounter probability questions involving changing information or strategic choices, you need to carefully track how new information affects the original probabilities. Let's think through this step-by-step. Initially, when you choose a door, there's a 13\frac{1}{3} chance the prize is behind your door and a 23\frac{2}{3} chance it's behind one of the other two doors. Here's the key insight: when the host opens an empty door, this doesn't change the probability that your original door has the prize—it's still 13\frac{1}{3}. However, the host's action concentrates all the probability from the two unchosen doors (23\frac{2}{3}) onto the single remaining door. If you switch, you win whenever your original choice was wrong, which happens 23\frac{2}{3} of the time. Choice A (13\frac{1}{3}) represents the probability of winning if you stay with your original choice, not if you switch. Choice B (12\frac{1}{2}) is a common misconception—many people think that with two doors remaining, each has a 50-50 chance, but this ignores the crucial information the host provided. Choice D (34\frac{3}{4}) has no basis in the problem's structure. The correct answer is C (23\frac{2}{3}). For probability problems involving revealed information, remember that new information doesn't always reset the probabilities equally—trace through how the revelation affects each outcome's likelihood.

Question 6

A coin is flipped three times. What is the probability that exactly two heads occur, given that at least one head occurs?

  1. 38\frac{3}{8}
  2. 37\frac{3}{7} (correct answer)
  3. 12\frac{1}{2}
  4. 47\frac{4}{7}
Explanation: When you encounter conditional probability questions, you need to recognize that you're finding the probability of one event given that another event has already occurred. The formula is P(A|B) = P(A and B) ÷ P(B). Let's identify all possible outcomes when flipping a coin three times. There are 23=82^3 = 8 total outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Since we're given that at least one head occurs, we exclude TTT from consideration. This leaves 7 favorable outcomes for our denominator: HHH, HHT, HTH, HTT, THH, THT, TTH. Now we need exactly two heads among these 7 outcomes. Counting carefully: HHT (2 heads), HTH (2 heads), HTT (1 head), THH (2 heads), THT (1 head), TTH (1 head), HHH (3 heads). We find exactly 3 outcomes with exactly two heads: HHT, HTH, and THH. Therefore, the probability is 37\frac{3}{7}, which is answer B. Answer A (38\frac{3}{8}) incorrectly uses all 8 possible outcomes in the denominator, failing to account for the given condition. Answer C (12\frac{1}{2}) likely results from incorrectly thinking about the problem as a simple either/or scenario. Answer D (47\frac{4}{7}) might occur if you mistakenly count outcomes with at least two heads instead of exactly two heads. Remember: conditional probability problems require you to restrict your sample space to only the outcomes that satisfy the given condition, then find your target event within that restricted space.

Question 7

Two dice are rolled simultaneously. What is the probability that their sum is greater than 8, given that both dice show even numbers?

  1. 13\frac{1}{3} (correct answer)
  2. 49\frac{4}{9}
  3. 12\frac{1}{2}
  4. 59\frac{5}{9}
Explanation: When you encounter conditional probability problems, you need to focus only on the outcomes that satisfy the given condition, then find what fraction of those meet your target criteria. Since both dice must show even numbers, each die can only show 2, 4, or 6. This gives us 3 × 3 = 9 total possible outcomes: (2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), and (6,6). These form our restricted sample space. Now let's find which of these 9 outcomes have a sum greater than 8:
  • (2,2) = 4 ✗
  • (2,4) = 6 ✗
  • (2,6) = 8 ✗
  • (4,2) = 6 ✗
  • (4,4) = 8 ✗
  • (4,6) = 10 ✓
  • (6,2) = 8 ✗
  • (6,4) = 10 ✓
  • (6,6) = 12 ✓
Only 3 outcomes have sums greater than 8, so the probability is 39=13\frac{3}{9} = \frac{1}{3}. This confirms answer A. Answer B (49\frac{4}{9}) might result from incorrectly counting outcomes or including sums equal to 8. Answer C (12\frac{1}{2}) could come from oversimplifying the problem without properly listing all cases. Answer D (59\frac{5}{9}) likely stems from calculation errors or misunderstanding which sums qualify. Remember: in conditional probability problems, always identify your restricted sample space first, then count favorable outcomes within that space only. Don't use the original 36 possible dice outcomes when a condition limits your possibilities.

Question 8

A deck of cards has 4 aces and 48 non-aces. Three cards are drawn without replacement. What is the probability that exactly one ace is drawn?

  1. 4⋅48⋅4752⋅51⋅50\frac{4 \cdot 48 \cdot 47}{52 \cdot 51 \cdot 50}
  2. 3⋅4⋅48⋅4752⋅51⋅50\frac{3 \cdot 4 \cdot 48 \cdot 47}{52 \cdot 51 \cdot 50} (correct answer)
  3. 4⋅482523\frac{4 \cdot 48^2}{52^3}
  4. 12⋅4⋅48252⋅51⋅50\frac{12 \cdot 4 \cdot 48^2}{52 \cdot 51 \cdot 50}
Explanation: When you encounter probability questions involving "without replacement," you're dealing with situations where each draw changes the conditions for subsequent draws. This requires careful attention to how the sample space shrinks with each selection. To find the probability of exactly one ace in three draws, you need to consider all possible arrangements where one card is an ace and two are non-aces. There are three distinct scenarios: ace on first draw, ace on second draw, or ace on third draw. For ace on first draw: 452×4851×4750\frac{4}{52} \times \frac{48}{51} \times \frac{47}{50} For ace on second draw: 4852×451×4750\frac{48}{52} \times \frac{4}{51} \times \frac{47}{50} For ace on third draw: 4852×4751×450\frac{48}{52} \times \frac{47}{51} \times \frac{4}{50} Since these are mutually exclusive events, you add them together. Notice that each scenario gives the same result: 4×48×4752×51×50\frac{4 \times 48 \times 47}{52 \times 51 \times 50}. Since there are 3 such scenarios, the total probability is 3×4×48×4752×51×50\frac{3 \times 4 \times 48 \times 47}{52 \times 51 \times 50}, which matches choice B. Choice A forgets to account for the three different positions where the ace can appear. Choice C incorrectly treats this as sampling with replacement (using 52352^3 in the denominator). Choice D incorrectly multiplies by 12 instead of 3, suggesting confusion about the number of arrangements. Remember: for "exactly one" problems in probability, always count all possible positions where that one success can occur, then multiply by the probability of each specific arrangement.

Question 9

A jar contains 6 red marbles and 4 green marbles. Marbles are drawn one at a time without replacement until a red marble is obtained. What is the probability that exactly 3 marbles are drawn?

  1. 16\frac{1}{6}
  2. 215\frac{2}{15}
  3. 15\frac{1}{5}
  4. 110\frac{1}{10} (correct answer)
Explanation: When you encounter probability questions involving drawing without replacement until a specific condition is met, you need to map out the exact sequence of events required. For exactly 3 marbles to be drawn, the first two draws must be green marbles, and the third draw must be red. Let's calculate this step by step. Initially, there are 10 marbles total (6 red + 4 green). First draw (green): The probability is 410=25\frac{4}{10} = \frac{2}{5} Second draw (green): Now there are 9 marbles left with 3 green remaining, so the probability is 39=13\frac{3}{9} = \frac{1}{3} Third draw (red): Now there are 8 marbles left with all 6 red marbles still available, so the probability is 68=34\frac{6}{8} = \frac{3}{4} The probability of this specific sequence is: 25×13×34=660=110\frac{2}{5} \times \frac{1}{3} \times \frac{3}{4} = \frac{6}{60} = \frac{1}{10} Choice A (16\frac{1}{6}) likely comes from incorrectly assuming replacement or miscounting the favorable outcomes. Choice B (215\frac{2}{15}) might result from forgetting to account for the changing number of marbles after each draw. Choice C (15\frac{1}{5}) could come from only considering the first draw's probability. Remember: In "without replacement" problems, the denominator decreases with each draw, and you must track which colors remain available. Always multiply the probabilities of each required event in sequence.

Question 10

A spinner has three equal sections colored red, blue, and green. The spinner is spun twice. What is the probability that the same color appears both times, given that red appears at least once?

  1. 15\frac{1}{5} (correct answer)
  2. 25\frac{2}{5}
  3. 13\frac{1}{3}
  4. 35\frac{3}{5}
Explanation: When you encounter conditional probability questions, you need to find the probability of one event happening given that another event has already occurred. The key formula is: P(A given B) = P(A and B) / P(B). Let's identify our events: A = "same color both times" and B = "red appears at least once." First, find P(B). It's easier to calculate the complement: P(no red) = P(blue or green both times) = 23×23=49\frac{2}{3} \times \frac{2}{3} = \frac{4}{9}. So P(red at least once) = 1−49=591 - \frac{4}{9} = \frac{5}{9}. Next, find P(A and B) = P(same color both times AND red appears at least once). The only way both conditions are met is if red appears both times: P(red, red) = 13×13=19\frac{1}{3} \times \frac{1}{3} = \frac{1}{9}. Therefore: P(same color | red at least once) = 1/95/9=19×95=15\frac{1/9}{5/9} = \frac{1}{9} \times \frac{9}{5} = \frac{1}{5}. Choice B (25\frac{2}{5}) likely comes from incorrectly including blue-blue and green-green outcomes in the numerator. Choice C (13\frac{1}{3}) represents the unconditional probability of getting the same color both times. Choice D (35\frac{3}{5}) might result from using the wrong denominator or misunderstanding the conditional relationship. The correct answer is A. For conditional probability problems, always clearly identify what you're given versus what you're finding, and remember that the "given" condition restricts your sample space to only those outcomes where that condition is true.

Question 11

A student answers a 5-question multiple-choice test where each question has 4 options. If the student guesses randomly on each question, what is the probability of getting exactly 3 questions correct?

  1. 901024\frac{90}{1024} (correct answer)
  2. 45512\frac{45}{512}
  3. 2701024\frac{270}{1024}
  4. 1351024\frac{135}{1024}
Explanation: When you see a question about getting exactly a certain number of successes in independent trials with fixed probability, you're dealing with binomial probability. This applies when each trial has the same probability of success and trials don't affect each other. For binomial probability, use the formula: P(X=k)=(nk)⋅pk⋅(1−p)n−kP(X = k) = \binom{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, n = 5 questions, k = 3 correct answers, and p = 14\frac{1}{4} (one correct option out of four). First, calculate (53)=5!3!⋅2!=10\binom{5}{3} = \frac{5!}{3! \cdot 2!} = 10. Then: P(X=3)=10⋅(14)3⋅(34)2=10⋅164⋅964=901024P(X = 3) = 10 \cdot \left(\frac{1}{4}\right)^3 \cdot \left(\frac{3}{4}\right)^2 = 10 \cdot \frac{1}{64} \cdot \frac{9}{64} = \frac{90}{1024} Choice A gives the correct answer: 901024\frac{90}{1024}. Choice B (45512\frac{45}{512}) likely results from using the wrong combination or making an arithmetic error in the calculation. Choice C (2701024\frac{270}{1024}) appears to triple the correct numerator, possibly from confusing the number of ways to arrange the correct and incorrect answers. Choice D (1351024\frac{135}{1024}) might come from incorrectly calculating the combination (53)\binom{5}{3} as 15 instead of 10. Remember: binomial probability questions always involve counting combinations, so practice calculating (nk)\binom{n}{k} quickly and accurately. Double-check your probability calculations by ensuring they're between 0 and 1.

Question 12

Two cards are drawn from a deck of 52 cards without replacement. What is the probability that both cards are face cards (Jacks, Queens, or Kings)?

  1. 11221\frac{11}{221} (correct answer)
  2. 33221\frac{33}{221}
  3. 44221\frac{44}{221}
  4. 66221\frac{66}{221}
Explanation: This problem tests conditional probability with replacement restrictions, a common concept in probability theory. When cards are drawn "without replacement," each draw changes the composition of the remaining deck. To find the probability that both cards are face cards, you need to multiply the probability of drawing a face card first by the probability of drawing a face card second, given that the first was already removed. First, identify how many face cards exist: there are 3 face cards (Jack, Queen, King) in each of the 4 suits, giving us 12 face cards total in a 52-card deck. For the first draw: P(face card) = 1252\frac{12}{52} For the second draw: Since one face card was already removed, there are now 11 face cards remaining out of 51 total cards. So P(face card | first was face card) = 1151\frac{11}{51} The probability of both events occurring is: 1252×1151=1322652=11221\frac{12}{52} \times \frac{11}{51} = \frac{132}{2652} = \frac{11}{221} This confirms answer choice A is correct. Answer choice B (33221\frac{33}{221}) likely results from incorrectly multiplying 1252×1151\frac{12}{52} \times \frac{11}{51} and making an arithmetic error. Answer choice C (44221\frac{44}{221}) might come from using 1252×1251\frac{12}{52} \times \frac{12}{51}, forgetting that the first face card is no longer available. Answer choice D (66221\frac{66}{221}) appears to involve more fundamental calculation errors. Remember: in "without replacement" problems, always adjust your numerator and denominator for subsequent draws based on what was previously removed.

Question 13

A bag contains 8 red marbles and 12 blue marbles. Two marbles are drawn without replacement. What is the probability that the first marble is red and the second marble is blue?

  1. 2495\frac{24}{95} (correct answer)
  2. 96400\frac{96}{400}
  3. 819\frac{8}{19}
  4. 25\frac{2}{5}
Explanation: When you encounter probability questions involving drawing items "without replacement," you're dealing with dependent events where each draw affects the next. The key is calculating the probability of each event in sequence, then multiplying them together. For this problem, you need the probability that the first marble is red AND the second marble is blue. Start with the first draw: there are 8 red marbles out of 20 total marbles, so P(first red) = 820=25\frac{8}{20} = \frac{2}{5}. After removing one red marble, 19 marbles remain: 7 red and 12 blue. For the second draw, P(second blue | first red) = 1219\frac{12}{19}. Since these are dependent events, multiply the probabilities: 25×1219=2495\frac{2}{5} \times \frac{12}{19} = \frac{24}{95}. This confirms answer A is correct. Looking at the wrong answers: Answer B (96400\frac{96}{400}) represents a common error where students multiply 820×1220\frac{8}{20} \times \frac{12}{20}, incorrectly treating the events as independent by keeping the denominator at 20 for both draws. Answer C (819\frac{8}{19}) only gives the probability of drawing red on the first attempt using the wrong total. Answer D (25\frac{2}{5}) is just the probability of the first event alone, ignoring the second draw entirely. Remember: "without replacement" always means the total count decreases with each draw. Always adjust your denominator after each selection, and multiply probabilities for sequential events connected by "and."

Question 14

Three fair coins are flipped simultaneously. Event A is "exactly two heads" and Event B is "at least one tail." What is P(A ∩ B)?

  1. 18\frac{1}{8}
  2. 38\frac{3}{8} (correct answer)
  3. 12\frac{1}{2}
  4. 58\frac{5}{8}
Explanation: When you encounter probability questions involving multiple events, you need to understand what intersection (∩) means: it's the probability that both events happen simultaneously. Let's map out all possible outcomes when flipping three coins. There are 23=82^3 = 8 equally likely outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Event A (exactly two heads) occurs in: HHT, HTH, THH. That's 3 outcomes. Event B (at least one tail) occurs in all outcomes except HHH: HHT, HTH, HTT, THH, THT, TTH, TTT. That's 7 outcomes. For A ∩ B, you need outcomes that satisfy both conditions - exactly two heads AND at least one tail. Looking at our list, the outcomes with exactly two heads (HHT, HTH, THH) all automatically have at least one tail. So A ∩ B includes all three outcomes: HHT, HTH, THH. Therefore, P(A ∩ B) = 38\frac{3}{8}, which is choice B. Choice A (18\frac{1}{8}) might come from incorrectly calculating the probability of one specific outcome. Choice C (12\frac{1}{2}) could result from confusing this with a simpler probability. Choice D (58\frac{5}{8}) might come from adding probabilities incorrectly instead of finding the intersection. Remember: when finding intersections, list all outcomes systematically and check which ones satisfy all conditions simultaneously. Don't just multiply individual probabilities unless the events are independent, which intersection problems typically aren't testing.

Question 15

Two independent events A and B have probabilities P(A) = 0.4 and P(B) = 0.3. What is the probability that exactly one of these events occurs?

  1. 0.42
  2. 0.46 (correct answer)
  3. 0.58
  4. 0.70
Explanation: When you encounter probability questions involving "exactly one" of two events, you're dealing with mutually exclusive outcomes that you need to add together. To find the probability that exactly one event occurs, you need to calculate: P(A occurs and B doesn't) + P(B occurs and A doesn't). Since the events are independent, you can multiply the individual probabilities. For event A occurring without B: P(A) × P(not B) = 0.4 × (1 - 0.3) = 0.4 × 0.7 = 0.28 For event B occurring without A: P(B) × P(not A) = 0.3 × (1 - 0.4) = 0.3 × 0.6 = 0.18 Adding these together: 0.28 + 0.18 = 0.46 Choice A (0.42) likely comes from incorrectly calculating P(A) × P(not B) + P(B) × P(not A) but making an arithmetic error, perhaps using 0.6 instead of 0.7 for P(not B). Choice C (0.58) represents the probability that at least one event occurs but not both, which would be calculated as P(A) + P(B) - 2×P(A)×P(B) = 0.4 + 0.3 - 2(0.12) = 0.46. This suggests a formula error. Choice D (0.70) is simply P(A) + P(B), which gives the probability of at least one event occurring if they were mutually exclusive, ignoring the overlap entirely. Remember: For "exactly one" problems with independent events, always calculate both scenarios separately (A without B, B without A) and add them. Don't forget to use the complement (1 - P) for the event that doesn't occur.

Question 16

A box contains 4 defective items and 16 non-defective items. Three items are selected at random without replacement. What is the probability that at least one item is defective?

  1. 1457\frac{14}{57}
  2. 4357\frac{43}{57} (correct answer)
  3. 4961140\frac{496}{1140}
  4. 6441140\frac{644}{1140}
Explanation: When you encounter probability questions asking for "at least one" of something, the most efficient approach is often to use the complement rule: find the probability that none of the events occur, then subtract from 1. Here, we want P(at least one defective) = 1 - P(no defective items). With 4 defective and 16 non-defective items (20 total), we need the probability that all 3 selected items are non-defective. For the first selection: 1620\frac{16}{20} chance of non-defective For the second selection: 1519\frac{15}{19} chance of non-defective (15 non-defective remain from 19 total) For the third selection: 1418\frac{14}{18} chance of non-defective P(all non-defective) = 1620×1519×1418=33606840=1457\frac{16}{20} \times \frac{15}{19} \times \frac{14}{18} = \frac{3360}{6840} = \frac{14}{57} Therefore, P(at least one defective) = 1−1457=43571 - \frac{14}{57} = \frac{43}{57} Choice A (1457\frac{14}{57}) represents the probability that NO items are defective—the complement of what we want. Choice C (4961140\frac{496}{1140}) and Choice D (6441140\frac{644}{1140}) likely come from attempting direct calculation of "at least one" scenarios or computational errors. Notice that 4961140\frac{496}{1140} simplifies to 31.672.5\frac{31.6}{72.5}, which doesn't match our cleaner fraction. The correct answer is B. Strategy tip: "At least one" probability problems are almost always easier using complements. Calculate "none," then subtract from 1—this avoids messy multiple-case calculations.