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