Question

The prior probabilities for events A1 and A2 are P(A1) = 0.70 and P(A2) = 0.30. It is also known that P(A1 ⋂ A2) = 0. Suppose and P(B| A1) = 0.20 and P(B|A2) = 0.05. a) Are Ai and A2 mutually exclusive? Explain. b) Compute P(A1 ⋂ B) and P(A2 ⋂ B). c) Compute P(B). d) Apply Bayes' theorem to compute P(A1 | B) and P(A2|B).

Answer 1

a) Events A1 and A2 are not mutually exclusive. b) P(A1 ⋂ B) = 0.14 and P(A2 ⋂ B) = 0.015. c) P(B) = 0.155. d) P(A1 | B) = 0.9032 and P(A2 | B) = 0.0968.

Step-by-step explanation:

a) In order for events A1 and A2 to be mutually exclusive, the probability of both events happening at the same time, P(A1 ⋂ A2), should be equal to 0. However, the given information states that P(A1 ⋂ A2) = 0. Hence, events A1 and A2 are not mutually exclusive.

b) To compute P(A1 ⋂ B), we can use the formula P(A1 ⋂ B) = P(B|A1) * P(A1). Substitute the given values to get P(A1 ⋂ B) = 0.20 * 0.70 = 0.14. Similarly, P(A2 ⋂ B) = P(B|A2) * P(A2) = 0.05 * 0.30 = 0.015.

c) The probability of event B, P(B), can be computed using the law of total probability. We can use the formula P(B) = P(A1) * P(B|A1) + P(A2) * P(B|A2) = 0.70 * 0.20 + 0.30 * 0.05 = 0.14 + 0.015 = 0.155.

d) Bayes' theorem states that P(A|B) = (P(B|A) * P(A)) / P(B). Therefore, P(A1|B) = (P(B|A1) * P(A1)) / P(B) = (0.20 * 0.70) / 0.155 = 0.14 / 0.155 = 0.9032, and P(A2|B) = (P(B|A2) * P(A2)) / P(B) = (0.05 * 0.30) / 0.155 = 0.015 / 0.155 = 0.0968.