Question

If A and B are two mutually exclusive events with P(A) =0.47 and P(B) =0.21, what is the probability of (A and B), (A or B), (not A), (not B), (not (A or B)) and (A and (not B))?

Answer 1

If A and B are mutually exclusive events with given probabilities, we can calculate the probabilities of various combinations of these events.

Step-by-step explanation:

Given that A and B are mutually exclusive events, we know that P(A AND B) = 0.

The probability of (A OR B) can be calculated using the equation P(A OR B) = P(A) + P(B) - P(A AND B). Therefore, P(A OR B) = 0.47 + 0.21 - 0 = 0.68.

The probability of (not A) is equal to 1 - P(A), which is equal to 1 - 0.47 = 0.53.

The probability of (not B) is equal to 1 - P(B), which is equal to 1 - 0.21 = 0.79.

The probability of (not (A OR B)) is equal to 1 - P(A OR B), which is equal to 1 - 0.68 = 0.32.

The probability of (A AND (not B)) can be calculated using the equation P(A AND (not B)) = P(A) - P(A AND B), which is equal to 0.47 - 0 = 0.47.