Q264 / 269 · NDA 2025 (Phase 1) · Mathematics
If $P(A) = \dfrac{1}{3}$, $P(B) = \dfrac{1}{2}$ and $P(A \cap B) = \dfrac{1}{4}$, then what is the value of $P(B | A^c)$?
- A$\dfrac{1}{8}$
- Correct answerB$\dfrac{3}{8}$
- C$\dfrac{5}{8}$
- D$\dfrac{7}{8}$
Solution
The conditional probability is given by:
$P(B | A^c) = \dfrac{P(B \cap A^c)}{P(A^c)}$
The probability of the complement of $A$ is:
$P(A^c) = 1 - P(A) = 1 - \dfrac{1}{3} = \dfrac{2}{3}$
The probability of $B \cap A^c$ is:
$P(B \cap A^c) = P(B) - P(A \cap B) = \dfrac{1}{2} - \dfrac{1}{4} = \dfrac{1}{4}$
Substituting these values into the formula:
$P(B | A^c) = \dfrac{\dfrac{1}{4}}{\dfrac{2}{3}} = \dfrac{3}{8}$
Answer: $\dfrac{3}{8}$
$P(B | A^c) = \dfrac{P(B \cap A^c)}{P(A^c)}$
The probability of the complement of $A$ is:
$P(A^c) = 1 - P(A) = 1 - \dfrac{1}{3} = \dfrac{2}{3}$
The probability of $B \cap A^c$ is:
$P(B \cap A^c) = P(B) - P(A \cap B) = \dfrac{1}{2} - \dfrac{1}{4} = \dfrac{1}{4}$
Substituting these values into the formula:
$P(B | A^c) = \dfrac{\dfrac{1}{4}}{\dfrac{2}{3}} = \dfrac{3}{8}$
Answer: $\dfrac{3}{8}$
Hence, the correct answer is option: 'B'.