Bluebook SAT Test Probability and Conditional Probability 6

Question 1

A rare disease affects 2% of a population. A diagnostic test for the disease has a sensitivity of 95% (correctly identifies 95% of those who have the disease) and a specificity of 90% (correctly identifies 90% of those who do not have the disease). If a randomly selected person tests positive, what is the probability — rounded to the nearest percent — that they actually have the disease?

Accepted Answer

About 16%

Answer

Use Bayes' theorem. P(disease) = 0.02, P(no disease) = 0.98. P(positive | disease) = 0.95, P(positive | no disease) = 0.10 (false positive rate = 1 − 0.90). P(positive) = (0.95)(0.02) + (0.10)(0.98) = 0.019 + 0.098 = 0.117. P(disease | positive) = 0.019 / 0.117 ≈ 0.162, or about 16%. The low disease prevalence causes most positives to be false positives — a classic base-rate effect.

Question 2

In a senior class, 70% of students passed the math exam, 60% passed the science exam, and 80% passed at least one of the two exams. A student who passed the math exam is selected at random. What is the probability that this student also passed the science exam?

Accepted Answer

5/7

Answer

First, find P(M ∩ S) using inclusion-exclusion: P(M ∪ S) = P(M) + P(S) − P(M ∩ S), so 0.80 = 0.70 + 0.60 − P(M ∩ S), giving P(M ∩ S) = 0.50. Then apply conditional probability: P(S | M) = P(M ∩ S) / P(M) = 0.50 / 0.70 = 5/7.

Question 3

Events A and B satisfy P(A) = 0.4, P(B) = 0.3, and P(A ∪ B) = 0.58. Which of the following correctly describes the relationship between A and B?

Accepted Answer

A and B are independent but not mutually exclusive

Answer

Find P(A ∩ B) via inclusion-exclusion: P(A ∩ B) = 0.4 + 0.3 − 0.58 = 0.12. Check independence: P(A) × P(B) = 0.4 × 0.3 = 0.12 = P(A ∩ B), so A and B ARE independent. Check mutual exclusivity: mutually exclusive events have P(A ∩ B) = 0, but here P(A ∩ B) = 0.12 ≠ 0, so they are NOT mutually exclusive. Independent events with nonzero probabilities can never be mutually exclusive.

Question 4

A bag contains 4 red, 3 blue, and 2 green marbles. Two marbles are drawn in sequence without replacement. Given that the first marble drawn is NOT green, what is the probability that the second marble drawn is red?

Accepted Answer

3/7

Answer

Condition on what the non-green first draw can be. P(first = red) = 4/9; if so, P(second = red) = 3/8. P(first = blue) = 3/9; if so, P(second = red) = 4/8. P(second red AND first not green) = (4/9)(3/8) + (3/9)(4/8) = 12/72 + 12/72 = 24/72. P(first not green) = 7/9 = 56/72. So P(second = red | first not green) = 24/72 ÷ 56/72 = 24/56 = 3/7.

Question 5

A biased coin has a probability of 2/5 of landing heads on any flip. The coin is flipped 4 times. What is the probability of getting heads on at least 3 of the 4 flips?

Accepted Answer

112/625

Answer

Use the binomial formula with n = 4, p = 2/5, q = 3/5. P(X = 3) = C(4,3)(2/5)³(3/5)¹ = 4 × (8/125) × (3/5) = 96/625. P(X = 4) = C(4,4)(2/5)⁴ = 1 × 16/625 = 16/625. P(X ≥ 3) = 96/625 + 16/625 = 112/625.

Bluebook SAT Practice Test

Bluebook SAT Practice Test

Bluebook SAT Test Probability and Conditional Probability 6