Bag A contains 2 white, 1 black and 3 red balls and bag B contains 3 black, 2 red and n wh…

Question

Bag A contains 2 white, 1 black and 3 red balls and bag B contains 3 black, 2 red and n white balls. One bag is chosen at random and 2 balls drawn from it at random, are found to be 1 red and 1 black. If the probability that both balls come from Bag A is $${6 \over {11}}$$, then n is equal to \_\_\_\_\_\_\_\_\_\_.

Accepted Answer

$$
\begin{aligned}
&P(1 R \text { and } 1 B)=P(A) \cdot P\left(\frac{1 R 1 B}{A}\right)+P(B) \cdot P\left(\frac{1 R 1 B}{B}\right) \\
&=\frac{1}{2} \cdot \frac{{ }^{3} C_{1} \cdot{ }^{1} C_{1}}{{ }^{6} C_{2}}+\frac{1}{2} \cdot \frac{{ }^{2} C_{1} \cdot{ }^{3} C_{1}}{{ }^{n+5} C_{2}} \\
&P\left(\frac{1 R 1 B}{A}\right)=\frac{\frac{1}{2} \cdot \frac{3}{15}}{\frac{1}{2} \cdot \frac{3}{15}+\frac{1}{2} \cdot \frac{6 \cdot 2}{(n+5)(n+4)}}=\frac{6}{11} \\
&\Rightarrow \frac{\frac{1}{10}}{\frac{1}{10}+\frac{6}{(n+5)(n+4)}}=\frac{6}{11}
\end{aligned}
$$

$$
\begin{aligned}
&\Rightarrow \frac{11}{10}=\frac{6}{10}+\frac{36}{(n+5)(n+4)} \\
&\Rightarrow \frac{5}{10 \times 36}=\frac{1}{(n+5)(n+4)} \\
&\Rightarrow n^{2}+9 n-52=0 \\
&\Rightarrow n=4 \text { is only possible value }
\end{aligned}
$$

Solutions · 1