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 \(\frac{29}{45},\) then n is equal to:

(1) 3

(2) 4

(3) 5

(4) 6

Answer 1

Correct option is (4) 6  

Bag 1 \(=\{4 \mathrm{W}, 5 \mathrm{B}\}\)

Bag 2 \(=\{\mathbf{n W}, \mathbf{3 B}\}\)

\(\mathrm{P}\left(\frac{\mathrm{W}}{\mathrm{Bag} \ 2}\right)=\frac{29}{45}\)

\(\Rightarrow \mathrm{P}\left(\frac{\mathrm{W}}{\mathrm{B}_{1}}\right) \times \mathrm{P}\left(\frac{\mathrm{W}}{\mathrm{B}_{2}}\right)+\mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{B}_{1}}\right) \times \mathrm{P}\left(\frac{\mathrm{W}}{\mathrm{B}_{2}}\right)=\frac{29}{45}\)

\(\frac{4}{9} \times \frac{n+1}{n+4}+\frac{5}{9} \times \frac{n}{n+4}=\frac{29}{45}\)

n = 6