Q94 / 150 · MHT CET 2025 (19 Apr Shift 2) · Mathematics

Let $A$ and $B$ are independent events with $P(B)=\frac{2}{5}, P(A \cup B)=\frac{11}{20}$, then $\mathrm{P}\left(\mathrm{A}^{\prime} \mid \mathrm{B}\right)$ is root of the equation

  • Correct answer
    A
    $4 x^2-7 x+3=0$
  • B
    $4 x^2+7 x+3=0$
  • C
    $4 x^2-3 x-7=0$
  • D
    $6 x^2-5 x+1=0$

Solution

Given independent events $A$ and $B$ with $P(B) = \frac{2}{5}$ and $P(A \cup B) = \frac{11}{20}$, the independence condition $P(A \cap B) = P(A)P(B)$ applies.

The probability of union is $P(A \cup B) = P(A) + P(B) - P(A)P(B)$. Substituting values yields $\frac{11}{20} = P(A) + \frac{2}{5} - \frac{2}{5}P(A)$.

Solving for $P(A)$: $\frac{11}{20} - \frac{8}{20} = P(A)\left(1 - \frac{2}{5}\right)$, so $\frac{3}{20} = P(A)\left(\frac{3}{5}\right)$ and $P(A) = \frac{1}{4}$.

For independent events, $P(A' \mid B) = P(A') = 1 - P(A) = \frac{3}{4}$.

Testing $x = \frac{3}{4}$ in the options shows it satisfies $4x^2 - 7x + 3 = 0$ since $4\left(\frac{9}{16}\right) - 7\left(\frac{3}{4}\right) + 3 = 0$.

Option A is correct.

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

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