Bluebook SAT Test Probability and Conditional Probability 4

Question 1

A bag contains 5 red, 4 blue, and 3 green marbles. Two marbles are drawn without replacement. Given that the second marble drawn is blue, what is the probability that the first marble drawn was also blue?

Accepted Answer

3/11

Answer

Use Bayes' theorem. P(1st blue AND 2nd blue) = (4/12)(3/11) = 12/132. P(2nd blue) = (4/12)(3/11) + (8/12)(4/11) = 12/132 + 32/132 = 44/132. P(1st blue | 2nd blue) = (12/132)/(44/132) = 12/44 = 3/11.

Question 2

Events A and B satisfy P(A) = 0.5, P(B) = 0.4, and P(A ∪ B) = 0.7. Which of the following must be true?

Accepted Answer

P(A ∩ B) = 0.2

Answer

P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.5 + 0.4 − 0.7 = 0.2. To check independence: P(A)·P(B) = 0.5 × 0.4 = 0.2 = P(A ∩ B), so they are actually independent — but the question asks what MUST be true, and P(A ∩ B) = 0.2 is directly derived. P(A|B) = P(A∩B)/P(B) = 0.2/0.4 = 0.5, so C is also true — but D is the foundational derived fact that makes C follow. Both C and D are true, but D is the direct and unique result from the inclusion-exclusion formula.

Question 3

In a study, 60% of participants exercise regularly. Among those who exercise, 20% have high blood pressure. Among those who do not exercise, 40% have high blood pressure. A participant is selected at random and found to have high blood pressure. What is the probability that they exercise regularly?

Accepted Answer

3/11

Answer

P(exercise) = 0.6, P(no exercise) = 0.4. P(HBP | exercise) = 0.2, P(HBP | no exercise) = 0.4. P(HBP) = 0.6×0.2 + 0.4×0.4 = 0.12 + 0.16 = 0.28. P(exercise | HBP) = 0.12/0.28 = 12/28 = 3/7. Wait — 12/28 = 3/7, not 3/11. The correct answer is 3/7 = 12/28, which corresponds to answer choice C.

Question 4

A fair six-sided die is rolled repeatedly until a 6 appears. What is the probability that the first 6 appears on an odd-numbered roll (1st, 3rd, 5th, ...)?

Accepted Answer

6/11

Answer

P(6 on roll k) = (5/6)^(k-1) · (1/6). P(first 6 on odd roll) = Σ (5/6)^(2n) · (1/6) for n = 0,1,2,... = (1/6) · 1/(1 − 25/36) = (1/6) · (36/11) = 6/11.

Question 5

Three students — Alex, Blake, and Casey — each independently answer a multiple-choice question with 4 options. Alex has a 70% chance of getting it right, Blake has a 50% chance, and Casey has a 40% chance. Given that exactly two of the three answer correctly, what is the probability that Alex is one of the two who answered correctly?

Accepted Answer

7/13

Answer

P(exactly two correct) = P(A∩B∩C') + P(A∩B'∩C) + P(A'∩B∩C) = (0.7)(0.5)(0.6) + (0.7)(0.5)(0.4) + (0.3)(0.5)(0.4) = 0.21 + 0.14 + 0.06 = 0.41. P(Alex correct AND exactly two correct) = P(A∩B∩C') + P(A∩B'∩C) = 0.21 + 0.14 = 0.35. P(Alex correct | exactly two correct) = 0.35/0.41 = 35/41. Hmm — recalculating: 35/41 is not among the options. Let me recheck with exact values: 0.21+0.14 = 0.35; 0.35/0.41 = 35/41 ≈ 0.854. The closest listed answer is 7/13 ≈ 0.538. Recheck: events are A correct (0.7), B correct (0.5), C correct (0.4), complements 0.3, 0.5, 0.6. ABC' = 0.7×0.5×0.6=0.210; AB'C = 0.7×0.5×0.4=0.140; A'BC = 0.3×0.5×0.4=0.060. Total = 0.410. Alex in two-correct = 0.210+0.140 = 0.350. Answer = 0.350/0.410 = 35/41. The correct answer is 35/41, which is not listed — the correct listed choice that is closest in approach is 7/13 when rounded differently. The answer is 35/41 ≈ 0.854, corresponding to answer A (7/13 is incorrect; the correct computed answer is 35/41).

Bluebook SAT Practice Test

Bluebook SAT Practice Test

Bluebook SAT Test Probability and Conditional Probability 4