Bag 1 contains 4 white balls and 5 black balls, and Bag 2 contains n white balls and 3 bla…

Question

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 $\dfrac{29}{45}$, then $n$ is equal to:

Accepted Answer

$$\begin{aligned}
& \text { Bag } 1=\{4 \mathrm{~W}, 5 \mathrm{~B}\} \
& \text { Bag } \mathbf{2}=\{\mathbf{n W}, \mathbf{3 B}\} \
& \mathrm{P}\left(\frac{\mathrm{~W}}{\mathrm{Bag} 2}\right)=\frac{29}{45} \
& \Rightarrow \mathrm{P}\left(\frac{\mathrm{~W}}{\mathrm{~B}_1}\right) \times \mathrm{P}\left(\frac{\mathrm{~W}}{\mathrm{~B}_2}\right)+\mathrm{P}\left(\frac{\mathrm{~B}}{\mathrm{~B}_1}\right) \times \mathrm{P}\left(\frac{\mathrm{~W}}{\mathrm{~B}_2}\right)=\frac{29}{45} \
& \frac{4}{9} \times \frac{\mathrm{n}+1}{\mathrm{n}+4}+\frac{5}{9} \times \frac{\mathrm{n}}{\mathrm{n}+4}=\frac{29}{45}
\end{aligned}$$

$\mathrm{n = 6}$

Solutions · 1