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

Answer 1

Answer: (A)

$\frac{7}{16}$

For the bag $A$ we can see that there are $2$ white, $3$ red and $2$ black balls. Similarly, from bag $B$ we have $4$ white, $2$ red and $3$ black balls.

Probability of choosing a white and then a red ball from bag $B$ is given by $=\frac{{}^4{C}_1\times {}^2{C}_1}{{}^9{C}_2}$

Probability of choosing a white ball then a red ball from bag $A$ is given by $=\frac{{}^2{C}_1\times {}^3{C}_1}{{}^7{C}_2}$

So, the probability of getting a white ball and then a red ball from bag $B$ is given by

$\frac{\frac{{}^4{C}_1\times {}^2{C}_1}{{}^9{C}_2}}{\frac{{}^4{C}_1{\times }^2{C}_1}{{}^9{C}_2}+\frac{{}^2{C}_1\times {}^3{C}_1}{{}^7{C}_2}}$ $=\frac{\frac{2}{9}}{\frac{2}{7}+\frac{2}{9}}=\frac{2\times 7}{18+14}=\frac{7}{16}$