Question-31
There are three boxes containing white and black balls.
- Box-1 contains 2 black and 1 white balls.
- Box-2 contains 1 black and 2 white balls.
- Box-3 contains 3 black and 3 white balls.
In a random experiment, one of these three boxes is selected, where the probability of choosing Box-1 is \(\cfrac{1}{2}\), Box-2 is \(\cfrac{1}{6}\), and Box-3 is \(\cfrac{1}{3}\). A ball is drawn at random from the selected box. Given that the ball drawn is white, the probability that it is drawn from Box-2 is ____ (Round off to two decimal places)
\(0.25\)
Classic Bayes’ theorem question. Let \(B_{1} ,B_{2} ,B_{3}\) be the events that boxes 1, 2 and 3 are chosen respectively. Let \(W\) be the event that a white ball is drawn. We need to find \(P( B_{2} \ |\ W)\). We have:
\[ \begin{aligned} P( B_{2} \ |\ W) & =\cfrac{P( B_{2} \cap W)}{P( W)}\\ & \\ & =\cfrac{P( B_{2}) \cdot P( W\ |\ B_{2})}{P( B_{1}) \cdot P( W\ |\ B_{1}) +P( B_{2}) \cdot P( W\ |\ B_{2}) +P( B_{3}) \cdot P( W\ |\ B_{3})}\\ & \\ & =\cfrac{\cfrac{1}{6} \times \cfrac{2}{3}}{\cfrac{1}{2} \times \cfrac{1}{3} +\cfrac{1}{6} \times \cfrac{2}{3} +\cfrac{1}{3} \times \cfrac{1}{2}}\\ & \\ & =\cfrac{\cfrac{1}{9}}{\cfrac{1}{6} +\cfrac{1}{9} +\cfrac{1}{6}}\\ & \\ & =\cfrac{1}{4} \end{aligned} \]