Business Statistics Quiz: Conditional Probability And Screening
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Conditional Probability And ScreeningQuestion 1 of 20

An automated quality control system on an assembly line is designed to catch improperly assembled units. The prevalence of such units is 4%. The system has a sensitivity of 99% and a specificity of 96%. If the system gives a 'pass' signal (negative result) for a unit, what is the probability that the unit is, in fact, improperly assembled?

0.0004
0.0100
0.0400
0.9996
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Business Statistics Quiz

Business Statistics Quiz: Conditional Probability And Screening

Practice Conditional Probability And Screening in Business Statistics 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 And Screening, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Statistics.

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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.

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Question 1

An automated quality control system on an assembly line is designed to catch improperly assembled units. The prevalence of such units is 4%. The system has a sensitivity of 99% and a specificity of 96%. If the system gives a 'pass' signal (negative result) for a unit, what is the probability that the unit is, in fact, improperly assembled?

  1. 0.0004 (correct answer)
  2. 0.0100
  3. 0.0400
  4. 0.9996
Explanation: This question asks for the probability of a defect given a negative test, or P(Defect∣Test Negative)P(\text{Defect} | \text{Test Negative}). This is also known as the False Omission Rate, or 1−NPV1 - \text{NPV}. Let D = Defect. We have P(D)=0.04P(D)=0.04, P(D′)=0.96P(D')=0.96, P(Test+∣D)=0.99P(\text{Test+}|D)=0.99, and P(Test-∣D′)=0.96P(\text{Test-}|D')=0.96. The false negative rate is P(Test-∣D)=1−0.99=0.01P(\text{Test-}|D) = 1 - 0.99 = 0.01. We can calculate P(D∣Test-)P(D|\text{Test-}) using Bayes' Theorem: P(D∣Test-)=P(Test-∣D)P(D)P(Test-∣D)P(D)+P(Test-∣D′)P(D′)=(0.01)(0.04)(0.01)(0.04)+(0.96)(0.96)=0.00040.0004+0.9216=0.00040.9220≈0.00043P(D|\text{Test-}) = \frac{P(\text{Test-}|D)P(D)}{P(\text{Test-}|D)P(D) + P(\text{Test-}|D')P(D')} = \frac{(0.01)(0.04)}{(0.01)(0.04) + (0.96)(0.96)} = \frac{0.0004}{0.0004 + 0.9216} = \frac{0.0004}{0.9220} \approx 0.00043.

Question 2

A data center has two independent backup power systems, A and B. The probability that system A fails is 0.10, and the probability that system B fails is 0.05. Given that at least one of the systems has failed, what is the probability that both systems have failed?

  1. 0.005
  2. 0.034 (correct answer)
  3. 0.050
  4. 0.145
Explanation: We want to find P(A∩B∣A∪B)P(A \cap B | A \cup B). The formula for this is P(A∩B)/P(A∪B)P(A \cap B) / P(A \cup B). Since the systems are independent: The probability both fail is P(A∩B)=P(A)×P(B)=0.10×0.05=0.005P(A \cap B) = P(A) \times P(B) = 0.10 \times 0.05 = 0.005. The probability at least one fails is P(A∪B)=P(A)+P(B)−P(A∩B)=0.10+0.05−0.005=0.145P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.10 + 0.05 - 0.005 = 0.145. Therefore, the conditional probability is 0.0050.145≈0.03448\frac{0.005}{0.145} \approx 0.03448.

Question 3

An e-commerce company knows that 30% of its customers are 'high-value'. For these high-value customers, the probability of responding to a targeted email ad is 0.6. What is the probability that a randomly selected customer is both high-value AND responds to the ad?

  1. 0.18 (correct answer)
  2. 0.30
  3. 0.60
  4. 0.80
Explanation: This question asks for a joint probability, P(A∩B)P(A \cap B). Let A be the event that a customer is high-value, and B be the event they respond to the ad. We are given P(A)=0.30P(A) = 0.30 and the conditional probability P(B∣A)=0.60P(B|A) = 0.60. The multiplication rule for conditional probability states P(A∩B)=P(B∣A)×P(A)P(A \cap B) = P(B|A) \times P(A). Therefore, the probability is 0.60×0.30=0.180.60 \times 0.30 = 0.18.

Question 4

A company screens all new hires for a specific skill using a proficiency test. The test has a sensitivity of 92% and a specificity of 85%. Historically, 20% of applicants possess the skill. For a new hire who tests positive for the skill, what is the probability that they do not actually possess it?

  1. 0.080
  2. 0.150
  3. 0.395 (correct answer)
  4. 0.605
Explanation: This question asks for P(No Skill∣Test Positive)P(\text{No Skill} | \text{Test Positive}), which is the false positive rate among those who test positive, or 1−PPV1 - \text{PPV}. Let S = has skill. We have P(S)=0.20P(S)=0.20, P(S′)=0.80P(S')=0.80, P(Test+∣S)=0.92P(\text{Test+}|S)=0.92, and P(Test-∣S′)=0.85P(\text{Test-}|S')=0.85. The false positive rate is P(Test+∣S′)=1−0.85=0.15P(\text{Test+}|S') = 1 - 0.85 = 0.15. First, calculate PPV = P(S∣Test+)=P(Test+∣S)P(S)P(Test+∣S)P(S)+P(Test+∣S′)P(S′)=(0.92)(0.20)(0.92)(0.20)+(0.15)(0.80)=0.1840.184+0.12=0.1840.304≈0.605P(S|\text{Test+}) = \frac{P(\text{Test+}|S)P(S)}{P(\text{Test+}|S)P(S) + P(\text{Test+}|S')P(S')} = \frac{(0.92)(0.20)}{(0.92)(0.20) + (0.15)(0.80)} = \frac{0.184}{0.184 + 0.12} = \frac{0.184}{0.304} \approx 0.605. The question asks for the probability they do not have the skill given a positive test, which is 1−PPV=1−0.605=0.3951 - \text{PPV} = 1 - 0.605 = 0.395.

Question 5

A medical diagnostics company reports that its new screening test for a certain condition has a sensitivity of 95% and a Positive Predictive Value (PPV) of 80%. Which of the following pieces of information is also required to calculate the test's specificity?

  1. The sample size of the study population.
  2. The total number of false negatives.
  3. The prevalence of the condition in the population. (correct answer)
  4. The cost of administering the test.
Explanation: Specificity is P(Test Negative∣No Condition)P(\text{Test Negative} | \text{No Condition}). The formula for PPV is PPV=Sens×PrevSens×Prev+(1−Spec)×(1−Prev)PPV = \frac{\text{Sens} \times \text{Prev}}{\text{Sens} \times \text{Prev} + (1-\text{Spec}) \times (1-\text{Prev})}. We are given PPV (0.80) and Sensitivity (0.95). In the formula, both Prevalence (Prev) and Specificity (Spec) are unknown. To solve for Specificity, we must be given the Prevalence. The other options are not sufficient. Sample size or number of false negatives are outcomes, not parameters of the same type, and cost is irrelevant.

Question 6

A performance review system categorizes employees as 'Below Expectations', 'Meets Expectations', or 'Exceeds Expectations'. In a given year, 15% of employees were rated 'Exceeds'. Of those rated 'Exceeds', 80% received a bonus. Of those not rated 'Exceeds', only 20% received a bonus. What is the probability that an employee who received a bonus was rated 'Exceeds Expectations'?

  1. 0.120
  2. 0.414 (correct answer)
  3. 0.800
  4. 0.850
Explanation: This is a Bayes' Theorem problem. Let E = 'Exceeds Expectations' and B = 'Received Bonus'. We want to find P(E∣B)P(E|B). We are given: P(E)=0.15P(E) = 0.15, so P(E′)=0.85P(E') = 0.85. We are also given P(B∣E)=0.80P(B|E) = 0.80 and P(B∣E′)=0.20P(B|E') = 0.20. First, we find the overall probability of receiving a bonus, P(B)P(B), using the Law of Total Probability: P(B)=P(B∣E)P(E)+P(B∣E′)P(E′)=(0.80)(0.15)+(0.20)(0.85)=0.12+0.17=0.29P(B) = P(B|E)P(E) + P(B|E')P(E') = (0.80)(0.15) + (0.20)(0.85) = 0.12 + 0.17 = 0.29. Now, we can find P(E∣B)=P(B∣E)P(E)P(B)=(0.80)(0.15)0.29=0.120.29≈0.4138P(E|B) = \frac{P(B|E)P(E)}{P(B)} = \frac{(0.80)(0.15)}{0.29} = \frac{0.12}{0.29} \approx 0.4138.

Question 7

A medical screening test for a rare disease has a sensitivity of 92% and a specificity of 85%. The disease prevalence in the population is 0.3%. If a randomly selected individual tests positive, what is the probability they actually have the disease?

  1. Approximately 1.8% (correct answer)
  2. Approximately 92%
  3. Approximately 85%
  4. Approximately 27.6%
Explanation: Using Bayes' theorem: P(Disease|Positive) = P(Positive|Disease) × P(Disease) / P(Positive). P(Positive) = 0.92 × 0.003 + 0.15 × 0.997 = 0.002760 + 0.14955 = 0.15231. Therefore P(Disease|Positive) = (0.92 × 0.003) / 0.15231 ≈ 0.018 or 1.8%. Choice B confuses sensitivity with posterior probability. Choice C confuses specificity with posterior probability. Choice D incorrectly uses P(Positive|Disease) / P(Positive|No Disease).

Question 8

In a quality control process, 3% of products are defective. A testing procedure correctly identifies 95% of defective products as defective, but also incorrectly classifies 8% of good products as defective. If a product is classified as defective by the test, what is the probability it is actually good?

  1. Approximately 73.1% (correct answer)
  2. Approximately 26.9%
  3. Approximately 92%
  4. Approximately 95%
Explanation: P(Good|Test Defective) = P(Test Defective|Good) × P(Good) / P(Test Defective). P(Test Defective) = 0.95 × 0.03 + 0.08 × 0.97 = 0.0285 + 0.0776 = 0.1061. Therefore P(Good|Test Defective) = (0.08 × 0.97) / 0.1061 ≈ 0.731 or 73.1%. Choice B gives P(Defective|Test Defective). Choice C confuses the complement of the false positive rate. Choice D confuses the sensitivity of the test.

Question 9

A corporation sources a particular microchip from two suppliers: Alpha and Beta. 60% of the chips come from Alpha, and 40% come from Beta. The defect rate for chips from Alpha is 1%, while the defect rate for chips from Beta is 3%. If a microchip is selected at random from the inventory, what is the overall probability that it is defective?

  1. 0.018 (correct answer)
  2. 0.020
  3. 0.025
  4. 0.040
Explanation: This requires using the Law of Total Probability. Let D be the event a chip is defective, A be the event it's from Alpha, and B be the event it's from Beta. We want to find P(D)P(D). We are given: P(A)=0.60P(A) = 0.60, P(B)=0.40P(B) = 0.40, P(D∣A)=0.01P(D|A) = 0.01, and P(D∣B)=0.03P(D|B) = 0.03. The formula is P(D)=P(D∣A)P(A)+P(D∣B)P(B)P(D) = P(D|A)P(A) + P(D|B)P(B). Plugging in the values: P(D)=(0.01)(0.60)+(0.03)(0.40)=0.006+0.012=0.018P(D) = (0.01)(0.60) + (0.03)(0.40) = 0.006 + 0.012 = 0.018.

Question 10

Using the information from the previous question (60% of chips from Alpha with a 1% defect rate, 40% from Beta with a 3% defect rate), suppose a chip is randomly selected and found to be defective. What is the probability that this defective chip came from supplier Beta?

  1. 0.012
  2. 0.030
  3. 0.400
  4. 0.667 (correct answer)
Explanation: This is a Bayes' Theorem problem. We want to find P(Beta∣Defective)P(\text{Beta} | \text{Defective}). From the previous question, we calculated the overall probability of a defect, P(D)=0.018P(D) = 0.018. We need the term P(D∩Beta)P(D \cap \text{Beta}), which is P(D∣Beta)P(Beta)=(0.03)(0.40)=0.012P(D|\text{Beta})P(\text{Beta}) = (0.03)(0.40) = 0.012. Now, apply the formula for conditional probability: P(Beta∣D)=P(D∩Beta)P(D)=0.0120.018=1218=23≈0.667P(\text{Beta}|D) = \frac{P(D \cap \text{Beta})}{P(D)} = \frac{0.012}{0.018} = \frac{12}{18} = \frac{2}{3} \approx 0.667.

Question 11

A pharmaceutical company is testing three different manufacturing processes. Process A produces 50% of the pills, Process B produces 30%, and Process C produces 20%. The contamination rates are 1% for Process A, 3% for Process B, and 2% for Process C.

If quality control finds that a randomly selected pill is contaminated, what is the probability it was produced by Process B, and how does this compare to Process B's share of total production?

  1. 18.8%; this is 0.63 times the production share
  2. 30%; this equals the production share exactly
  3. 3%; this is much lower than production share
  4. 56.3%; this is 1.88 times higher than production share (correct answer)
Explanation: This question tests Bayes' theorem, which helps you find the probability of a cause given an observed effect. When you see quality control scenarios where you need to trace a defective item back to its source, think about how contamination rates and production volumes both matter. To find the probability that a contaminated pill came from Process B, you need to calculate what fraction of all contaminated pills originated from Process B. First, find the total contamination rate: (0.50×0.01)+(0.30×0.03)+(0.20×0.02)=0.005+0.009+0.004=0.018(0.50 × 0.01) + (0.30 × 0.03) + (0.20 × 0.02) = 0.005 + 0.009 + 0.004 = 0.018 or 1.8%. Next, find what portion of contaminated pills came from Process B: 0.30×0.030.018=0.0090.018=0.563\frac{0.30 × 0.03}{0.018} = \frac{0.009}{0.018} = 0.563 or 56.3%. Compared to Process B's 30% production share, this is 56.3%30%=1.88\frac{56.3\%}{30\%} = 1.88 times higher. Choice A incorrectly calculates the probability as 18.8% (likely confusing it with the total contamination rate) and gets the wrong ratio. Choice B assumes the contamination probability equals the production share, ignoring that Process B has a higher contamination rate than average. Choice C gives just the contamination rate of Process B (3%), which isn't what the question asks for. Remember: when tracing defects back to their source, processes with higher defect rates will be overrepresented among the defective items compared to their production share. Always use Bayes' theorem to account for both the prior probability (production share) and the likelihood (contamination rate).

Question 12

A financial fraud detection system flags 0.5% of legitimate transactions and fails to flag 5% of fraudulent transactions. If 0.1% of all transactions are fraudulent, what is the probability that a flagged transaction is actually fraudulent?

  1. Approximately 84.0%
  2. Approximately 95%
  3. Approximately 0.1%
  4. Approximately 16.0% (correct answer)
Explanation: This is a classic Bayes' theorem problem in conditional probability. When you see questions about diagnostic tests, fraud detection, or any scenario involving false positives and false negatives, you need to work backwards from the test result to find the true probability. Let's set up the problem systematically. Define the events: F = fraudulent transaction, L = legitimate transaction, Flag = transaction is flagged. Given information: P(Flag|L) = 0.5% = 0.005, P(not Flag|F) = 5%, so P(Flag|F) = 95% = 0.95, and P(F) = 0.1% = 0.001, so P(L) = 99.9% = 0.999. Using Bayes' theorem: P(F∣Flag)=P(Flag∣F)×P(F)P(Flag)P(F|Flag) = \frac{P(Flag|F) \times P(F)}{P(Flag)} First, find P(Flag) using the law of total probability: P(Flag)=P(Flag∣F)×P(F)+P(Flag∣L)×P(L)P(Flag) = P(Flag|F) \times P(F) + P(Flag|L) \times P(L) P(Flag)=0.95×0.001+0.005×0.999=0.00095+0.004995=0.005945P(Flag) = 0.95 \times 0.001 + 0.005 \times 0.999 = 0.00095 + 0.004995 = 0.005945 Now calculate: P(F∣Flag)=0.000950.005945≈0.160=16.0%P(F|Flag) = \frac{0.00095}{0.005945} ≈ 0.160 = 16.0\% Answer D is correct. Answer A (84%) likely comes from calculating P(Flag|F) instead of P(F|Flag). Answer B (95%) is the detection rate for fraudulent transactions, not the probability a flagged transaction is fraudulent. Answer C (0.1%) is simply the base rate of fraud, ignoring the flagging information entirely. Study tip: In Bayes' problems, the counterintuitive result often occurs when the base rate is very low—even accurate tests produce many false positives. Always distinguish between P(A|B) and P(B|A).

Question 13

A company is considering two different tests to screen for a critical flaw in a high-value component. The cost of a false negative (failing to detect a flaw) is extremely high, leading to catastrophic failure. The cost of a false positive (discarding a good component) is significant but much lower. Test A has 99% sensitivity and 90% specificity. Test B has 95% sensitivity and 98% specificity. Which test should the company choose and why?

  1. Test A, because its higher sensitivity minimizes the very costly false negatives. (correct answer)
  2. Test B, because its higher specificity minimizes the number of discarded good components.
  3. Test B, because its overall accuracy, (0.95+0.98)/2, is higher than Test A's.
  4. Either test is acceptable, as the choice depends on the unknown prevalence of the flaw.
Explanation: The decision depends on the relative costs of the two types of errors. A false negative (a flawed component is passed as good) is described as having an 'extremely high' cost. Sensitivity is the probability that a truly flawed component is correctly identified (P(Test+∣Flaw)P(\text{Test+}|\text{Flaw})). A higher sensitivity means a lower false negative rate (1−Sensitivity1 - \text{Sensitivity}). Therefore, to minimize the most costly error, the company should choose the test with the highest sensitivity. Test A has 99% sensitivity, which is higher than Test B's 95%.

Question 14

Following up on the previous scenario, suppose the company manufactures a different, low-value component where the primary concern is minimizing waste. The cost of a false positive (unnecessarily discarding a good component) is now considered more significant than the cost of a false negative (letting a flawed component pass). Using the same two tests (Test A: 99% sens, 90% spec; Test B: 95% sens, 98% spec), which test would be preferable in this new context?

  1. Test A, because its high sensitivity ensures the highest quality output overall.
  2. Test B, because its higher specificity minimizes the number of costly false positives. (correct answer)
  3. Test A, because its false positive rate (10%) is still acceptably low.
  4. The choice is still ambiguous without knowing the component's production volume.
Explanation: In this scenario, the cost of a false positive (discarding a good item) is the primary concern. Specificity is the probability that a good component is correctly identified as good (P(Test-∣No Flaw)P(\text{Test-}|\text{No Flaw})). The false positive rate is 1−Specificity1 - \text{Specificity}. To minimize false positives, the company needs the test with the highest specificity. Test B has a specificity of 98%, which is higher than Test A's 90%. Therefore, Test B is the better choice to minimize the waste of good components.

Question 15

In a certain company, 40% of employees have a postgraduate degree. Of the employees with a postgraduate degree, 70% are in management positions. Of the employees without a postgraduate degree, 20% are in management positions. If an employee is selected at random and is found to be in a management position, what is the probability that they have a postgraduate degree?

  1. 0.28
  2. 0.40
  3. 0.60
  4. 0.70 (correct answer)
Explanation: Let PG = Postgraduate Degree and M = Management. We want to find P(PG∣M)P(PG | M). We are given: P(PG)=0.40P(PG) = 0.40, so P(PG′)=0.60P(PG') = 0.60. P(M∣PG)=0.70P(M | PG) = 0.70. P(M∣PG′)=0.20P(M | PG') = 0.20. First, find the overall probability of being in management, P(M)P(M), using the Law of Total Probability: P(M)=P(M∣PG)P(PG)+P(M∣PG′)P(PG′)=(0.70)(0.40)+(0.20)(0.60)=0.28+0.12=0.40P(M) = P(M|PG)P(PG) + P(M|PG')P(PG') = (0.70)(0.40) + (0.20)(0.60) = 0.28 + 0.12 = 0.40. Now use Bayes' Theorem: P(PG∣M)=P(M∣PG)P(PG)P(M)=(0.70)(0.40)0.40=0.70P(PG|M) = \frac{P(M|PG)P(PG)}{P(M)} = \frac{(0.70)(0.40)}{0.40} = 0.70.

Question 16

A loan approval algorithm has different error rates for different risk categories. For low-risk applicants (70% of all applicants), it incorrectly denies 3% of qualified candidates. For high-risk applicants (30% of all applicants), it incorrectly approves 12% of unqualified candidates. If 85% of low-risk and 40% of high-risk applicants are actually qualified, what is the probability that a randomly selected denied applicant was actually qualified?

  1. Approximately 8.5%
  2. Approximately 40.4% (correct answer)
  3. Approximately 59.6%
  4. Approximately 12%
Explanation: Qualified applicants wrongly denied: Low-risk = 0.70 × 0.85 × 0.03 = 0.01785; High-risk = 0.30 × 0.40 × 0.88 = 0.1056. Unqualified applicants correctly denied: Low-risk = 0.70 × 0.15 × 1 = 0.105; High-risk = 0.30 × 0.60 × 1 = 0.18. Total denied = 0.01785 + 0.1056 + 0.105 + 0.18 = 0.40905. P(Qualified|Denied) = (0.01785 + 0.1056) / 0.40905 = 0.12345 / 0.40905 ≈ 0.404 or 40.4%. Choice A uses only the low-risk wrongly denied rate. Choice C gives P(Unqualified|Denied). Choice D uses the high-risk incorrect approval rate.

Question 17

A credit scoring model classifies loan applicants as 'low risk' or 'high risk'. A recent analysis showed that P(Default∣Classified as High Risk)=0.40P(\text{Default} | \text{Classified as High Risk}) = 0.40. Which of the following is the correct interpretation of this statistic?

  1. Forty percent of all applicants who default are classified as high risk by the model.
  2. Forty percent of all applicants are classified as high risk and will default on their loan.
  3. Of the applicants that the model classifies as high risk, forty percent are expected to default. (correct answer)
  4. The model correctly predicts default for forty percent of the applicants it classifies as high risk.
Explanation: The notation P(A∣B)P(A|B) represents the conditional probability of event A occurring given that event B has already occurred. In this context, P(Default∣Classified as High Risk)=0.40P(\text{Default} | \text{Classified as High Risk}) = 0.40 means that for the specific subgroup of applicants who were classified as high risk, there is a 40% probability of default. Choice C accurately describes this. Choice A describes P(Classified as High Risk∣Default)P(\text{Classified as High Risk} | \text{Default}). Choice B describes the joint probability P(Classified as High Risk∩Default)P(\text{Classified as High Risk} \cap \text{Default}). Choice D is an imprecise statement about accuracy; the statistic only concerns the default rate within the high-risk group, not the correctness of the classification itself for non-defaulters.

Question 18

A manufacturer uses a screening test to detect defective products. The test has a sensitivity of 90% and a specificity of 95%. The true defect rate is 10%. If a product is tested and the result is negative (not defective), what is the probability that the product is actually not defective?

  1. 0.989 (correct answer)
  2. 0.950
  3. 0.900
  4. 0.855
Explanation: This question asks for the Negative Predictive Value (NPV), or P(Not Defective∣Test Negative)P(\text{Not Defective} | \text{Test Negative}). Let D be the event a product is defective. We are given: P(D)=0.10P(D) = 0.10, Sensitivity P(Test +∣D)=0.90P(\text{Test +}|D) = 0.90, Specificity P(Test -∣D′)=0.95P(\text{Test -}|D') = 0.95. From this, we can derive: P(D′)=0.90P(D') = 0.90, False Negative Rate P(Test -∣D)=1−0.90=0.10P(\text{Test -}|D) = 1 - 0.90 = 0.10. Using the formula for NPV: NPV=P(Test -∣D′)P(D′)P(Test -∣D′)P(D′)+P(Test -∣D)P(D)NPV = \frac{P(\text{Test -}|D')P(D')}{P(\text{Test -}|D')P(D') + P(\text{Test -}|D)P(D)}. Plugging in the values: NPV=(0.95)(0.90)(0.95)(0.90)+(0.10)(0.10)=0.8550.855+0.01=0.8550.865≈0.9884NPV = \frac{(0.95)(0.90)}{(0.95)(0.90) + (0.10)(0.10)} = \frac{0.855}{0.855 + 0.01} = \frac{0.855}{0.865} \approx 0.9884.

Question 19

A marketing team is analyzing customer data. They find the probability that a customer buys product A is P(A)=0.5P(A) = 0.5, and the probability that a customer buys product B is P(B)=0.4P(B) = 0.4. They also find that the probability a customer buys product B given that they have bought product A is P(B∣A)=0.6P(B|A) = 0.6. Which of the following statements is correct?

  1. The events of buying A and buying B are independent, and the probability of buying both is 0.20.
  2. The events of buying A and buying B are independent, and the probability of buying both is 0.30.
  3. The events of buying A and buying B are not independent, and the probability of buying both is 0.20.
  4. The events of buying A and buying B are not independent, and the probability of buying both is 0.30. (correct answer)
Explanation: First, check for independence. Two events A and B are independent if P(B∣A)=P(B)P(B|A) = P(B). Here, P(B∣A)=0.6P(B|A) = 0.6 and P(B)=0.4P(B) = 0.4. Since 0.6≠0.40.6 \neq 0.4, the events are not independent. Next, calculate the probability of buying both, which is the joint probability P(A∩B)P(A \cap B). Using the multiplication rule, P(A∩B)=P(B∣A)×P(A)=0.6×0.5=0.30P(A \cap B) = P(B|A) \times P(A) = 0.6 \times 0.5 = 0.30. Therefore, the events are not independent and the probability of buying both is 0.30.

Question 20

A cybersecurity firm develops a new system to detect malware. The system's sensitivity and specificity are held constant. If the system is deployed in a new environment where the prevalence of malware is significantly lower than the environment it was tested in, how will the Positive Predictive Value (PPV) of the system be affected?

  1. The PPV will increase, because there are fewer instances of malware to detect.
  2. The PPV will remain the same, because the test's intrinsic accuracy is unchanged.
  3. The PPV will decrease, because the ratio of false positives to true positives will increase. (correct answer)
  4. The PPV will become equal to the sensitivity, as prevalence becomes negligible.
Explanation: The Positive Predictive Value (PPV) is heavily dependent on the prevalence of the condition. The formula is PPV=Sens×PrevSens×Prev+(1−Spec)×(1−Prev)PPV = \frac{\text{Sens} \times \text{Prev}}{\text{Sens} \times \text{Prev} + (1-\text{Spec}) \times (1-\text{Prev})}. As prevalence (Prev) decreases, the numerator (Sens×Prev\text{Sens} \times \text{Prev}, representing true positives) decreases. The second term in the denominator ((1−Spec)×(1−Prev)(1-\text{Spec}) \times (1-\text{Prev}), representing false positives) increases, as (1−Prev)(1-\text{Prev}) gets larger. This means the proportion of positive tests that are true positives goes down, so the PPV decreases. A lower prevalence of malware means that a positive flag is more likely to be a false alarm.