JEE Main 2025 — Probability Question with Solution

To solve this problem, start by considering the equation given for the probabilities:

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

The roots of this equation are:

$x=\frac{1}{3},\frac{1}{4}$

Assume $P(A\mid B)=\frac{1}{3}$ and $P(B\mid A)=\frac{1}{4}$.

From the definitions of conditional probability, we have:

$P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{1}{3}$

$P(B\mid A)=\frac{P(A\cap B)}{P(A)}=\frac{1}{4}$

Given that $P(A\cap B)=0.1$, we can use these equations to find $P(B)$ and $P(A)$:

From $\frac{0.1}{P(B)}=\frac{1}{3}$, we find $P(B)=0.3$.

From $\frac{0.1}{P(A)}=\frac{1}{4}$, we find $P(A)=0.4$.

Now, calculate $P(A\cup B)$ using the formula for the union of two events:

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

The task is to find the value of:

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

Using De Morgan's laws and the complements:

$P(\overline{A}\cup \overline{B})=P(\overline{A\cap B})=1-P(A\cap B)=1-0.1=0.9$

$P(\overline{A}\cap \overline{B})=1-P(A\cup B)=1-0.6=0.4$

Finally, compute the ratio:

$\frac{P(\overline{A}\cup \overline{B})}{P(\overline{A}\cap \overline{B})}=\frac{0.9}{0.4}=\frac{9}{4}$