Question

Events A1, A2 and A3 form a partiton of the sample space S with probabilities P(A1) = 0.3, P(A2) = 0.5, P(A3) = 0.2.

Answer 1

Answer:

Step-by-step explanation:

Given that events A1, A2 and A3 form a partiton of the sample space S with probabilities P(A1) = 0.3, P(A2) = 0.5, P(A3) = 0.2.

i.e. A1, A2, and A3 are mutually exclusive and exhaustive

E is an event such that

P(E|A1) = 0.1, P(E|A2) = 0.6, P(E|A3) = 0.8,

$P(E) = P(A_1E)+P(A_2E)+P(A_3E)\\= \Sigma P(E/A_1) P(A_1) \\= 0.1(0.3)+0.5(0.6)+0.8(0.2)\\= 0.03+0.3+0.16\\= 0.49$

$P(A_1/E) = P(A_1E)/P(E) = \frac{0.3(0.1)}{0.49} \\=0.061224$

$P(A_2/E) = P(A_2E)/P(E) = \frac{0.5)(0.6)}{0.49} \\=0.61224$

$P(A_3/E) = P(A_3E)/P(E) = \frac{0.2)(0.8)}{0.49} \\=0.3265$