Question

Box I contains two white and three black balls. Box II contains four white and one black balls and box III contains three white and four black balls. A dice having three red, two yellow and one green face, is thrown to select the box. If red face turns up, we pick up box I, if a yellow face turns up we pick up box II, otherwise, we pick up box III. Then, we draw a ball from the selected box. If the ball drawn is white, what is the probability that the dice had turned up with a red face?

Answer 1

Let $E_1,E_2,E_3$ and $A$ be the events defined as follows :
$E_1$ : red face turns up
$E_2$ : yellow face turns up
$E_3$ : green face turns up
$A$ : white ball is drawn
We wish to calculate the probability that dice had turn up with red face when A has occured (white ball is selected)
i.e., $\left(E_1|A\right)$

By Bayes' Theorem,
$P\left(E_1|A\right)=\frac{P\left(E_1\right)\times P\left(A|E_1\right)}{P\left(E_1\right)\times P\left(A|E_1\right)+P\left(E_2\right)\times P\left(A|E_2\right)+P\left(E_3\right)\times P\left(A|E_3\right)}$

$=\frac{\frac{1}{2}\times \frac{2}{5}}{\frac{1}{2}\times \frac{2}{5}+\frac{1}{3}\times \frac{4}{5}+\frac{1}{6}\times \frac{3}{7}}$

$=\frac{\frac{1}{5}}{\frac{1}{5}+\frac{4}{15}+\frac{1}{14}}$$=\frac{\frac{1}{5}}{\frac{113}{210}}=\frac{1}{5}\times \frac{210}{113}=\frac{42}{113}$.