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 B₂, is

Given a box contains k red balls and (k+1) white balls i.e.
B₁ contains 1R,2W;B₂ contain 2R,3W and B₃ contains 3R, 4W
Also given that, $P\left(B_1\right)=\frac{1}{2},P\left(B_2\right)=\frac{1}{3}$ and $P\left(B_3\right)=\frac{1}{6}$

$\therefore P\left(\frac{B_2}{R}\right)$

$=\frac{P\left(B_2\right)P\left(\frac{R}{B_2}\right)}{P\left(B_1\right)P\left(\frac{R}{B_1}\right)+P\left(B_2\right)P\left(\frac{R}{B_2}\right)+P\left(B_3\right)P\left(\frac{R}{B_3}\right)}$

$=\frac{\frac{1}{3}\times \frac{2}{5}}{\frac{1}{2}\times \frac{1}{3}+\frac{1}{3}\times \frac{2}{5}+\frac{1}{6}\times \frac{3}{7}}$

$=\frac{\frac{2}{15}}{\frac{1}{6}+\frac{2}{15}+\frac{1}{14}}$

$=\frac{\frac{2}{15}}{\frac{35+28+15}{210}}$

$=\frac{2}{15}\times \frac{210}{78}$

$=\frac{14}{39}$