Question

A box ${}^{\prime }{A}^{\prime }$ contains $2$ white, $3$ red and $2$ black balls. Another box ${}^{\prime }{B}^{\prime }$ contains $4$ white, $2$ red and $3$ black balls. If two balls are drawn at random, without replacement, from a randomly selected box and one ball turns out to be white while the other ball turns out to be red, then the probability that both balls are drawn from box ${}^{\prime }{B}^{\prime }$ is :

• A. $\frac{9}{16}$
• B. $\frac{7}{16}$
• C. $\frac{9}{32}$
• D. $\frac{7}{8}$

Answer 1

The correct option is B $\frac{7}{16}$

$P\left(E\right):$ Probability of choosing $1$ red ball and $1$ white ball.
$P\left(BoxA\right):$ Probability of choosing box $A$.
$P\left(BoxB\right):$ Probability of choosing box $B$.
Hence the required probabiity will be,
$P\left(\frac{BoxB}{E}\right)=\frac{P\left(BoxB\right)\times P\left(\frac{E}{BoxB}\right)}{P\left(BoxA\right)\times P\left(\frac{E}{BoxA}\right)+P\left(BoxB\right)\times P\left(\frac{E}{BoxB}\right)}\Rightarrow P\left(\frac{BoxB}{E}\right)=\frac{\frac{1}{2}\times \frac{{}^4{C}_1\times {}^2{C}_1}{{}^9{C}_2}}{\frac{1}{2}\times \frac{{}^2{C}_1\times {}^3{C}_1}{{}^7{C}_2}+\frac{1}{2}\times \frac{{}^4{C}_1\times {}^2{C}_1}{{}^9{C}_2}}\Rightarrow P\left(\frac{BoxB}{E}\right)=\frac{\frac{1}{9}}{\frac{1}{7}+\frac{1}{9}}\Rightarrow P\left(\frac{BoxB}{E}\right)=\frac{7}{16}$