The prior probabilities for events A1 and A2 are P(A1) = 0.40 and P(A2) = 0.60. It is also known that P(A1 ∩ A2) = 0. Suppose P(B | A1) = 0.20 and P(B | A2) = 0.10. (a) Are A1 and A2 mutually exclusive? They mutually exclusive. How could you tell whether or not they are mutually exclusive? P(B | A1) ≠ P(B | A2) P(A1) ≠ P(A1 | A2) P(A1) + P(A2) = 1 P(A1 ∩ A2) = 0 P(A2) ≠ P(A2 | A1)

Check Explanation

Step-by-step explanation:

a) Yes, the two events are mutually exclusive. The true meaning of mutually exclusive is that both A1 and A2 have no sample points in common and cannot both occur at the same time.

The only sure mathematical way to tell is from P(A1 n A2) = 0.

Note that this is different from when two events are independent. Two events are independent when the chances of one occurring does not depend on the chances of the other event occurring at all. Independent events can both occur at the same time. Mathematically, independent events are described as P(A1|A2) = P(A1) and P(A2|A1) = P(A2)

b) To solve for P(B)

We use the mathematical definition of the conditional statement

P(B|A1) = P(B n A1)/P(A1)

P(B n A1) = P(B|A1) × P(A1) = 0.2 × 0.4 = 0.08

P(B|A2) = P(B n A2)/P(A2)

P(B n A2) = P(B|A2) × P(A2) = 0.05 × 0.6 = 0.03

P(B) = P(B n A1) + P(B n A2) (especially because A1 and A2 are mutually exclusive)

P(B) = 0.08 + 0.03 = 0.11

Hope this helps!!

step-by-step explanation:

a rhombus is a parallelogram enjoying his symmetry.