Probability and Conditional Probability Flashcards

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?

Answer: About 16%

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.

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?

Answer: 5/7

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.

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?

Answer: A and B are independent but not mutually exclusive

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.

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?

Answer: 3/7

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.

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?

Answer: 112/625

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.

Factory A produces 60% of all light bulbs sold by a company, and Factory B produces the remaining 40%. Factory A has a 3% defect rate, while Factory B has a 5% defect rate. A randomly chosen bulb is found to be defective. What is the probability it was produced by Factory B?

Answer: 10/19

P(A) = 0.6, P(B) = 0.4. P(defective | A) = 0.03, P(defective | B) = 0.05. P(defective) = (0.6)(0.03) + (0.4)(0.05) = 0.018 + 0.020 = 0.038. By Bayes' theorem: P(B | defective) = (0.4 × 0.05) / 0.038 = 0.020 / 0.038 = 20/38 = 10/19 ≈ 52.6%. Even though B produces fewer bulbs, its higher defect rate means a defective bulb is more likely to come from B than its 40% share would suggest.