Q265 / 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 $P(A^c \cap B^c)$?
- A$\dfrac{1}{4}$
- Correct answerB$\dfrac{5}{12}$
- C$\dfrac{7}{12}$
- D$\dfrac{11}{12}$
Solution
Using the formula for the probability of the union of two events:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Substituting the given values:
$P(A \cup B) = \dfrac{1}{3} + \dfrac{1}{2} - \dfrac{1}{4}$
$P(A \cup B) = \dfrac{4 + 6 - 3}{12} = \dfrac{7}{12}$
By De Morgan's laws, $A^c \cap B^c = (A \cup B)^c$.
Therefore, the probability is:
$P(A^c \cap B^c) = 1 - P(A \cup B)$
$P(A^c \cap B^c) = 1 - \dfrac{7}{12} = \dfrac{5}{12}$
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Substituting the given values:
$P(A \cup B) = \dfrac{1}{3} + \dfrac{1}{2} - \dfrac{1}{4}$
$P(A \cup B) = \dfrac{4 + 6 - 3}{12} = \dfrac{7}{12}$
By De Morgan's laws, $A^c \cap B^c = (A \cup B)^c$.
Therefore, the probability is:
$P(A^c \cap B^c) = 1 - P(A \cup B)$
$P(A^c \cap B^c) = 1 - \dfrac{7}{12} = \dfrac{5}{12}$
Hence, the correct answer is option: 'B'.