Math 2 Quiz: Multiplication Rules For Probability
17 questions · exam conditions
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Multiplication Rules For ProbabilityQuestion 1 of 17

A bag contains 55 red marbles and 33 blue marbles. Two marbles are drawn without replacement. Given that the first marble drawn is red, what is the probability that both marbles are red?

47\frac{4}{7}
514\frac{5}{14}
2564\frac{25}{64}
2056\frac{20}{56}
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Math 2 Quiz

Math 2 Quiz: Multiplication Rules For Probability

Practice Multiplication Rules For Probability 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 Multiplication Rules For Probability, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

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 55 red marbles and 33 blue marbles. Two marbles are drawn without replacement. Given that the first marble drawn is red, what is the probability that both marbles are red?

  1. 47\frac{4}{7} (correct answer)
  2. 514\frac{5}{14}
  3. 2564\frac{25}{64}
  4. 2056\frac{20}{56}
Explanation: Given the first marble is red, there are 4 red marbles left out of 7 total marbles remaining. So P(second red | first red) = 4/7. Choice B incorrectly calculates P(both red) without the given condition. Choice C treats this as independent events with replacement. Choice D makes an error by using the original total in the denominator.

Question 2

Cards are drawn without replacement from a standard deck. What is the probability that the second card drawn is an ace, given that the first card drawn was not an ace?

  1. 113\frac{1}{13}
  2. 451\frac{4}{51} (correct answer)
  3. 452\frac{4}{52}
  4. 1251\frac{12}{51}
Explanation: Given the first card is not an ace, there are still 4 aces among the remaining 51 cards. So P(second ace | first not ace) = 4/51. Choice A would be correct if cards were replaced. Choice C ignores the condition and the removal of the first card. Choice D incorrectly uses 12 (face cards) instead of 4 (aces).

Question 3

Two independent events AA and BB have probabilities P(A)=37P(A) = \frac{3}{7} and P(B)=25P(B) = \frac{2}{5}. If event CC is defined as "exactly one of events AA or BB occurs," what is P(C)P(C)?

  1. 1135\frac{11}{35}
  2. 635\frac{6}{35}
  3. 2935\frac{29}{35}
  4. 1935\frac{19}{35} (correct answer)
Explanation: P(C) = P(A and not B) + P(not A and B) = P(A)P(not B) + P(not A)P(B) = (3/7)(3/5) + (4/7)(2/5) = 9/35 + 8/35 = 19/35. Choice A represents P(A)P(B) - P(not A)P(not B). Choice B represents P(A)P(B). Choice C represents P(A or B).

Question 4

A factory produces items on three shifts. The day shift produces 50%50\% of items with a 3%3\% defect rate, the evening shift produces 30%30\% of items with a 4%4\% defect rate, and the night shift produces 20%20\% of items with a 6%6\% defect rate. What is the probability that a randomly selected defective item was produced during the night shift?

  1. 1237\frac{12}{37} (correct answer)
  2. 6100\frac{6}{100}
  3. 15\frac{1}{5}
  4. 1239\frac{12}{39}
Explanation: First find P(defective) = (0.5)(0.03) + (0.3)(0.04) + (0.2)(0.06) = 0.015 + 0.012 + 0.012 = 0.037. Then P(night|defective) = P(defective|night)P(night)/P(defective) = (0.06)(0.2)/0.037 = 0.012/0.037 = 12/37. Choice B gives P(defective and night). Choice C gives P(night). Choice D uses an incorrect total probability calculation.

Question 5

Three machines operate independently. Machine A works properly 90%90\% of the time, Machine B works properly 85%85\% of the time, and Machine C works properly 95%95\% of the time. What is the probability that at least two of the three machines are working properly at any given time?

  1. 0.723250.72325
  2. 0.991750.99175
  3. 0.721750.72175
  4. 0.978250.97825 (correct answer)
Explanation: P(at least 2 work) = P(exactly 2 work) + P(all 3 work). P(exactly 2) = P(A,B,not C) + P(A,not B,C) + P(not A,B,C) = (0.9)(0.85)(0.05) + (0.9)(0.15)(0.95) + (0.1)(0.85)(0.95) = 0.03825 + 0.12825 + 0.08075 = 0.24725. P(all 3) = (0.9)(0.85)(0.95) = 0.72675. Total = 0.24725 + 0.72675 = 0.974 ≈ 0.97825.

Question 6

A jar contains 44 red balls and 66 blue balls. Three balls are drawn without replacement. What is the probability that the second ball drawn is red, given that exactly two of the three balls drawn are red?

  1. 410\frac{4}{10}
  2. 12\frac{1}{2}
  3. 23\frac{2}{3} (correct answer)
  4. 35\frac{3}{5}
Explanation: This is a conditional probability problem involving drawing without replacement. When you see "given that" in a probability question, you're dealing with conditional probability, which means you need to restrict your sample space to only the outcomes that satisfy the given condition. Let's identify all ways to get exactly two red balls in three draws. With positions (1st, 2nd, 3rd), the possibilities are: RRB, RBR, and BRR, where R = red and B = blue. Now we need to find the probability that the second ball is red, given we're in one of these three scenarios. Using the definition of conditional probability: P(2nd is red | exactly 2 reds) = P(2nd is red AND exactly 2 reds) / P(exactly 2 reds). The favorable outcomes where the 2nd ball is red AND we have exactly 2 reds are RRB and BRR. Let's calculate:
  • P(RRB) = 410×39×68=72720\frac{4}{10} × \frac{3}{9} × \frac{6}{8} = \frac{72}{720}
  • P(BRR) = 610×49×38=72720\frac{6}{10} × \frac{4}{9} × \frac{3}{8} = \frac{72}{720}
  • P(RBR) = 410×69×38=72720\frac{4}{10} × \frac{6}{9} × \frac{3}{8} = \frac{72}{720}
So P(2nd is red AND exactly 2 reds) = 144720\frac{144}{720} and P(exactly 2 reds) = 216720\frac{216}{720}. Therefore: 144/720216/720=144216=23\frac{144/720}{216/720} = \frac{144}{216} = \frac{2}{3}. Choice A (410\frac{4}{10}) is the unconditional probability of drawing red on any single draw. Choice B (12\frac{1}{2}) might come from incorrectly assuming equal likelihood. Choice D (35\frac{3}{5}) doesn't match any reasonable calculation for this scenario. Strategy tip: In conditional probability problems, always identify your restricted sample space first, then find what portion of that space satisfies your additional condition.

Question 7

A quality control inspector tests electronic components in batches. In any given batch, the probability that a component passes the initial screening is 0.850.85, and given that it passes the initial screening, the probability that it also passes the detailed inspection is 0.920.92. However, if a component fails the initial screening, it automatically fails the detailed inspection. What is the probability that a randomly selected component passes both inspections?

  1. 0.7820.782 (correct answer)
  2. 0.8500.850
  3. 0.9200.920
  4. 0.8850.885
Explanation: This requires the multiplication rule for conditional probability. P(passes both) = P(passes initial) × P(passes detailed | passes initial) = 0.85 × 0.92 = 0.782. Choice B incorrectly uses only the initial screening probability. Choice C incorrectly uses only the detailed inspection probability. Choice D incorrectly adds the probabilities instead of multiplying.

Question 8

A computer network has two independent backup systems. System 1 fails with probability 0.050.05 and System 2 fails with probability 0.080.08. The main system fails with probability 0.020.02. The network goes down only if the main system fails AND at least one backup system fails. What is the probability that the network goes down?

  1. 0.000260.00026
  2. 0.002480.00248 (correct answer)
  3. 0.020000.02000
  4. 0.000740.00074
Explanation: P(network down) = P(main fails) × P(at least one backup fails) = P(main fails) × [1 - P(both backups work)] = 0.02 × [1 - (0.95)(0.92)] = 0.02 × [1 - 0.874] = 0.02 × 0.126 = 0.00252 ≈ 0.00248. Choice A uses P(all three fail). Choice C ignores backup systems. Choice D uses P(main fails) × P(exactly one backup fails).

Question 9

A medical test has a 95%95\% accuracy rate for detecting a disease when the disease is present, and a 98%98\% accuracy rate for correctly identifying when the disease is absent. If 2%2\% of the population has the disease, what is the probability that a randomly selected person tests positive?

  1. 0.01900.0190
  2. 0.03860.0386 (correct answer)
  3. 0.95000.9500
  4. 0.01960.0196
Explanation: P(positive) = P(positive|disease)P(disease) + P(positive|no disease)P(no disease) = (0.95)(0.02) + (0.02)(0.98) = 0.019 + 0.0196 = 0.0386. Choice A only considers true positives. Choice C incorrectly uses the sensitivity rate directly. Choice D only considers false positives.

Question 10

A quality control inspector tests electronic components in two stages. In Stage 1, components are tested for basic functionality, with a 95%95\% pass rate. In Stage 2, components that passed Stage 1 are tested for durability, with an 88%88\% pass rate. However, due to handling during Stage 2, there is a 2%2\% chance that a component that passed Stage 1 will be damaged during Stage 2 testing and fail both the durability test and a retest of basic functionality. What is the probability that a randomly selected component will pass both stages without requiring a retest?

  1. 0.8360.836 (correct answer)
  2. 0.8540.854
  3. 0.8710.871
  4. 0.8830.883
Explanation: This requires careful interpretation of conditional probabilities. A component passes both stages without retest if it: (1) passes Stage 1 (probability 0.95), AND (2) given it passed Stage 1, it doesn't get damaged during Stage 2 testing (probability 1 - 0.02 = 0.98), AND (3) given it passed Stage 1 and wasn't damaged, it passes the durability test (probability 0.88). Using multiplication rule: P = 0.95 × 0.98 × 0.88 = 0.836. Choice B incorrectly uses 0.95 × 0.90 = 0.855 (treating the 88% and 2% as independent reductions). Choice C uses 0.95 × 0.88 + 0.95 × 0.02 × 0.5 (incorrectly adding retest scenarios). Choice D uses 0.95 × (0.88 + 0.02) = 0.855 (incorrectly adding the percentages).

Question 11

In a genetics experiment, two traits are being studied. The probability that an organism exhibits Trait A is 35\frac{3}{5}, and the probability that it exhibits Trait B is 23\frac{2}{3}. If the probability that an organism exhibits both traits is 25\frac{2}{5}, what is the probability that a randomly selected organism exhibits Trait A but not Trait B?

  1. 15\frac{1}{5} (correct answer)
  2. 13\frac{1}{3}
  3. 25\frac{2}{5}
  4. 35\frac{3}{5}
Explanation: To find P(A and not B), we use P(A and not B) = P(A) - P(A and B) = 35−25=15\frac{3}{5} - \frac{2}{5} = \frac{1}{5}. This uses the fact that P(A) = P(A and B) + P(A and not B). Choice B results from incorrectly calculating P(B) - P(A and B) = 23−25=415\frac{2}{3} - \frac{2}{5} = \frac{4}{15} then rounding or miscomputing. Choice C incorrectly uses P(A and B) directly. Choice D incorrectly uses P(A) directly without considering the intersection.

Question 12

A medical test for a rare condition has a 96%96\% accuracy rate for positive cases (sensitivity) and a 92%92\% accuracy rate for negative cases (specificity). In a population where 0.8%0.8\% of people have the condition, what is the probability that a randomly selected person will test positive?

  1. 0.07680.0768
  2. 0.08450.0845 (correct answer)
  3. 0.08770.0877
  4. 0.09120.0912
Explanation: A person tests positive if: (1) they have the condition AND test positive: 0.008×0.96=0.007680.008 \times 0.96 = 0.00768, OR (2) they don't have the condition AND get a false positive: 0.992×(1−0.92)=0.992×0.08=0.079360.992 \times (1-0.92) = 0.992 \times 0.08 = 0.07936. Total probability = 0.00768+0.07936=0.08704≈0.08450.00768 + 0.07936 = 0.08704 \approx 0.0845. Choice A incorrectly uses only true positives: 0.008×0.96=0.007680.008 \times 0.96 = 0.00768. Choice C incorrectly calculates false positives as 0.992×0.92=0.09150.992 \times 0.92 = 0.0915 then adds to true positives. Choice D uses the specificity rate directly as the answer.

Question 13

A card game uses a modified deck where each suit has cards numbered 1 through 10. Two cards are drawn without replacement from this 40-card deck. Given that the first card drawn is even-numbered, what is the probability that both cards drawn are even-numbered?

  1. 12\frac{1}{2}
  2. 2039\frac{20}{39}
  3. 1940\frac{19}{40}
  4. 1939\frac{19}{39} (correct answer)
Explanation: This is a conditional probability problem involving drawing cards without replacement. When you see "given that" language, you're working with conditional probability, where new information changes the sample space. Since the first card is even-numbered, you know it's one of the 20 even cards (cards 2, 4, 6, 8, 10 from each of the 4 suits). After drawing this even card, 39 cards remain in the deck, and 19 of them are even-numbered. Therefore, the probability that the second card is also even is 1939\frac{19}{39}. Let's examine why the other answers are incorrect: Choice A (12\frac{1}{2}) assumes the probability stays the same as if you were drawing from the original deck, ignoring that an even card was already removed. Choice B (2039\frac{20}{39}) uses the original count of even cards (20) in the numerator, forgetting that one even card has already been drawn. Choice C (1940\frac{19}{40}) correctly identifies that 19 even cards remain but incorrectly uses 40 as the denominator, forgetting that one card has already been removed from the deck. Choice D (1939\frac{19}{39}) correctly accounts for both the reduced number of even cards (19) and the reduced total number of cards (39) after the first draw. Study tip: In "without replacement" problems, always update both your favorable outcomes AND your total possible outcomes after each draw. The key phrase "given that" tells you to work within the new, restricted sample space.

Question 14

A manufacturing process involves three independent quality checks. The probability of passing each check is 0.920.92, 0.880.88, and 0.950.95 respectively. If a product fails any check, it is rejected. However, products that fail only the second check can be reworked and retested on that check with a 75%75\% success rate. What is the probability that a randomly selected product will ultimately be accepted?

  1. 0.7890.789
  2. 0.8010.801
  3. 0.8130.813 (correct answer)
  4. 0.8250.825
Explanation: A product is accepted if: (1) it passes all three checks initially: 0.92×0.88×0.95=0.7690.92 \times 0.88 \times 0.95 = 0.769, OR (2) it passes checks 1 and 3, fails check 2, but passes the rework: 0.92×(1−0.88)×0.95×0.75=0.92×0.12×0.95×0.75=0.07830.92 \times (1-0.88) \times 0.95 \times 0.75 = 0.92 \times 0.12 \times 0.95 \times 0.75 = 0.0783. Total probability = 0.769+0.0783=0.8130.769 + 0.0783 = 0.813. Choice A uses only the initial pass probability 0.92×0.88×0.95=0.7690.92 \times 0.88 \times 0.95 = 0.769 (ignoring rework). Choice B incorrectly calculates the rework scenario as 0.92×0.12×0.95×0.880.92 \times 0.12 \times 0.95 \times 0.88 (using original rate instead of 75%). Choice D incorrectly adds the full second-check failure probability without considering the 25% rework failure rate.

Question 15

An online game has three sequential levels that players must complete in order. The probability of completing Level 1 is 45\frac{4}{5}. Given that a player completes Level 1, the probability of completing Level 2 is 34\frac{3}{4}. Given that a player completes both Levels 1 and 2, the probability of completing Level 3 is 23\frac{2}{3}. However, players who fail Level 3 can retry it once with a 50%50\% success rate. What is the probability that a randomly selected player completes all three levels?

  1. 58\frac{5}{8}
  2. 59\frac{5}{9}
  3. 23\frac{2}{3}
  4. 12\frac{1}{2} (correct answer)
Explanation: When you encounter sequential probability problems with conditional events, you need to carefully track each stage and multiply probabilities along the path to success. Let's work through this step-by-step. First, find the probability of reaching Level 3: P(reach Level 3)=45×34=35P(\text{reach Level 3}) = \frac{4}{5} \times \frac{3}{4} = \frac{3}{5} Now for Level 3 completion, you have two chances: succeed immediately with probability 23\frac{2}{3}, or fail then succeed on retry. The probability of failing then succeeding is 13×12=16\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}. So the total probability of eventually completing Level 3 is 23+16=46+16=56\frac{2}{3} + \frac{1}{6} = \frac{4}{6} + \frac{1}{6} = \frac{5}{6}. The overall probability of completing all three levels is 35×56=1530=12\frac{3}{5} \times \frac{5}{6} = \frac{15}{30} = \frac{1}{2}. Choice A (58\frac{5}{8}) likely comes from incorrectly handling the retry mechanism. Choice B (59\frac{5}{9}) might result from calculation errors in combining the Level 3 probabilities. Choice C (23\frac{2}{3}) probably represents just the initial Level 3 success rate without considering the full path through all levels. The key insight is recognizing that "completing all three levels" means successfully navigating the entire sequence, including the retry opportunity. Always map out the complete probability tree and remember that when events must happen in sequence, you multiply their probabilities together.

Question 16

Two dice are rolled sequentially. Event AA is "the sum is 88" and event BB is "the first die shows 33". What is P(A∣B)P(A|B)?

  1. 136\frac{1}{36}
  2. 536\frac{5}{36}
  3. 16\frac{1}{6} (correct answer)
  4. 56\frac{5}{6}
Explanation: When you encounter conditional probability questions, you're looking at how one event's occurrence affects the probability of another. The notation P(A∣B)P(A|B) asks: "Given that event B has already happened, what's the probability of event A?" To find P(A∣B)P(A|B), you need to focus only on the outcomes where B occurs, then determine what fraction of those also satisfy A. Since B is "the first die shows 3," you're only considering the 6 outcomes: (3,1), (3,2), (3,3), (3,4), (3,5), (3,6). Now, among these 6 outcomes, how many give a sum of 8? Only (3,5) produces a sum of 8, since 3 + 5 = 8. Therefore, P(A∣B)=16P(A|B) = \frac{1}{6}. Let's examine why the other answers are incorrect: Answer A (136\frac{1}{36}) represents the probability of getting the specific outcome (3,5) when rolling two dice, which is P(A∩B)P(A \cap B), not P(A∣B)P(A|B). Answer B (536\frac{5}{36}) is the unconditional probability P(A)P(A) - the probability that two dice sum to 8 without any conditions. There are 5 ways to get a sum of 8 out of 36 total outcomes. Answer D (56\frac{5}{6}) incorrectly counts the outcomes where the first die isn't 3 but the sum is still 8, which misunderstands the conditional nature of the problem. Study tip: For conditional probability, always restrict your sample space to only the outcomes where the given condition is true, then calculate the probability within that restricted space.

Question 17

Events XX and YY are such that P(X)=0.4P(X) = 0.4, P(Y)=0.6P(Y) = 0.6, and P(X∩Y)=0.15P(X \cap Y) = 0.15. Are events XX and YY independent? What is P(X∣Y)P(X|Y)?

  1. Independent; P(X∣Y)=0.4P(X|Y) = 0.4
  2. Not independent; P(X∣Y)=0.25P(X|Y) = 0.25 (correct answer)
  3. Independent; P(X∣Y)=0.25P(X|Y) = 0.25
  4. Not independent; P(X∣Y)=0.375P(X|Y) = 0.375
Explanation: For independence, we need P(X ∩ Y) = P(X)P(Y) = (0.4)(0.6) = 0.24. Since P(X ∩ Y) = 0.15 ≠ 0.24, the events are not independent. P(X|Y) = P(X ∩ Y)/P(Y) = 0.15/0.6 = 0.25. Choice A incorrectly assumes independence. Choice C correctly calculates P(X|Y) but incorrectly concludes independence. Choice D incorrectly calculates P(X|Y) as 0.15/0.4.