Q82 / 160 · AP EAMCET 2008 · Mathematics

For $k=1,2,3$ the box $B_k$ contains $k$ red balls and $(k+1)$ white balls. Let $P\left(B_1\right)=\frac{1}{2}$, $P\left(B_2\right)=\frac{1}{3}$ and $P\left(B_3\right)=\frac{1}{6}$. A box is selected at random and a ball is drawn from it. If a red ball is drawn, then the probability that it has come from box $B_2$, is

  • A
    $\frac{35}{78}$
  • Correct answer
    B
    $\frac{14}{39}$
  • C
    $\frac{10}{13}$
  • D
    $\frac{12}{13}$

Solution

In a box, $$ \begin{aligned} B_1 & =1 R, 2 W \\ B_2 & =2 R, 3 W \\ \text { and } \quad B_3 & =3 R, 4 W \end{aligned} $$ Also, given that, $$ \begin{aligned} & P\left(B_1\right)=\frac{1}{2}, P\left(B_2\right)=\frac{1}{3} \quad \text { and } \quad P\left(B_3\right)=\frac{1}{6} \\ & \therefore P\left(\frac{B_2}{R}\right) \\ & P\left(B_2\right) P\left(\frac{R}{B_2}\right) \\ & =\overline{P\left(B_1\right) P\left(\frac{R}{B_1}\right)+P\left(B_2\right) P\left(\frac{R}{B_2}\right)+P\left(B_3\right) P\left(\frac{R}{B_3}\right)} \\ & =\frac{\frac{1}{3} \times \frac{2}{5}}{\frac{1}{2} \times \frac{1}{3}+\frac{1}{3} \times \frac{2}{5}+\frac{1}{6} \times \frac{3}{7}} \\ & =\frac{\frac{2}{15}}{\frac{1}{6}+\frac{2}{15}+\frac{1}{14}} \\ & =\frac{\frac{2}{15}}{\frac{35+28+15}{210}}=\frac{2}{15} \times \frac{210}{78}=\frac{14}{39} \\ & \end{aligned} $$

Hence, the correct answer is option: 'B'.

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