The prior probabilities for events A1 and A2 are P(A1) = 0.30 and P(A2) = 0.55. 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? 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).

Using probability concepts, it is found that:

a) Since , events A1 and A2 are mutually exclusive.

b) P(A1 ∩ B) = 0.06, P(A2 ∩ B) = 0.0275.

c) P(B) = 0.0875.

d) P(A1|B) = 0.6857 and P(A2|B) = 0.3143.

-----------

Conditional Probability  

In which

-----------

Bayes Theorem:

Two events, A and B.

In which

-----------

Item a:

-----------

Item b:

To compute these probabilities, we use conditional probability.

A1 and B:

Since

Thus P(A1 ∩ B) = 0.06.

A2 and B:

Since

Thus P(A2 ∩ B) = 0.0275.

-----------

Item c:

P(B) can be written as:

Thus P(B) = 0.0875.

-----------

Item d:

Applying Bayes Theorem, first for A1 given B.

Then for A2 given B.

Thus P(A1|B) = 0.6857 and P(A2|B) = 0.3143.

A similar problem is given at link

We're given the following probabilities:

(a) Yes, and are mutually exclusive. This is exactly what zero probability of their intersection means. The two events cannot occur simultaneously.

(b) Use the definition of conditional probability to expand:

(c) By the law of total probability,

(d) Bayes' theorem says

no

step-by-step explanation:

m∠5 + m∠3 = m∠1   is true.

m∠5 + m∠6 = 180 is true

m∠2 +m∠3 = m∠6   is true

m∠2 +m∠3 +m∠5 = 180   is true