JEE Main 2025 — Probability Question with Solution

Probability that a Red ball comes from Bag I ($p$):

$p(B_1/R)=\frac{p(B_1)\cdot p(R/B_1)}{p(R)}$

Substituting the values,

$p(B_1/R)=\frac{\frac{1}{3}\times \frac{3}{10}}{\frac{1}{3}\times \frac{3}{10}+\frac{1}{3}\times \frac{4}{10}+\frac{1}{3}\times \frac{5}{10}}$

Simplify to find $p$:

$=\frac{\frac{1}{10}}{\frac{1}{3}(1.2)}=\frac{1}{4}$

So, $p=\frac{1}{4}$.

Probability that a Green ball comes from Bag III ($q$):

$p(B_3/G)=\frac{p(B_3)\cdot p(G/B_3)}{p(G)}$

Substitute the values,

$p(B_3/G)=\frac{\frac{1}{3}\times \frac{4}{10}}{\frac{1}{3}\times \frac{5}{10}+\frac{1}{3}\times \frac{3}{10}+\frac{1}{3}\times \frac{4}{10}}$

Simplify to find $q$:

$=\frac{\frac{2}{15}}{\frac{1}{3}\times 1.2}=\frac{1}{3}$

So, $q=\frac{1}{3}$.

Calculation of $\left(\frac{1}{p}+\frac{1}{q}\right)$:

$\frac{1}{p}=4,\frac{1}{q}=3$

Therefore,

$\left(\frac{1}{p}+\frac{1}{q}\right)=4+3=\boxed{7}$