SOLUTION: Two events A and B are such that P(A) = 9/16, P(B) = 3/8, and P(A | B) = 1/4. Find the probability that: a. Both events will happen b. Only one of the event shows will happen C

Question 1117696: Two events A and B are such that P(A) = 9/16, P(B) = 3/8, and P(A | B) = 1/4. Find the probability that:
a. Both events will happen
b. Only one of the event shows will happen
C. Neither event will happen
Answer by greenestamps(13216) (Show Source): You can put this solution on YOUR website!

If you go with the definition of P(A|B), then

P(A|B) = P(A AND B)/P(B)

One way to look at this definition is that the only part of the sample space you are considering is B, so P(B) is the denominator; and you are concerned with only the part of B that is also A, which makes P(A AND B) the numerator.

For your problem, the calculation is

1/4 = P(A AND B)/(3/8)
P(A AND B) = (1/4)(3/8) = 3/32

That calculation is in the form

P(A AND B) = P(B)*{P(A|B)

That form of the definition of P(A|B) makes more sense to me personally; it says that to find the probability of both A and B, you start with the probability of B and multiply it by the probability that A is true if B is true.

So we have P(A AND B) = 3/32, which is the first answer we are looking for.

Then P(A only) is P(A) minus P(A AND B) = 9/16-3/32 = 15/32 and P(B only) = 3/8 - 3/32 = 9/32; so the answer to the second question is 15/32+9/32 = 24/32 = 3/4.

The probabilities from the first two parts of the problem cover all the cases where at least one of A or B happens; so the probability that neither event will happen is 1 - 27/32 = 5/32.

Answers: a. 3/32 b. 3/4 c. 5/32
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