The prior probabilities for events A1 and A2 are P(A1) = 0.35 and P(A2) = 0.50. It is also known that P(A1 ∩ A2) = 0. Suppose P(B | A1) = 0.20 and P(B | A2) = 0.05. If needed, round your answers to three decimal digits. (a) Are A1 and A2 mutually exclusive? Explain your answer. The input in the box below will not be graded, but may be reviewed and considered by your instructor. (b) Compute P(A1 ∩ B) and P(A2 ∩ B). P(A1 ∩ B) = 0.2 P(A2 ∩ B) = 0.05 (c) Compute P(B). P(B) = (d) Apply Bayes’ theorem to compute P(A1 | B) and P(A2 | B). P(A1 | B) = P(A2 | B) =

Two events are mutually exclusive when the occurrence of one of them prevents the occurrence of the other in one repetition of the trial, this means that both events cannot occur at the same time and therefore they'll intersection is void (and its probability zero)

Considering that P(A₁∩A₂)= 0, we can assume that both events are mutually exclusive.

b)

Considering that [tex]P(BIA)= \frac{P(AnB)}{P(A)}[/tex] you can clear the intersection from the formula [tex]P(AnB)= P(B/A)*P(A)[/tex] and apply it for the given events:

[tex]P(A_1nB)= P(B/A_1) * P(A_1)= 0.20*0.35= 0.07[/tex]

[tex]P(A_2nB)= P(B/A_2)*P(A_2)= 0.05*0.50= 0.025[/tex]

c)

The probability of "B" is marginal, to calculate it you have to add all intersections where it occurs:

P(B)= (A₁∩B) + P(A₂∩B)= 0.07 + 0.025= 0.095

d)

The Bayes' theorem states that:

[tex]P(Ai/B)= \frac{P(B/Ai)*P(A)}{P(B)}[/tex]

Then:

[tex]P(A_1/B)= \frac{P(B/A_1)*P(A_1)}{P(B)}= \frac{0.20*0.35}{0.095}= 0.737 = 0.74[/tex]

[tex]P(A_2/B)= \frac{P(B/A_2)*P(A_2)}{P(B)} = \frac{0.05*0.50}{0.095} = 0.26[/tex]

I hope it helps!