Given three indentical bags each containing 10 balls, whose colours are as follows : $\begin{array}{cccc} &…
$\begin{array}{cccc} & \text{Red} & \text{Blue} & \text{Green} \\ \text{Bag I} & 3 & 2 & 5 \\ \text{Bag II} & 4 & 3 & 3 \\ \text{Bag III} & 5 & 1 & 4\end{array}$
A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from bag I is p and if the balls is Green, the probability that it is from bag III is q , then the value of $\left(\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}\right)$ is :
- $6$
- $9$
- $7$
- $8$
Solution
$\mathrm{p}=\mathrm{P}\left(\frac{\mathrm{B}_{\mathrm{I}}}{\mathrm{R}}\right)=\frac{\frac{1}{3}\left(\frac{3}{10}\right)}{\frac{1}{3}\left(\frac{3}{10}+\frac{4}{10}+\frac{5}{10}\right)}=\frac{1}{4}$
$\begin{aligned} & \mathrm{q}=\left(\frac{\mathrm{B}_{\mathrm{III}}}{\mathrm{G}}\right)=\frac{\frac{1}{3}\left(\frac{4}{10}\right)}{\frac{1}{3}\left(\frac{5}{10}+\frac{3}{10}+\frac{4}{10}\right)}=\frac{1}{3} \\ & \therefore \frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=7\end{aligned}$ ,
Asked in: JEE Main 2025 (02 Apr Shift 2)