Mathematics · Probability

JEE Main 2025 — 22 January, Evening Shift — Question 17

If A and B are two events such that $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.1$ , and $P(A \mid B)$ and $P(B \mid A)$ are the roots of the equation $12 x^{2}-7 x+1=0$ , then the value of $\frac{P(\overline{\mathrm{~A}} \cup \overline{\mathrm{~B}})}{\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})}$ is:

Option A:
$\frac{5}{3}$
Option B:
$\frac{4}{3}$
Option C:
$\frac{9}{4}$
Correct
Option D:
$\frac{7}{4}$

Answer: C

Step-by-step solution

\begin{aligned} &\text{Given\; } P(A \cap B) = 0.1, \text{ and } P(A|B),\, P(B|A)\; \text{ are\; roots\; of \;} 12x^2 - 7x + 1 = 0.\\[6pt] &\Rightarrow x_1 + x_2 = \frac{7}{12}, \quad x_1x_2 = \frac{1}{12}.\\[6pt] &P(A|B) = \frac{0.1}{P(B)} = x_1 \Rightarrow P(B) = \frac{0.1}{x_1},\quad P(B|A) = \frac{0.1}{P(A)} = x_2 \Rightarrow P(A) = \frac{0.1}{x_2}.\\[6pt] &P(A\cup B) = P(A) + P(B) - P(A\cap B) = 0.1\!\left(\frac{1}{x_1} + \frac{1}{x_2} - 1\right).\\[6pt] &P(\overline{A}\cap\overline{B}) = 1 - P(A\cup B) = 1 - 0.1\!\left(\frac{1}{x_1} + \frac{1}{x_2} - 1\right),\\ &P(\overline{A}\cup\overline{B}) = 1 - P(A\cap B) = 0.9.\\[6pt] &\therefore \frac{P(\overline{A}\cup\overline{B})}{P(\overline{A}\cap\overline{B})} = \frac{0.9}{ 1 - 0.1\!\left(\frac{1}{x_1} + \frac{1}{x_2} - 1\right)}.\\[6pt] &\text{Since } \frac{1}{x_1} + \frac{1}{x_2} = \frac{x_1+x_2}{x_1x_2} = \frac{7/12}{1/12} = 7,\\[6pt] &\Rightarrow \frac{P(\overline{A}\cup\overline{B})}{P(\overline{A}\cap\overline{B})} = \frac{0.9}{1 - 0.1(7 - 1)} = \frac{0.9}{0.4} = \frac{9}{4}.\\[6pt] &\boxed{\frac{9}{4}} \end{aligned}

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 22 January, Evening Shift
Subject: Mathematics
Chapter: Probability
Topic: Conditional Probability and Multiplication Theorem