Math 2 Quiz: Conditional Probability
13 questions · exam conditions
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Conditional ProbabilityQuestion 1 of 13

A manufacturing process produces items that are inspected by two quality checkers working independently. Checker 1 approves 85% of all items, and Checker 2 approves 90% of all items. An item is accepted only if both checkers approve it. If an item is known to be accepted, what is the probability that Checker 1 would have approved it even if working alone?

1720\frac{17}{20}
153200\frac{153}{200}
85100\frac{85}{100}
8577\frac{85}{77}
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Math 2 Quiz

Math 2 Quiz: Conditional Probability

Practice Conditional 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 Conditional 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 manufacturing process produces items that are inspected by two quality checkers working independently. Checker 1 approves 85% of all items, and Checker 2 approves 90% of all items. An item is accepted only if both checkers approve it. If an item is known to be accepted, what is the probability that Checker 1 would have approved it even if working alone?

  1. 1720\frac{17}{20}
  2. 153200\frac{153}{200}
  3. 85100\frac{85}{100} (correct answer)
  4. 8577\frac{85}{77}
Explanation: Since the checkers work independently, P(Checker 1 approves|both approve) = P(Checker 1 approves) = 0.85. The condition of acceptance doesn't change Checker 1's individual approval probability due to independence.

Question 2

A survey of 200 college students found that 120 own smartphones, 80 own tablets, and 50 own both devices. Given that a randomly selected student from this survey owns a tablet, what is the probability that this student does not own a smartphone?

  1. 38\frac{3}{8} (correct answer)
  2. 58\frac{5}{8}
  3. 3080\frac{30}{80}
  4. 70200\frac{70}{200}
Explanation: P(no smartphone|tablet) = P(tablet only)/P(tablet) = 30/80 = 3/8. Of the 80 tablet owners, 50 also own smartphones, so 30 own only tablets.

Question 3

In a game, players draw cards from a deck where 60% are red and 40% are blue. Red cards have a 70% chance of being winning cards, while blue cards have a 30% chance of being winning cards. If a player draws a winning card, what is the probability it was red?

  1. 2135\frac{21}{35}
  2. 4254\frac{42}{54}
  3. 79\frac{7}{9} (correct answer)
  4. 57\frac{5}{7}
Explanation: Using Bayes' theorem: P(red|winning) = P(winning|red)×P(red)/P(winning) = (0.7×0.6)/(0.7×0.6 + 0.3×0.4) = 0.42/(0.42 + 0.12) = 0.42/0.54 = 7/9.

Question 4

A medical study tracks patients with two conditions: diabetes (D) and high blood pressure (H). Among all patients, 30% have diabetes, 40% have high blood pressure, and 15% have both conditions. If a patient is selected at random and found to have high blood pressure, what is the probability that this patient also has diabetes?

  1. 18\frac{1}{8}
  2. 38\frac{3}{8} (correct answer)
  3. 12\frac{1}{2}
  4. 58\frac{5}{8}
Explanation: We need to find P(D|H) = P(D and H)/P(H) = 0.15/0.40 = 3/8. This represents the proportion of high blood pressure patients who also have diabetes.

Question 5

A quality control system has two stages. Stage 1 detects defects with probability 0.8, and Stage 2 detects defects with probability 0.7. If a product passes Stage 1 (no defect detected), it goes to Stage 2. Given that a defective product was not detected by either stage, what was the probability it passed Stage 1?

  1. 13\frac{1}{3}
  2. 25\frac{2}{5}
  3. 35\frac{3}{5}
  4. 23\frac{2}{3} (correct answer)
Explanation: P(passed Stage 1 | undetected) = P(undetected via Stage 1)/P(total undetected) = 0.2/(0.2 + 0.8×0.3) = 0.2/0.44 = 2/3. The product could be undetected by failing Stage 1 (prob 0.2) or passing Stage 1 but failing Stage 2 (prob 0.8×0.3 = 0.24).

Question 6

A company's hiring process has three rounds. Candidates have a 70% chance of passing Round 1, and those who pass Round 1 have a 60% chance of passing Round 2, and those who pass Round 2 have an 80% chance of passing Round 3. Given that a candidate failed to get hired (didn't pass all three rounds), what is the probability they failed in Round 1?

  1. 3067\frac{30}{67}
  2. 310\frac{3}{10}
  3. 3050\frac{30}{50}
  4. 3066\frac{30}{66} (correct answer)
Explanation: P(fail Round 1|not hired) = P(fail Round 1)/P(not hired) = 0.3/0.664 ≈ 30/66. P(not hired) = 1 - P(pass all) = 1 - (0.7×0.6×0.8) = 1 - 0.336 = 0.664. Of all candidates who don't get hired, 30/66 failed at the very first round.

Question 7

A diagnostic test for a disease has a 95% accuracy rate for sick patients (sensitivity) and an 88% accuracy rate for healthy patients (specificity). In a population where 2% of people have the disease, if someone tests positive, what is the probability they actually have the disease?

  1. 19118\frac{19}{118}
  2. 19137\frac{19}{137} (correct answer)
  3. 9598\frac{95}{98}
  4. 19100\frac{19}{100}
Explanation: When you encounter a medical testing problem, you're dealing with conditional probability and Bayes' theorem. The key insight is that a positive test result doesn't guarantee disease presence—you must account for the disease's rarity in the population. Let's work through this systematically. In a population of 10,000 people, 2% (200 people) have the disease and 9,800 are healthy. Among the 200 sick patients, 95% test positive, giving us 190 true positives. Among the 9,800 healthy patients, 88% test negative, meaning 12% test positive—that's 1,176 false positives. Total positive tests: 190 + 1,176 = 1,366 True positives: 190 So the probability of actually having the disease given a positive test is 1901,366=19136.6\frac{190}{1,366} = \frac{19}{136.6}. Since we need an exact fraction, this simplifies to 19137\frac{19}{137}. Choice A (19118\frac{19}{118}) likely comes from incorrectly calculating the false positives—perhaps using 88% instead of 12% for healthy patients who test positive. Choice C (9598\frac{95}{98}) represents the common misconception of thinking the answer is simply the sensitivity rate, ignoring the population's disease prevalence. Choice D (19100\frac{19}{100}) appears to incorrectly use percentages without proper population weighting. Remember: rare diseases combined with imperfect tests often yield surprisingly low positive predictive values. Always consider both the test's accuracy AND the disease's prevalence in the population—this combination frequently appears on standardized exams.

Question 8

A security system uses two independent sensors. Sensor A triggers 90% of the time when there's an intrusion and 5% of the time when there's no intrusion. Sensor B triggers 85% of the time when there's an intrusion and 8% of the time when there's no intrusion. Intrusions occur 1% of the time. If both sensors trigger simultaneously, what is the probability there is actually an intrusion?

  1. 0.7651.161\frac{0.765}{1.161} (correct answer)
  2. 0.7650.765\frac{0.765}{0.765}
  3. 85100\frac{85}{100}
  4. 0.7651.000\frac{0.765}{1.000}
Explanation: P(intrusion|both trigger) = P(both trigger|intrusion)×P(intrusion)/P(both trigger) = (0.9×0.85×0.01)/((0.9×0.85×0.01) + (0.05×0.08×0.99)) = 0.00765/(0.00765 + 0.00396) = 0.765/1.161.

Question 9

A company's hiring process involves two interviews. Historical data shows that 60% of applicants pass the first interview. Of those who pass the first interview, 40% also pass the second interview. If an applicant is selected at random from those who attempted both interviews, what is the probability they passed the second interview given that they failed the first interview?

  1. 00 (correct answer)
  2. 0.240.24
  3. 0.400.40
  4. 0.600.60
Explanation: If an applicant fails the first interview, they cannot proceed to the second interview, so the probability of passing the second interview given failure in the first is 0. Choice B represents P(pass both interviews). Choice C represents P(pass second | pass first). Choice D represents P(pass first interview).

Question 10

A cybersecurity system flags suspicious network activity based on two criteria: unusual data volume (V) and irregular access patterns (A). The system's performance data shows P(V) = 0.15, P(A) = 0.25, and P(V ∩ A) = 0.08. If the system detects irregular access patterns, what is the probability that there is also unusual data volume?

  1. 0.080.15≈0.533\frac{0.08}{0.15} ≈ 0.533
  2. 0.080.08
  3. 0.080.25=0.32\frac{0.08}{0.25} = 0.32 (correct answer)
  4. 0.15×0.250.08≈0.469\frac{0.15 \times 0.25}{0.08} ≈ 0.469
Explanation: When you encounter questions about finding probabilities "given that" something has occurred, you're dealing with conditional probability. The key phrase here is "if the system detects irregular access patterns" - this tells you we want P(V|A), the probability of unusual data volume given irregular access patterns. The conditional probability formula is P(V|A) = P(V ∩ A) / P(A). You're given P(V ∩ A) = 0.08 (both events occur together) and P(A) = 0.25 (irregular access patterns occur). Substituting these values: P(V|A) = 0.08 / 0.25 = 0.32. Choice A calculates 0.080.15\frac{0.08}{0.15} - this would be P(A|V), the probability of irregular access given unusual volume. This reverses the conditional relationship and uses P(V) in the denominator instead of P(A). Choice B simply states 0.08, which is P(V ∩ A) - the probability both events occur simultaneously. This ignores the conditional nature entirely and doesn't account for the given information. Choice D calculates 0.15×0.250.08\frac{0.15 \times 0.25}{0.08}, which resembles Bayes' theorem but applies it incorrectly. This formula would only be relevant if we were trying to update a prior probability, which isn't the case here. Remember: conditional probability questions always follow the pattern P(Event|Given) = P(Event ∩ Given) / P(Given). The denominator should always be the probability of whatever comes after "given that" in the question.

Question 11

An online retailer analyzes customer behavior regarding two actions: adding items to wishlist (W) and making a purchase within 30 days (P). Data shows that 45% of customers add items to wishlist, 30% make purchases, and 20% do both. What is the probability that a customer who adds items to their wishlist will make a purchase within 30 days?

  1. 0.200.45≈0.444\frac{0.20}{0.45} ≈ 0.444 (correct answer)
  2. 0.200.30≈0.667\frac{0.20}{0.30} ≈ 0.667
  3. 0.250.45≈0.556\frac{0.25}{0.45} ≈ 0.556
  4. 0.300.45≈0.667\frac{0.30}{0.45} ≈ 0.667
Explanation: We need P(P | W) = P(P ∩ W) / P(W) = 0.20 / 0.45 ≈ 0.444. Choice B gives P(W | P) instead by using P(P) in denominator. Choice C incorrectly calculates the numerator as 0.45 - 0.20 = 0.25 (customers with wishlist but no purchase). Choice D uses P(P) in the numerator instead of P(P ∩ W).

Question 12

A medical study tracks patients with two conditions: high blood pressure (H) and diabetes (D). The data shows that 35% of all patients have high blood pressure, 20% have diabetes, and 12% have both conditions. If a randomly selected patient is known to have diabetes, what is the probability that this patient also has high blood pressure?

  1. 1235=0.343\frac{12}{35} = 0.343
  2. 1220=0.60\frac{12}{20} = 0.60 (correct answer)
  3. 12100=0.12\frac{12}{100} = 0.12
  4. 3520=1.75\frac{35}{20} = 1.75
Explanation: We need to find P(H|D) = P(H and D)/P(D). Given that P(H and D) = 0.12 and P(D) = 0.20, we get P(H|D) = 0.12/0.20 = 0.60. Choice A incorrectly uses P(D) in the numerator and P(H) in the denominator. Choice C gives the joint probability, not the conditional probability. Choice D incorrectly divides P(H) by P(D).

Question 13

A survey of 500 college students found that 180 students own a car, 220 own a bicycle, and 90 own both a car and a bicycle. Given that a randomly selected student owns a bicycle, what is the probability that they do NOT own a car?

  1. 90220=0.409\frac{90}{220} = 0.409
  2. 320500=0.64\frac{320}{500} = 0.64
  3. 130500=0.26\frac{130}{500} = 0.26
  4. 130220≈0.591\frac{130}{220} ≈ 0.591 (correct answer)
Explanation: When you encounter conditional probability problems with overlapping sets, you need to focus on the specific group mentioned in the condition. This question asks for the probability of NOT owning a car, given that the student owns a bicycle. Since we know the student owns a bicycle, our sample space is now just the 220 bicycle owners, not all 500 students. Among these 220 bicycle owners, 90 also own a car. Therefore, 220−90=130220 - 90 = 130 bicycle owners do NOT own a car. The conditional probability is: P(no car | bicycle)=bicycle owners without carstotal bicycle owners=130220≈0.591P(\text{no car | bicycle}) = \frac{\text{bicycle owners without cars}}{\text{total bicycle owners}} = \frac{130}{220} ≈ 0.591 Answer choice A (90220\frac{90}{220}) gives you the probability that a bicycle owner DOES own a car - this is the complement of what we want. Answer choice B (320500\frac{320}{500}) incorrectly uses the entire population of 500 students as the denominator, ignoring the conditional aspect. The 320 represents all students who don't own cars, but we only care about bicycle owners. Answer choice C (130500\frac{130}{500}) makes the same error with the denominator - it finds the probability that any randomly selected student owns a bicycle but not a car, rather than the conditional probability we need. Remember: in conditional probability problems, always restrict your sample space to the given condition. The phrase "given that" tells you to focus only on that subset of the population.