Answer to Question #136344 in Statistics and Probability for Tunra Joy

Question

Let A and B be two events. Suppose the probability that at least one of them occurs is 2/3. What
is the probability that neither A nor B occurs?

Answer 1

Solution :

Here, there are only four possibilities i.e

$$P(A \cap B) \cup P(A \cap B') \cup P(A' \cap B) \cup P (A' \cap B')$$

Whose probability sums to one.

We know that

$$P(A \cap B) \cup P(A \cap B') \cup P(A' \cap B)= {2 \over 3}$$

i.e

$$P(A \cap B) +P(A \cap B') +P(A' \cap B)= {2 \over 3}$$

Thus;

$$P(A' \cap B') =$$ $$1- \big( P(A \cap B) + P(A \cap B') + P(A' \cap B) \big)$$

$$=1 - {2 \over 3} = {1 \over 3}$$

Answer: 1/3