Question

Let A and B be two events. Suppose that P (A) = 0.41 and P (B) = 0.46.

(a) Find P (A or B), given that A and B are independent.
(b) Find P (A or B), given that A and B are mutually exclusive.

Answer 1

When A and B are independent, P (A or B) = 0.6814.

When A and B are mutually exclusive, P (A or B) = 0.87

Here, A and B are two events. The given probabilities are P(A) = 0.41 and P(B) = 0.46.

When A and B are independent, P (A or B) = P(A) + P(B) - P (A and B).

This shows three possibilities -

1. Only event A happens

2. Only event B happens

3. Neither A nor B happens.

When the events A and B are independent, it means that the probability of occurrence of both events do not affect each other. Hence, here we remove the probability of both events happening together which is P (A and B).

Therefore,
First, we find P (A and B) which is, P(A)*P(B).

P (A and B) = 0.41*0.46 = 0.1886

Now,

P (A or B) = 0.41 + 0.46 - 0.1886 = 0.6814

When A and B are mutually exclusive, P (A or B) = P(A) + P(B)

When two events are mutually exclusive, it means that those two events cannot occur at the same time.

Hence, here P (A and B) = NULL (Zero)

This is how we arrive at the formula,

P (A or B) = P(A) + P(B).

Now,

P (A or B) = 0.41 + 0.46 = 0.87