Discussion Forum

A box B₁ contains 1 white ball, 3 red balls and 2 black balls. Another box B₂ contains 2 white balls, 3 red balls and 4 black balls. A third box B₃ contains 3 white balls, 4 red balls and 5 black balls.

If 2 balls are drawn (without replacement)   from a randomly selected box and one of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box B₂ is 

Answer:

Explanation:

 P(Ball drawn from box 2/one is white and one is red)

$=\frac{p(A\cap B)}{p(B)}$

  $=\frac{\frac{1}{3}\times\frac{2\times3}{^{9}C_{2}}}{\frac{1}{3}\left[\frac{1\times3}{^{6}C_{2}}+\frac{2\times3}{^{9}C_{2}}+\frac{3\times4}{^{12}C_{2}}\right]}$

 =   $\frac{\frac{1}{6}}{\frac{1}{5}+\frac{1}{6}+\frac{2}{11}}=\frac{\frac{1}{6}}{\frac{181}{55\times60}}=\frac{55}{181}$