Q54 / 150 · MHT CET 2026 (15 April Shift 1) · Mathematics

If A and B are two events such that $P(A) = \dfrac{2}{3}$, $P(B) = \dfrac{1}{2}$ and $P(A \mid B) = \dfrac{2}{3}$, then $P(A' \cup B) + P(A \cup B') = $

  • A
    $\dfrac{2}{3}$
  • B
    $1$
  • Correct answer
    C
    $\dfrac{3}{2}$
  • D
    $\dfrac{5}{6}$

Solution

Given $P(A) = \dfrac{2}{3}$, $P(B) = \dfrac{1}{2}$, $P(A \mid B) = \dfrac{2}{3}$

$P(A \cap B) = P(A \mid B) P(B) = \dfrac{2}{3} \times \dfrac{1}{2} = \dfrac{1}{3}$

$P(A' \cup B) = 1 - P(A \cap B') = 1 - (P(A) - P(A \cap B))$

$P(A' \cup B) = 1 - \left(\dfrac{2}{3} - \dfrac{1}{3}\right) = \dfrac{2}{3}$

$P(A \cup B') = 1 - P(A' \cap B) = 1 - (P(B) - P(A \cap B))$

$P(A \cup B') = 1 - \left(\dfrac{1}{2} - \dfrac{1}{3}\right) = \dfrac{5}{6}$

$P(A' \cup B) + P(A \cup B') = \dfrac{2}{3} + \dfrac{5}{6} = \dfrac{9}{6} = \dfrac{3}{2}$

Hence, the correct answer is option: 'C'.

Relevant Questions