The prior probabilities for events A1 and A2 are P(A1) = 0.20 and P(A2) = 0.80. It is also known that P(A1 ∩ A2) = 0. Suppose P(B | A1) = 0.25 and P(B | A2) = 0.05. (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(A2) = 1 P(A1 ∩ A2) = 0 P(A1) ≠ P(A1 | A2) P(A2) ≠ P(A2 | A1) (b) Compute P(A1 ∩ B) and P(A2 ∩ B). P(A1 ∩ B) = P(A2 ∩ B) = (c) Compute P(B). (d) Apply Bayes' theorem to compute P(A1 | B) and P(A2 | B). (Round your answers to four decimal places.) P(A1 | B) = P(A2 | B) =

Using probability concepts, along with conditional probability and Bayes Theorem, it is found that:

a) , thus yes, A1 and A2 are mutually exclusive.

b) The probabilities are: P(A1 ∩ B) = 0.05, P(A2 ∩ B) = 0.04.

c) The probability of event B is: P(B) = 0.09.

d) The probabilities are: P(A1|B) = 0.5555 and P(A2|B) = 0.4445.

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Conditional Probability  

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Bayes Theorem:  

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Item a:

Item b:

Using conditional probability.

A1 and B:

A2 and B:

Item c:

P(B) is given by:

Item d:

Applying Bayes Theorem

A similar problem is given at link

(a) and are indeed mutually-exclusive.

(b) , whereas .

(c) .

(d) , whereas

Step-by-step explanation:

means that it is impossible for events and to happen at the same time. Therefore, event and are mutually-exclusive.

By the definition of conditional probability:

.

Rearrange to obtain:

.

Similarly:

.

Note that:

.

In other words, and are collectively-exhaustive. Since and are collectively-exhaustive and mutually-exclusive at the same time:

.

By Bayes' Theorem:

.

Similarly:

.

5.6%

step-by-step explanation:

sine

step-by-step explanation: