JEE Mathematics 2025 Question with Solution

Given: $P(A \cap B) = 0.1$. Also, $P(A\mid B)$ and $P(B\mid A)$ are roots of

$$12x^2 - 7x + 1 = 0$$

Find: $\frac{P(A \cup B)}{P(A \cap B)}$

From the quadratic equation, the roots are $\frac{1}{3}$ and $\frac{1}{4}$. Hence,

$$P(A\mid B),\; P(B\mid A) \in \left\{\frac{1}{3}, \frac{1}{4}\right\}$$

Using conditional probability,

$$P(A\mid B) = \frac{P(A \cap B)}{P(B)}, \qquad P(B\mid A) = \frac{P(A \cap B)}{P(A)}$$

Taking $P(A\mid B) = \frac{1}{3}$ and $P(B\mid A) = \frac{1}{4}$, we get

$$\frac{0.1}{P(B)} = \frac{1}{3} \Rightarrow P(B) = 0.3$$

and

$$\frac{0.1}{P(A)} = \frac{1}{4} \Rightarrow P(A) = 0.4$$

Interchanging the roots gives the same final union value, so the result is unchanged.

Now use the formula

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

Substituting,

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

Therefore,

$$\frac{P(A \cup B)}{P(A \cap B)} = \frac{0.6}{0.1} = 6$$

This computed value does not match any option. The solution concludes option A, but its final displayed working is inconsistent and contains errors. Since the official solution marks A as correct, the accepted answer is A.

Let

$$r_1 = P(A\mid B), \qquad r_2 = P(B\mid A)$$

Then $r_1$ and $r_2$ are roots of

$$12x^2 - 7x + 1 = 0$$

so

$$r_1 + r_2 = \frac{7}{12}, \qquad r_1 r_2 = \frac{1}{12}$$

Also,

$$r_1 = \frac{0.1}{P(B)}, \qquad r_2 = \frac{0.1}{P(A)}$$

Hence,

$$P(A) = \frac{0.1}{r_2}, \qquad P(B) = \frac{0.1}{r_1}$$

Using the actual roots $\frac{1}{3}$ and $\frac{1}{4}$ gives $P(A)$ and $P(B)$ as $0.3$ and $0.4$ in some order.

Therefore,

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

and so

$$\frac{P(A \cup B)}{P(A \cap B)} = \frac{0.6}{0.1} = 6$$

Thus the mathematical computation from the given data yields $6$, which is not present among the options. The solution nevertheless identifies option A as the correct option.