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 $BagA$ 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
$P\left(\frac{A}{E_1}\right)=\frac{{}^3{C}_1\times {}^1{C}_1}{{}^6{C}_2}=\frac{1}{5}$
$P\left(\frac{A}{E_1}\right)=\frac{{}^3{C}_1\times {}^2{C}_1}{(n+5{)}_{C_2}}=\frac{12}{(n+5)(n+4)}$
$P\left(\frac{E_1}{A}\right)=\frac{P\left(E_1\right)\times P\left(\frac{A}{E_1}\right)}{P\left(E_1\right)\times P\left(\frac{A}{E_1}\right)+P\left(E_2\right)\times P\left(\frac{A}{E_2}\right)}$
$=\frac{\frac{1}{10}}{\frac{1}{10}+\frac{6}{(n+5)(n+4)}}=\frac{6}{11}$
$\Rightarrow n=4$