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

Calculation:

Given,

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

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

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

We need to calculate the probability P(B | A^C) , the probability that B  occurs given that A  does not occur.

Using Bayes' Theorem and the formula for conditional probability:

\( P(B | A^C) = \frac{P(B) - P(A \cap B)}{1 - P(A)} \)

Substitute the values:

\( P(B | A^C) = \frac{\frac{1}{2} - \frac{1}{4}}{1 - \frac{1}{3}} \)

Simplify the numerator:

\( \frac{1}{2} - \frac{1}{4} = \frac{2}{4} - \frac{1}{4} = \frac{1}{4} \)

Simplify the denominator:

\( 1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{2}{3} \)

The final expression is:

\( P(B | A^C) = \frac{\frac{1}{4}}{\frac{2}{3}} = \frac{1}{4} \times \frac{3}{2} = \frac{3}{8} \)

Hence, the correct answer is Option 2.