If $P(A)=1/3$, $P(B)=1/2$ and $P(A\cap B)=1/4$, then what is the value of $P(A{}^C\cap B{}^C)$?

Calculation:

Given,

\( P(A) = \frac{1}{3} \)

\( P(B) = \frac{1}{2} \)

\( P(A \cap B) = \frac{1}{4} \)

We need to calculate\( P(A^C \cap B^C) \), the probability that neither A  nor B  occurs.

Using the complement rule:

\( P(A^C \cap B^C) = 1 - P(A \cup B) \)

Now, apply the inclusion-exclusion formula to find \(( P(A \cup B) \):

\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)

Substitute the given values:

\( P(A \cup B) = \frac{1}{3} + \frac{1}{2} - \frac{1}{4} \)

To simplify:

\( P(A \cup B) = \frac{4}{12} + \frac{6}{12} - \frac{3}{12} = \frac{7}{12} \)

Now, calculate \(P(A^C \cap B^C) \):

\( P(A^C \cap B^C) = 1 - \frac{7}{12} = \frac{5}{12} \)

Hence, the correct answer is Option 2.