JEE Main 2025 Jan 22 Shift 2, Mathematics Q17: Conditional Probability

JEE Main2025Jan 22, Shift 2Mathematics

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

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

Solution

Solve $12 x^{2} - 7 x + 1 = 0$ : factor as $(3 x - 1) (4 x - 1) = 0 \Rightarrow x = \frac{1}{3}$ or $x = \frac{1}{4}$ .

Assign $P (A ∣ B) = \frac{1}{3}$ and $P (B ∣ A) = \frac{1}{4}$ . Using definitions:

$\frac{P ( A \cap B )}{P ( B )} = \frac{1}{3} \Rightarrow P (B) = 3 \cdot 0.1 = 0.3$ .

$\frac{P ( A \cap B )}{P ( A )} = \frac{1}{4} \Rightarrow P (A) = 4 \cdot 0.1 = 0.4$ .

Then $P (A \cup B) = P (A) + P (B) - P (A \cap B) = 0.4 + 0.3 - 0.1 = 0.6$ .

By De Morgan, $\overset{ˉ}{A} \cup \overset{ˉ}{B} = \overline{A \cap B}$ and $\overset{ˉ}{A} \cap \overset{ˉ}{B} = \overline{A \cup B}$ .

Hence: $\frac{P ( A ˉ \cup B ˉ )}{P ( A ˉ \cap B ˉ )} = \frac{1 - P ( A \cap B )}{1 - P ( A \cup B )} = \frac{1 - 0.1}{1 - 0.6} = \frac{0.9}{0.4} = \frac{9}{4}$ .

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JEE Main 2025 Jan 22 Shift 2, Mathematics Q17: Conditional Probability

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