College Statistics Quiz: Bayes Rule
20 questions · exam conditions
0:00
Bayes RuleQuestion 1 of 20

In a study, a screening test for a side effect of a new drug is evaluated. The test has a sensitivity of 95% and a specificity of 90%. It is known that 5% of patients taking the drug actually experience the side effect. Which expression correctly calculates the probability that a patient who tests positive actually has the side effect?

(0.90)(0.05)(0.90)(0.05)+(0.10)(0.95)\frac{(0.90)(0.05)}{(0.90)(0.05) + (0.10)(0.95)}
(0.95)(0.05)(0.95)(0.05)+(0.90)(0.95)\frac{(0.95)(0.05)}{(0.95)(0.05) + (0.90)(0.95)}
(0.95)(0.95)(0.95)(0.95)+(0.10)(0.05)\frac{(0.95)(0.95)}{(0.95)(0.95) + (0.10)(0.05)}
(0.95)(0.05)(0.95)(0.05)+(0.10)(0.95)\frac{(0.95)(0.05)}{(0.95)(0.05) + (0.10)(0.95)}
← Back to quizzes

College Statistics Quiz

College Statistics Quiz: Bayes Rule

Practice Bayes Rule in College 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 Bayes Rule, giving you a quick way to practice the rules, question types, and explanations that matter most for College Statistics.

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 study, a screening test for a side effect of a new drug is evaluated. The test has a sensitivity of 95% and a specificity of 90%. It is known that 5% of patients taking the drug actually experience the side effect. Which expression correctly calculates the probability that a patient who tests positive actually has the side effect?

  1. (0.90)(0.05)(0.90)(0.05)+(0.10)(0.95)\frac{(0.90)(0.05)}{(0.90)(0.05) + (0.10)(0.95)}
  2. (0.95)(0.05)(0.95)(0.05)+(0.90)(0.95)\frac{(0.95)(0.05)}{(0.95)(0.05) + (0.90)(0.95)}
  3. (0.95)(0.95)(0.95)(0.95)+(0.10)(0.05)\frac{(0.95)(0.95)}{(0.95)(0.95) + (0.10)(0.05)}
  4. (0.95)(0.05)(0.95)(0.05)+(0.10)(0.95)\frac{(0.95)(0.05)}{(0.95)(0.05) + (0.10)(0.95)} (correct answer)
Explanation: When you encounter a medical screening test problem, you're dealing with Bayes' theorem and the concept of positive predictive value (PPV) - the probability that someone who tests positive actually has the condition. To find PPV, you need to set up the formula: PPV=True PositivesAll Positive Tests\text{PPV} = \frac{\text{True Positives}}{\text{All Positive Tests}}. This translates to P(positive test and has condition)P(positive test)\frac{P(\text{positive test and has condition})}{P(\text{positive test})}. The correct calculation uses Bayes' theorem. You have sensitivity = 95% (probability of testing positive given you have the side effect), specificity = 90% (probability of testing negative given you don't have the side effect), and prevalence = 5% (probability of having the side effect). The numerator represents true positives: (0.95)(0.05) - those who have the condition AND test positive. The denominator includes all positive tests: true positives (0.95)(0.05) plus false positives (0.10)(0.95). Note that false positives use (1 - specificity) = 0.10 for the probability of testing positive when you don't have the condition, multiplied by 0.95 (the probability of not having the condition). Choice A incorrectly uses specificity (0.90) in the numerator instead of sensitivity. Choice B uses specificity (0.90) multiplied by the complement of prevalence in the denominator, which doesn't represent false positives. Choice C uses (0.95)(0.95) in the numerator, incorrectly squaring the sensitivity, and (0.10)(0.05) in the denominator, which represents the wrong probability combination. Remember: for PPV problems, always identify sensitivity, specificity, and prevalence first, then carefully distinguish between true positives and false positives in your Bayes' formula.

Question 2

Of the students at a university, 60% live on-campus and 40% live off-campus. 30% of on-campus students have a meal plan, while 80% of off-campus students have a meal plan. If a student with a meal plan is selected at random, what is the probability that they live on-campus?

  1. 0.500
  2. 0.180
  3. 0.300
  4. 0.360 (correct answer)
Explanation: This is a conditional probability problem that requires Bayes' theorem. When you see a question asking for the probability of one condition given another has occurred, you need to work backwards from the given information to find what portion of the specific group came from each source. First, let's find the total probability of having a meal plan. Using the law of total probability: P(meal plan)=P(meal plan∣on-campus)×P(on-campus)+P(meal plan∣off-campus)×P(off-campus)P(\text{meal plan}) = P(\text{meal plan}|\text{on-campus}) \times P(\text{on-campus}) + P(\text{meal plan}|\text{off-campus}) \times P(\text{off-campus}) This equals: (0.30)(0.60)+(0.80)(0.40)=0.18+0.32=0.50(0.30)(0.60) + (0.80)(0.40) = 0.18 + 0.32 = 0.50 Now we can use Bayes' theorem to find the probability a student with a meal plan lives on-campus: P(on-campus∣meal plan)=P(meal plan∣on-campus)×P(on-campus)P(meal plan)=0.180.50=0.360P(\text{on-campus}|\text{meal plan}) = \frac{P(\text{meal plan}|\text{on-campus}) \times P(\text{on-campus})}{P(\text{meal plan})} = \frac{0.18}{0.50} = 0.360 Choice A (0.500) represents the overall probability of having a meal plan, not the conditional probability we're seeking. Choice B (0.180) is the probability of being both on-campus AND having a meal plan, but doesn't account for conditioning on having a meal plan. Choice C (0.300) is simply the percentage of on-campus students with meal plans, ignoring the off-campus contribution entirely. When tackling conditional probability problems, always identify what you're conditioning on (here, "has a meal plan"), then use Bayes' theorem to work backwards. Draw a tree diagram if it helps visualize the different pathways to the given condition.

Question 3

A manufacturing process has a 10% defect rate. To catch defects, items pass through two independent inspection stations. Station 1 catches 90% of defective items that pass through it. Any item (defective or not) that passes Station 1 moves to Station 2, which catches 80% of the remaining defective items. A defective item is caught by one of the stations. What is the probability it was caught at Station 2?

  1. 0.082 (correct answer)
  2. 0.100
  3. 0.800
  4. 0.471
Explanation: Let D be the event an item is defective, so P(D)=0.10P(D)=0.10. Let C1 be the event an item is caught at Station 1, and C2 be the event it is caught at Station 2. We are looking for the probability that an item was caught at Station 2, given that it was caught. This is P(C2∣C1∪C2)P(C2 | C1 \cup C2). Since an item can only be caught once, C1 and C2 are mutually exclusive. Therefore, P(C2∣C1∪C2)=P(C2)P(C1)+P(C2)P(C2 | C1 \cup C2) = \frac{P(C2)}{P(C1) + P(C2)}. First, let's find the unconditional probabilities of being caught at each station: P(C1)=P(caught at S1∣D)P(D)=0.90×0.10=0.09P(C1) = P(\text{caught at S1} | D)P(D) = 0.90 \times 0.10 = 0.09. For an item to be caught at Station 2, it must be defective AND pass Station 1. The probability of a defective item passing S1 is 1−0.90=0.101-0.90 = 0.10. The probability of an item being defective and passing S1 is P(D∩Pass S1)=P(Pass S1∣D)P(D)=0.10×0.10=0.01P(D \cap \text{Pass S1}) = P(\text{Pass S1} | D)P(D) = 0.10 \times 0.10 = 0.01. Station 2 catches 80% of these. So, P(C2)=0.80×P(D∩Pass S1)=0.80×0.01=0.008P(C2) = 0.80 \times P(D \cap \text{Pass S1}) = 0.80 \times 0.01 = 0.008. Now, we can calculate the desired conditional probability: P(caught at S2∣caught)=P(C2)P(C1)+P(C2)=0.0080.09+0.008=0.0080.098≈0.0816≈0.082P(\text{caught at S2} | \text{caught}) = \frac{P(C2)}{P(C1) + P(C2)} = \frac{0.008}{0.09 + 0.008} = \frac{0.008}{0.098} \approx 0.0816 \approx 0.082. (B) 0.100 is the proportion of defective items that reach Station 2. (C) 0.800 is the conditional probability that Station 2 catches a defective item that reaches it. (D) 0.471 is calculated as 0.80/(0.90+0.80)≈0.4710.80 / (0.90 + 0.80) \approx 0.471, which ignores the sequential nature of the process.

Question 4

An email spam filter is being tested. It is known that 20% of all incoming emails are spam. The filter uses the presence of the word 'sale' to help classify emails. The word 'sale' appears in 10% of spam emails and in 1% of non-spam emails. If an incoming email contains the word 'sale', what is the probability that it is spam?

  1. 0.100
  2. 0.200
  3. 0.714 (correct answer)
  4. 0.909
Explanation: Let S be the event that an email is spam, and W be the event that the email contains the word 'sale'. We want to find P(S∣W)P(S|W). We are given: P(S)=0.20P(S) = 0.20, P(W∣S)=0.10P(W|S) = 0.10, and P(W∣Sc)=0.01P(W|S^c) = 0.01. Also, P(Sc)=1−0.20=0.80P(S^c) = 1 - 0.20 = 0.80. Using Bayes' Rule: P(S∣W)=P(W∣S)P(S)P(W)P(S|W) = \frac{P(W|S)P(S)}{P(W)}. The denominator, P(W)P(W), is found using the Law of Total Probability: P(W)=P(W∣S)P(S)+P(W∣Sc)P(Sc)P(W) = P(W|S)P(S) + P(W|S^c)P(S^c). Numerator: P(W∣S)P(S)=0.10×0.20=0.020P(W|S)P(S) = 0.10 \times 0.20 = 0.020. Denominator: (0.10×0.20)+(0.01×0.80)=0.020+0.008=0.028(0.10 \times 0.20) + (0.01 \times 0.80) = 0.020 + 0.008 = 0.028. So, P(S∣W)=0.0200.028=2028=57≈0.714P(S|W) = \frac{0.020}{0.028} = \frac{20}{28} = \frac{5}{7} \approx 0.714. (A) 0.100 is the conditional probability of seeing the word 'sale' in a spam email, P(W∣S)P(W|S). (B) 0.200 is the prior probability of an email being spam, P(S)P(S). (D) 0.909 is the result of ignoring the prior probabilities, calculated as 0.10/(0.10+0.01)≈0.9090.10 / (0.10 + 0.01) \approx 0.909, a common mistake.

Question 5

A geologist's initial estimate for the probability of finding oil at a drilling site is 30%. The company performs a seismic test that is 80% reliable if oil is present (i.e., it gives a positive result with 80% probability) and 60% reliable if no oil is present (i.e., it gives a negative result with 60% probability). If the seismic test result is negative, what is the geologist's revised probability of finding oil?

  1. 0.125 (correct answer)
  2. 0.200
  3. 0.060
  4. 0.538
Explanation: Let O be the event that there is oil, and T- be the event the test is negative. We want to find P(O∣T−)P(O|T-). We are given: P(O)=0.30P(O) = 0.30, so P(Oc)=0.70P(O^c) = 0.70. We are given P(T+∣O)=0.80P(T+|O) = 0.80 and P(T−∣Oc)=0.60P(T-|O^c) = 0.60. We need the probabilities of a negative test. From the given info: P(T−∣O)=1−P(T+∣O)=1−0.80=0.20P(T-|O) = 1 - P(T+|O) = 1 - 0.80 = 0.20. And P(T+∣Oc)=1−P(T−∣Oc)=1−0.60=0.40P(T+|O^c) = 1 - P(T-|O^c) = 1 - 0.60 = 0.40. Using Bayes' Rule for a negative result: P(O∣T−)=P(T−∣O)P(O)P(T−∣O)P(O)+P(T−∣Oc)P(Oc)P(O|T-) = \frac{P(T-|O)P(O)}{P(T-|O)P(O) + P(T-|O^c)P(O^c)}. Numerator: P(T−∣O)P(O)=0.20×0.30=0.06P(T-|O)P(O) = 0.20 \times 0.30 = 0.06. Denominator: (0.20×0.30)+(0.60×0.70)=0.06+0.42=0.48(0.20 \times 0.30) + (0.60 \times 0.70) = 0.06 + 0.42 = 0.48. So, P(O∣T−)=0.060.48=648=18=0.125P(O|T-) = \frac{0.06}{0.48} = \frac{6}{48} = \frac{1}{8} = 0.125. (B) 0.200 is P(T−∣O)P(T-|O), the probability of a negative test given oil. (C) 0.060 is the numerator, P(O∩T−)P(O \cap T-). (D) 0.538 is approximately 1−P(O∣T+)1 - P(O|T+). A common mistake is to assume P(O∣T−)=1−P(O∣T+)P(O|T-) = 1 - P(O|T+). P(O∣T+)P(O|T+) would be 0.8×0.3(0.8×0.3)+(0.4×0.7)=0.240.52≈0.462\frac{0.8 \times 0.3}{(0.8 \times 0.3) + (0.4 \times 0.7)} = \frac{0.24}{0.52} \approx 0.462, and 1−0.462=0.5381-0.462 = 0.538.

Question 6

In a certain city, 80% of taxis are yellow and 20% are blue. A witness to a hit-and-run accident identified the taxi involved as blue. In nighttime visibility tests, the witness correctly identified the color of a taxi 75% of the time, regardless of its color. What is the probability that the taxi involved was actually blue, given the witness's testimony?

  1. 0.429 (correct answer)
  2. 0.750
  3. 0.200
  4. 0.600
Explanation: Let B be the event that the taxi is blue, and Y be the event that the taxi is yellow. Let W_B be the event that the witness identifies the taxi as blue. We want to find P(B∣WB)P(B|W_B). We are given: P(B)=0.20P(B) = 0.20 and P(Y)=0.80P(Y) = 0.80. The witness's reliability is 75%, so P(WB∣B)=0.75P(W_B|B) = 0.75 (correctly identifying blue as blue) and P(WY∣Y)=0.75P(W_Y|Y) = 0.75 (correctly identifying yellow as yellow). We need the probability of misidentification: P(WB∣Y)=1−P(WY∣Y)=1−0.75=0.25P(W_B|Y) = 1 - P(W_Y|Y) = 1 - 0.75 = 0.25. Using Bayes' Rule: P(B∣WB)=P(WB∣B)P(B)P(WB)P(B|W_B) = \frac{P(W_B|B)P(B)}{P(W_B)}. Denominator P(WB)=P(WB∣B)P(B)+P(WB∣Y)P(Y)P(W_B) = P(W_B|B)P(B) + P(W_B|Y)P(Y). Numerator: P(WB∣B)P(B)=0.75×0.20=0.15P(W_B|B)P(B) = 0.75 \times 0.20 = 0.15. Denominator: (0.75×0.20)+(0.25×0.80)=0.15+0.20=0.35(0.75 \times 0.20) + (0.25 \times 0.80) = 0.15 + 0.20 = 0.35. So, P(B∣WB)=0.150.35=1535=37≈0.429P(B|W_B) = \frac{0.15}{0.35} = \frac{15}{35} = \frac{3}{7} \approx 0.429. (B) 0.750 is the reliability of the witness, P(WB∣B)P(W_B|B), not accounting for the base rate of taxi colors. (C) 0.200 is the prior probability of a taxi being blue, P(B)P(B). (D) 0.600 is P(Y∣WB)=0.20/0.35≈0.571P(Y|W_B) = 0.20 / 0.35 \approx 0.571, this distractor is close but not the correct calculation.

Question 7

An insurance company classifies 10% of its drivers as high-risk and 90% as low-risk. The probability of a high-risk driver having an accident in a year is 0.25. The probability of a low-risk driver having an accident in a year is 0.05. A driver is randomly selected and has an accident. Which of the following statements is the most accurate?

  1. The driver is more likely to be high-risk than low-risk.
  2. The driver is more likely to be low-risk than high-risk. (correct answer)
  3. The driver is about equally likely to be high-risk as low-risk.
  4. This cannot be determined without knowing the total number of drivers.
Explanation: Let H be high-risk, L be low-risk, and A be having an accident. We need to compare P(H∣A)P(H|A) and P(L∣A)P(L|A). We are given: P(H)=0.10,P(L)=0.90,P(A∣H)=0.25,P(A∣L)=0.05P(H)=0.10, P(L)=0.90, P(A|H)=0.25, P(A|L)=0.05. First, find the overall probability of an accident, P(A)P(A): P(A)=P(A∣H)P(H)+P(A∣L)P(L)=(0.25×0.10)+(0.05×0.90)=0.025+0.045=0.070P(A) = P(A|H)P(H) + P(A|L)P(L) = (0.25 \times 0.10) + (0.05 \times 0.90) = 0.025 + 0.045 = 0.070. Now find the posterior probabilities: P(H∣A)=P(A∣H)P(H)P(A)=0.0250.070=2570=514≈0.357P(H|A) = \frac{P(A|H)P(H)}{P(A)} = \frac{0.025}{0.070} = \frac{25}{70} = \frac{5}{14} \approx 0.357. P(L∣A)=P(A∣L)P(L)P(A)=0.0450.070=4570=914≈0.643P(L|A) = \frac{P(A|L)P(L)}{P(A)} = \frac{0.045}{0.070} = \frac{45}{70} = \frac{9}{14} \approx 0.643. Since P(L∣A)>P(H∣A)P(L|A) > P(H|A), the driver is more likely to be low-risk. This is because the low-risk group is so much larger that they account for more of the total accidents, despite their lower individual accident rate. (A) is incorrect; this reflects an intuitive guess based on the higher accident rate for high-risk drivers, ignoring the base rates. (C) is incorrect as the probabilities are not close. (D) is incorrect because the proportions are sufficient.

Question 8

A system relies on a backup power source. The main power fails with a probability of 0.01. If the main power fails, the backup source has a 99% chance of activating correctly. There is also a 0.5% chance the backup source will activate unnecessarily when the main power is working. Given that the backup power source is observed to be active, what is the probability that the main power has failed?

  1. 0.500
  2. 0.990
  3. 0.010
  4. 0.668 (correct answer)
Explanation: This is a classic conditional probability problem that requires Bayes' theorem. When you see a question asking "given that X is observed, what's the probability that Y occurred," you're dealing with conditional probability where you need to consider all possible ways the observed event could happen. Let's define our events: M = main power fails, B = backup activates. We need P(M|B) - the probability main power failed given that backup is active. Using Bayes' theorem: P(M∣B)=P(B∣M)×P(M)P(B)P(M|B) = \frac{P(B|M) \times P(M)}{P(B)} First, find P(B) using the law of total probability. The backup can activate in two scenarios:
  • Main power fails AND backup activates: P(M) × P(B|M) = 0.01 × 0.99 = 0.0099
  • Main power works AND backup activates unnecessarily: P(M') × P(B|M') = 0.99 × 0.005 = 0.00495
So P(B) = 0.0099 + 0.00495 = 0.01485 Therefore: P(M∣B)=0.99×0.010.01485=0.00990.01485≈0.667P(M|B) = \frac{0.99 \times 0.01}{0.01485} = \frac{0.0099}{0.01485} ≈ 0.667 Answer D (0.668) is correct. Answer A (0.500) likely comes from ignoring the different activation rates. Answer B (0.990) incorrectly uses P(B|M) instead of P(M|B) - a classic confusion between conditional probabilities. Answer C (0.010) simply uses the original failure probability, ignoring the conditional information entirely. Remember: In Bayes' problems, always account for all ways the observed event can occur, not just the "obvious" one.

Question 9

A car alarm has a 98% chance of triggering during an actual break-in. However, on any given night without a break-in, it has a 1% chance of triggering due to other causes (a false alarm). The probability of a car in this neighborhood being targeted for a break-in on a given night is estimated to be 0.2%. If you hear the alarm from this car, what is the approximate probability that a break-in is actually occurring?

  1. 0.980
  2. 0.662
  3. 0.164 (correct answer)
  4. 0.010
Explanation: Let B be the event of a break-in, and A be the event the alarm triggers. We want to find P(B∣A)P(B|A). We are given: P(B)=0.002P(B) = 0.002, P(A∣B)=0.98P(A|B) = 0.98 (sensitivity), and P(A∣Bc)=0.01P(A|B^c) = 0.01 (false positive rate). We also have P(Bc)=1−0.002=0.998P(B^c) = 1 - 0.002 = 0.998. Using Bayes' Rule: P(B∣A)=P(A∣B)P(B)P(A)P(B|A) = \frac{P(A|B)P(B)}{P(A)}. Denominator P(A)=P(A∣B)P(B)+P(A∣Bc)P(Bc)P(A) = P(A|B)P(B) + P(A|B^c)P(B^c). Numerator: P(A∣B)P(B)=0.98×0.002=0.00196P(A|B)P(B) = 0.98 \times 0.002 = 0.00196. Denominator: (0.98×0.002)+(0.01×0.998)=0.00196+0.00998=0.01194(0.98 \times 0.002) + (0.01 \times 0.998) = 0.00196 + 0.00998 = 0.01194. So, P(B∣A)=0.001960.01194≈0.164P(B|A) = \frac{0.00196}{0.01194} \approx 0.164. (A) 0.980 is the sensitivity, P(A∣B)P(A|B), a common distractor representing the prosecutor's fallacy. (B) 0.662 is calculated if the prior probability of break-in is assumed to be much higher, for example 5%. (D) 0.010 is the false alarm rate, P(A∣Bc)P(A|B^c).

Question 10

An analyst models a company's stock performance. The probability of the company reporting positive earnings is 70%. If the company reports positive earnings, its stock price has an 80% chance of increasing. If the company does not report positive earnings, its stock price has a 30% chance of increasing. Given that the company's stock price increased, what is the probability that it reported positive earnings?

  1. 0.800
  2. 0.862 (correct answer)
  3. 0.560
  4. 0.727
Explanation: Let E be the event of positive earnings, and I be the event of a stock price increase. We want to find P(E∣I)P(E|I). We are given: P(E)=0.70P(E) = 0.70, P(I∣E)=0.80P(I|E) = 0.80, and P(I∣Ec)=0.30P(I|E^c) = 0.30. We can deduce P(Ec)=1−0.70=0.30P(E^c) = 1 - 0.70 = 0.30. Using Bayes' Rule: P(E∣I)=P(I∣E)P(E)P(I)P(E|I) = \frac{P(I|E)P(E)}{P(I)}. Denominator P(I)=P(I∣E)P(E)+P(I∣Ec)P(Ec)P(I) = P(I|E)P(E) + P(I|E^c)P(E^c). Numerator: P(I∣E)P(E)=0.80×0.70=0.56P(I|E)P(E) = 0.80 \times 0.70 = 0.56. Denominator: (0.80×0.70)+(0.30×0.30)=0.56+0.09=0.65(0.80 \times 0.70) + (0.30 \times 0.30) = 0.56 + 0.09 = 0.65. So, P(E∣I)=0.560.65≈0.862P(E|I) = \frac{0.56}{0.65} \approx 0.862. (A) 0.800 is the conditional probability of an increase given positive earnings, P(I∣E)P(I|E). (C) 0.560 is the joint probability of positive earnings and an increase, P(I∩E)P(I \cap E), which is the numerator of the Bayes' formula. (D) 0.727 is the result of incorrectly ignoring priors: 0.80/(0.80+0.30)≈0.7270.80 / (0.80 + 0.30) \approx 0.727.

Question 11

A test for a particular condition has a sensitivity of P(T+∣D)=0.95P(T+|D) = 0.95 and a prevalence of P(D)=0.05P(D) = 0.05. When a randomly selected person tests positive, the updated probability that they have the condition is P(D∣T+)=0.50P(D|T+) = 0.50. Based on this information, what is the false positive rate of the test, P(T+∣Dc)P(T+|D^c)?

  1. 0.950
  2. 0.050 (correct answer)
  3. 0.500
  4. 0.047
Explanation: When you encounter diagnostic test problems involving sensitivity, specificity, and prevalence, you're working with Bayes' theorem and conditional probabilities. The key is systematically organizing the given information and using the relationship between these probabilities. You're given: sensitivity P(T+∣D)=0.95P(T+|D) = 0.95, prevalence P(D)=0.05P(D) = 0.05, and posterior probability P(D∣T+)=0.50P(D|T+) = 0.50. You need to find the false positive rate P(T+∣Dc)P(T+|D^c). Using Bayes' theorem: P(D∣T+)=P(T+∣D)⋅P(D)P(T+)P(D|T+) = \frac{P(T+|D) \cdot P(D)}{P(T+)} Since P(T+)=P(T+∣D)⋅P(D)+P(T+∣Dc)⋅P(Dc)P(T+) = P(T+|D) \cdot P(D) + P(T+|D^c) \cdot P(D^c), you can substitute: 0.50=0.95×0.050.95×0.05+P(T+∣Dc)×0.950.50 = \frac{0.95 \times 0.05}{0.95 \times 0.05 + P(T+|D^c) \times 0.95} Solving: 0.50=0.04750.0475+0.95⋅P(T+∣Dc)0.50 = \frac{0.0475}{0.0475 + 0.95 \cdot P(T+|D^c)} Cross-multiplying and solving gives P(T+∣Dc)=0.05P(T+|D^c) = 0.05. Choice A (0.950) confuses the false positive rate with the sensitivity. Choice C (0.500) mistakes the posterior probability for the false positive rate. Choice D (0.047) represents a calculation error, possibly confusing the numerator in Bayes' theorem with the false positive rate. Choice B (0.050) correctly identifies the false positive rate. Study tip: In diagnostic test problems, always set up Bayes' theorem systematically and remember that the false positive rate P(T+∣Dc)P(T+|D^c) is different from sensitivity, specificity, and posterior probabilities. Drawing a 2×2 table can help visualize these relationships.

Question 12

A manufacturer receives 70% of its microchips from Supplier A and 30% from Supplier B. 98% of the chips from Supplier A are functional, while 95% of the chips from Supplier B are functional. If a randomly selected chip is found to be non-functional, what is the probability it came from Supplier B?

  1. 0.050
  2. 0.300
  3. 0.517 (correct answer)
  4. 0.714
Explanation: Let A and B be the events that a chip is from Supplier A and B, respectively. Let F be the event a chip is functional, and F' be the event it is non-functional. We want to find P(B∣F′)P(B|F'). Given: P(A)=0.70,P(B)=0.30P(A)=0.70, P(B)=0.30. P(F∣A)=0.98  ⟹  P(F′∣A)=1−0.98=0.02P(F|A)=0.98 \implies P(F'|A) = 1 - 0.98 = 0.02. P(F∣B)=0.95  ⟹  P(F′∣B)=1−0.95=0.05P(F|B)=0.95 \implies P(F'|B) = 1 - 0.95 = 0.05. Using Bayes' Rule: P(B∣F′)=P(F′∣B)P(B)P(F′)P(B|F') = \frac{P(F'|B)P(B)}{P(F')}. Denominator P(F′)=P(F′∣A)P(A)+P(F′∣B)P(B)P(F') = P(F'|A)P(A) + P(F'|B)P(B). Numerator: P(F′∣B)P(B)=0.05×0.30=0.015P(F'|B)P(B) = 0.05 \times 0.30 = 0.015. Denominator: (0.02×0.70)+(0.05×0.30)=0.014+0.015=0.029(0.02 \times 0.70) + (0.05 \times 0.30) = 0.014 + 0.015 = 0.029. So, P(B∣F′)=0.0150.029≈0.517P(B|F') = \frac{0.015}{0.029} \approx 0.517. (A) 0.050 is the defect rate for Supplier B, P(F′∣B)P(F'|B). (B) 0.300 is the prior probability for Supplier B, P(B)P(B). (D) 0.714 is calculated as 0.05/(0.05+0.02)≈0.7140.05 / (0.05+0.02) \approx 0.714, which ignores the supplier proportions.

Question 13

The probability that a new startup will succeed is 0.20. If it succeeds, the probability that its valuation will exceed $10 million is 0.75. If it fails, the probability that its valuation will exceed $10 million (e.g., through an acquisition of its assets) is 0.10. Given that the startup's valuation exceeds $10 million, what is the probability that it succeeded?

  1. 0.150
  2. 0.750
  3. 0.882
  4. 0.652 (correct answer)
Explanation: This is a classic conditional probability problem that requires Bayes' theorem. When you see a question asking "given that [event B occurred], what's the probability that [event A caused it]," you need to work backwards from the observed outcome. Let's define our events: S = startup succeeds, V = valuation exceeds $10 million. We want P(S|V) - the probability the startup succeeded given that its valuation exceeds $10 million. Using Bayes' theorem: $P(S∣V)=P(V∣S)×P(S)P(V)P(S|V) = \frac{P(V|S) \times P(S)}{P(V)} $ First, find P(V) using the law of total probability: P(V) = P(V|S) \times P(S) + P(V|S^c) \times P(S^c) P(V) = 0.75 \times 0.20 + 0.10 \times 0.80 = 0.15 + 0.08 = 0.23 Now calculate P(S|V): P(S|V) = \frac{0.75 \times 0.20}{0.23} = \frac{0.15}{0.23} = 0.652 The answer is D) 0.652. A) 0.150 is simply P(V|S) × P(S), which gives you the probability of both success and high valuation, not the conditional probability we need. B) 0.750 is P(V|S), the probability of high valuation given success - but we need the reverse conditional probability. C) 0.882 likely comes from incorrectly calculating the denominator or mixing up the conditional probabilities. Remember: Bayes' theorem problems always require you to find the total probability of the observed outcome first, then use it as your denominator. Practice identifying the "given" information versus what you're solving for.

Question 14

In a certain city, 80% of taxis are yellow and 20% are blue. A witness to a hit-and-run accident identified the taxi involved as blue. In nighttime visibility tests, the witness correctly identified the color of a taxi 75% of the time, regardless of its color. What is the probability that the taxi involved was actually blue, given the witness's testimony?

  1. 0.429 (correct answer)
  2. 0.750
  3. 0.200
  4. 0.600
Explanation: Let B be the event that the taxi is blue, and Y be the event that the taxi is yellow. Let W_B be the event that the witness identifies the taxi as blue. We want to find P(B∣WB)P(B|W_B). We are given: P(B)=0.20P(B) = 0.20 and P(Y)=0.80P(Y) = 0.80. The witness's reliability is 75%, so P(WB∣B)=0.75P(W_B|B) = 0.75 (correctly identifying blue as blue) and P(WY∣Y)=0.75P(W_Y|Y) = 0.75 (correctly identifying yellow as yellow). We need the probability of misidentification: P(WB∣Y)=1−P(WY∣Y)=1−0.75=0.25P(W_B|Y) = 1 - P(W_Y|Y) = 1 - 0.75 = 0.25. Using Bayes' Rule: P(B∣WB)=P(WB∣B)P(B)P(WB)P(B|W_B) = \frac{P(W_B|B)P(B)}{P(W_B)}. Denominator P(WB)=P(WB∣B)P(B)+P(WB∣Y)P(Y)P(W_B) = P(W_B|B)P(B) + P(W_B|Y)P(Y). Numerator: P(WB∣B)P(B)=0.75×0.20=0.15P(W_B|B)P(B) = 0.75 \times 0.20 = 0.15. Denominator: (0.75×0.20)+(0.25×0.80)=0.15+0.20=0.35(0.75 \times 0.20) + (0.25 \times 0.80) = 0.15 + 0.20 = 0.35. So, P(B∣WB)=0.150.35=1535=37≈0.429P(B|W_B) = \frac{0.15}{0.35} = \frac{15}{35} = \frac{3}{7} \approx 0.429. (B) 0.750 is the reliability of the witness, P(WB∣B)P(W_B|B), not accounting for the base rate of taxi colors. (C) 0.200 is the prior probability of a taxi being blue, P(B)P(B). (D) 0.600 is P(Y∣WB)=0.20/0.35≈0.571P(Y|W_B) = 0.20 / 0.35 \approx 0.571, this distractor is close but not the correct calculation.

Question 15

An analyst models a company's stock performance. The probability of the company reporting positive earnings is 70%. If the company reports positive earnings, its stock price has an 80% chance of increasing. If the company does not report positive earnings, its stock price has a 30% chance of increasing. Given that the company's stock price increased, what is the probability that it reported positive earnings?

  1. 0.800
  2. 0.862 (correct answer)
  3. 0.560
  4. 0.727
Explanation: Let E be the event of positive earnings, and I be the event of a stock price increase. We want to find P(E∣I)P(E|I). We are given: P(E)=0.70P(E) = 0.70, P(I∣E)=0.80P(I|E) = 0.80, and P(I∣Ec)=0.30P(I|E^c) = 0.30. We can deduce P(Ec)=1−0.70=0.30P(E^c) = 1 - 0.70 = 0.30. Using Bayes' Rule: P(E∣I)=P(I∣E)P(E)P(I)P(E|I) = \frac{P(I|E)P(E)}{P(I)}. Denominator P(I)=P(I∣E)P(E)+P(I∣Ec)P(Ec)P(I) = P(I|E)P(E) + P(I|E^c)P(E^c). Numerator: P(I∣E)P(E)=0.80×0.70=0.56P(I|E)P(E) = 0.80 \times 0.70 = 0.56. Denominator: (0.80×0.70)+(0.30×0.30)=0.56+0.09=0.65(0.80 \times 0.70) + (0.30 \times 0.30) = 0.56 + 0.09 = 0.65. So, P(E∣I)=0.560.65≈0.862P(E|I) = \frac{0.56}{0.65} \approx 0.862. (A) 0.800 is the conditional probability of an increase given positive earnings, P(I∣E)P(I|E). (C) 0.560 is the joint probability of positive earnings and an increase, P(I∩E)P(I \cap E), which is the numerator of the Bayes' formula. (D) 0.727 is the result of incorrectly ignoring priors: 0.80/(0.80+0.30)≈0.7270.80 / (0.80 + 0.30) \approx 0.727.

Question 16

A car alarm has a 98% chance of triggering during an actual break-in. However, on any given night without a break-in, it has a 1% chance of triggering due to other causes (a false alarm). The probability of a car in this neighborhood being targeted for a break-in on a given night is estimated to be 0.2%. If you hear the alarm from this car, what is the approximate probability that a break-in is actually occurring?

  1. 0.980
  2. 0.662
  3. 0.164 (correct answer)
  4. 0.010
Explanation: Let B be the event of a break-in, and A be the event the alarm triggers. We want to find P(B∣A)P(B|A). We are given: P(B)=0.002P(B) = 0.002, P(A∣B)=0.98P(A|B) = 0.98 (sensitivity), and P(A∣Bc)=0.01P(A|B^c) = 0.01 (false positive rate). We also have P(Bc)=1−0.002=0.998P(B^c) = 1 - 0.002 = 0.998. Using Bayes' Rule: P(B∣A)=P(A∣B)P(B)P(A)P(B|A) = \frac{P(A|B)P(B)}{P(A)}. Denominator P(A)=P(A∣B)P(B)+P(A∣Bc)P(Bc)P(A) = P(A|B)P(B) + P(A|B^c)P(B^c). Numerator: P(A∣B)P(B)=0.98×0.002=0.00196P(A|B)P(B) = 0.98 \times 0.002 = 0.00196. Denominator: (0.98×0.002)+(0.01×0.998)=0.00196+0.00998=0.01194(0.98 \times 0.002) + (0.01 \times 0.998) = 0.00196 + 0.00998 = 0.01194. So, P(B∣A)=0.001960.01194≈0.164P(B|A) = \frac{0.00196}{0.01194} \approx 0.164. (A) 0.980 is the sensitivity, P(A∣B)P(A|B), a common distractor representing the prosecutor's fallacy. (B) 0.662 is calculated if the prior probability of break-in is assumed to be much higher, for example 5%. (D) 0.010 is the false alarm rate, P(A∣Bc)P(A|B^c).

Question 17

A geologist's initial estimate for the probability of finding oil at a drilling site is 30%. The company performs a seismic test that is 80% reliable if oil is present (i.e., it gives a positive result with 80% probability) and 60% reliable if no oil is present (i.e., it gives a negative result with 60% probability). If the seismic test result is negative, what is the geologist's revised probability of finding oil?

  1. 0.125 (correct answer)
  2. 0.200
  3. 0.060
  4. 0.538
Explanation: Let O be the event that there is oil, and T- be the event the test is negative. We want to find P(O∣T−)P(O|T-). We are given: P(O)=0.30P(O) = 0.30, so P(Oc)=0.70P(O^c) = 0.70. We are given P(T+∣O)=0.80P(T+|O) = 0.80 and P(T−∣Oc)=0.60P(T-|O^c) = 0.60. We need the probabilities of a negative test. From the given info: P(T−∣O)=1−P(T+∣O)=1−0.80=0.20P(T-|O) = 1 - P(T+|O) = 1 - 0.80 = 0.20. And P(T+∣Oc)=1−P(T−∣Oc)=1−0.60=0.40P(T+|O^c) = 1 - P(T-|O^c) = 1 - 0.60 = 0.40. Using Bayes' Rule for a negative result: P(O∣T−)=P(T−∣O)P(O)P(T−∣O)P(O)+P(T−∣Oc)P(Oc)P(O|T-) = \frac{P(T-|O)P(O)}{P(T-|O)P(O) + P(T-|O^c)P(O^c)}. Numerator: P(T−∣O)P(O)=0.20×0.30=0.06P(T-|O)P(O) = 0.20 \times 0.30 = 0.06. Denominator: (0.20×0.30)+(0.60×0.70)=0.06+0.42=0.48(0.20 \times 0.30) + (0.60 \times 0.70) = 0.06 + 0.42 = 0.48. So, P(O∣T−)=0.060.48=648=18=0.125P(O|T-) = \frac{0.06}{0.48} = \frac{6}{48} = \frac{1}{8} = 0.125. (B) 0.200 is P(T−∣O)P(T-|O), the probability of a negative test given oil. (C) 0.060 is the numerator, P(O∩T−)P(O \cap T-). (D) 0.538 is approximately 1−P(O∣T+)1 - P(O|T+). A common mistake is to assume P(O∣T−)=1−P(O∣T+)P(O|T-) = 1 - P(O|T+). P(O∣T+)P(O|T+) would be 0.8×0.3(0.8×0.3)+(0.4×0.7)=0.240.52≈0.462\frac{0.8 \times 0.3}{(0.8 \times 0.3) + (0.4 \times 0.7)} = \frac{0.24}{0.52} \approx 0.462, and 1−0.462=0.5381-0.462 = 0.538.

Question 18

A manufacturer receives 70% of its microchips from Supplier A and 30% from Supplier B. 98% of the chips from Supplier A are functional, while 95% of the chips from Supplier B are functional. If a randomly selected chip is found to be non-functional, what is the probability it came from Supplier B?

  1. 0.050
  2. 0.300
  3. 0.517 (correct answer)
  4. 0.714
Explanation: Let A and B be the events that a chip is from Supplier A and B, respectively. Let F be the event a chip is functional, and F' be the event it is non-functional. We want to find P(B∣F′)P(B|F'). Given: P(A)=0.70,P(B)=0.30P(A)=0.70, P(B)=0.30. P(F∣A)=0.98  ⟹  P(F′∣A)=1−0.98=0.02P(F|A)=0.98 \implies P(F'|A) = 1 - 0.98 = 0.02. P(F∣B)=0.95  ⟹  P(F′∣B)=1−0.95=0.05P(F|B)=0.95 \implies P(F'|B) = 1 - 0.95 = 0.05. Using Bayes' Rule: P(B∣F′)=P(F′∣B)P(B)P(F′)P(B|F') = \frac{P(F'|B)P(B)}{P(F')}. Denominator P(F′)=P(F′∣A)P(A)+P(F′∣B)P(B)P(F') = P(F'|A)P(A) + P(F'|B)P(B). Numerator: P(F′∣B)P(B)=0.05×0.30=0.015P(F'|B)P(B) = 0.05 \times 0.30 = 0.015. Denominator: (0.02×0.70)+(0.05×0.30)=0.014+0.015=0.029(0.02 \times 0.70) + (0.05 \times 0.30) = 0.014 + 0.015 = 0.029. So, P(B∣F′)=0.0150.029≈0.517P(B|F') = \frac{0.015}{0.029} \approx 0.517. (A) 0.050 is the defect rate for Supplier B, P(F′∣B)P(F'|B). (B) 0.300 is the prior probability for Supplier B, P(B)P(B). (D) 0.714 is calculated as 0.05/(0.05+0.02)≈0.7140.05 / (0.05+0.02) \approx 0.714, which ignores the supplier proportions.

Question 19

In a study, a screening test for a side effect of a new drug is evaluated. The test has a sensitivity of 95% and a specificity of 90%. It is known that 5% of patients taking the drug actually experience the side effect. Which expression correctly calculates the probability that a patient who tests positive actually has the side effect?

  1. (0.90)(0.05)(0.90)(0.05)+(0.10)(0.95)\frac{(0.90)(0.05)}{(0.90)(0.05) + (0.10)(0.95)}
  2. (0.95)(0.05)(0.95)(0.05)+(0.90)(0.95)\frac{(0.95)(0.05)}{(0.95)(0.05) + (0.90)(0.95)}
  3. (0.95)(0.95)(0.95)(0.95)+(0.10)(0.05)\frac{(0.95)(0.95)}{(0.95)(0.95) + (0.10)(0.05)}
  4. (0.95)(0.05)(0.95)(0.05)+(0.10)(0.95)\frac{(0.95)(0.05)}{(0.95)(0.05) + (0.10)(0.95)} (correct answer)
Explanation: When you encounter a medical screening test problem, you're dealing with Bayes' theorem and the concept of positive predictive value (PPV) - the probability that someone who tests positive actually has the condition. To find PPV, you need to set up the formula: PPV=True PositivesAll Positive Tests\text{PPV} = \frac{\text{True Positives}}{\text{All Positive Tests}}. This translates to P(positive test and has condition)P(positive test)\frac{P(\text{positive test and has condition})}{P(\text{positive test})}. The correct calculation uses Bayes' theorem. You have sensitivity = 95% (probability of testing positive given you have the side effect), specificity = 90% (probability of testing negative given you don't have the side effect), and prevalence = 5% (probability of having the side effect). The numerator represents true positives: (0.95)(0.05) - those who have the condition AND test positive. The denominator includes all positive tests: true positives (0.95)(0.05) plus false positives (0.10)(0.95). Note that false positives use (1 - specificity) = 0.10 for the probability of testing positive when you don't have the condition, multiplied by 0.95 (the probability of not having the condition). Choice A incorrectly uses specificity (0.90) in the numerator instead of sensitivity. Choice B uses specificity (0.90) multiplied by the complement of prevalence in the denominator, which doesn't represent false positives. Choice C uses (0.95)(0.95) in the numerator, incorrectly squaring the sensitivity, and (0.10)(0.05) in the denominator, which represents the wrong probability combination. Remember: for PPV problems, always identify sensitivity, specificity, and prevalence first, then carefully distinguish between true positives and false positives in your Bayes' formula.

Question 20

Of the students at a university, 60% live on-campus and 40% live off-campus. 30% of on-campus students have a meal plan, while 80% of off-campus students have a meal plan. If a student with a meal plan is selected at random, what is the probability that they live on-campus?

  1. 0.500
  2. 0.180
  3. 0.300
  4. 0.360 (correct answer)
Explanation: This is a conditional probability problem that requires Bayes' theorem. When you see a question asking for the probability of one condition given another has occurred, you need to work backwards from the given information to find what portion of the specific group came from each source. First, let's find the total probability of having a meal plan. Using the law of total probability: P(meal plan)=P(meal plan∣on-campus)×P(on-campus)+P(meal plan∣off-campus)×P(off-campus)P(\text{meal plan}) = P(\text{meal plan}|\text{on-campus}) \times P(\text{on-campus}) + P(\text{meal plan}|\text{off-campus}) \times P(\text{off-campus}) This equals: (0.30)(0.60)+(0.80)(0.40)=0.18+0.32=0.50(0.30)(0.60) + (0.80)(0.40) = 0.18 + 0.32 = 0.50 Now we can use Bayes' theorem to find the probability a student with a meal plan lives on-campus: P(on-campus∣meal plan)=P(meal plan∣on-campus)×P(on-campus)P(meal plan)=0.180.50=0.360P(\text{on-campus}|\text{meal plan}) = \frac{P(\text{meal plan}|\text{on-campus}) \times P(\text{on-campus})}{P(\text{meal plan})} = \frac{0.18}{0.50} = 0.360 Choice A (0.500) represents the overall probability of having a meal plan, not the conditional probability we're seeking. Choice B (0.180) is the probability of being both on-campus AND having a meal plan, but doesn't account for conditioning on having a meal plan. Choice C (0.300) is simply the percentage of on-campus students with meal plans, ignoring the off-campus contribution entirely. When tackling conditional probability problems, always identify what you're conditioning on (here, "has a meal plan"), then use Bayes' theorem to work backwards. Draw a tree diagram if it helps visualize the different pathways to the given condition.