Bag 1 contains 4 white balls and 5 black balls, and Bag 2 contains $n$ white balls and 3 black balls. One…

Bag 1 contains 4 white balls and 5 black balls, and Bag 2 contains $n$ white balls and 3 black balls. One ball is drawn randomly from Bag 1 and transferred to Bag 2. A ball is then drawn randomly from Bag 2. If the probability, that the ball drawn is white, is $29 / 45$, then $n$ is equal to :
  1. 6
  2. 3
  3. 5
  4. 4

Solution

Bag $1 \rightarrow 4 w, 5 B$
Bag $2 \rightarrow n w, 3 B$
$(\mathrm{I}) \rightarrow$ Transferred ball is white $P(w)=\frac{n+1}{n+4} \cdot \frac{4}{9}$
(II) $\rightarrow$ Transferred ball is black $P(w)=\frac{5}{9} \cdot \frac{n}{n+4}$
$\begin{aligned}
& \frac{4 n+4}{9 n+36}+\frac{5 n}{9 n+36}=\frac{29}{45} \\ & \frac{9 n+4}{9 n+36}=\frac{29}{45} \Rightarrow n=6
\end{aligned}$ ,

Asked in: JEE Main 2025 (29 Jan Shift 2)

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