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 $\frac{6}{11}$, then $n$ is equal to ___

• A. 13
• B. 6
• C. 4
• D. 3

Answer 1

Answer: (C)

$E _{1}=$ denotes selection for $1^{\text {st }}$ bag
$E _{2}=$ denotes selection for $2^{\text {nd }}$ bag
$P \left( E _{1}\right)=\frac{1}{2}, P \left( E _{2}\right)=\frac{1}{2}$
$A =$ selected balls are $1$ red & $1 $ black
$P \left(\frac{ A }{ E _{1}}\right)=\frac{{ }^{3} C _{1} \times{ }^{1} C _{1}}{{ }^{6} C _{2}}=\frac{1}{5}$
$P \left(\frac{ A }{ E _{1}}\right)=\frac{{ }^{3} C _{1} \times{ }^{2} C _{1}}{( n +5)_{ C _{2}}}=\frac{12}{( n +5)( n +4)}$
$P\left(\frac{E_{1}}{A}\right)=\frac{P\left(E_{1}\right) \times P\left(\frac{A}{E_{1}}\right)}{P\left(E_{1}\right) \times P\left(\frac{A}{E_{1}}\right)+P\left(E_{2}\right) \times P\left(\frac{A}{E_{2}}\right)}$
$=\frac{\frac{1}{10}}{\frac{1}{10}+\frac{6}{(n+5)(n+4)}}=\frac{6}{11}$
$\Rightarrow n =4$