Probability

Three urns are given each containing red and white chips as given below:

Urn $I$ : $6$ red $4$ white ; Urn $II$ : $3$ red $5$ white ; Urn $III$ : $4$ red $6$ white

An urn is chosen at random and a chip is drawn from the urn.

1. Find the probability that it is white.
2. If the chip is white find the probability that it is from urn $II$.

Let $A_1$ be the event of selecting urn $I$.

Let $A_2$ be the event of selecting urn $II$.

Let $A_3$ be the event of selecting urn $III$.

Let $B$ be the event of selecting a white chip.

1. To find: $\;$ P(of selecting a white chip) $= P \left(B\right)$
   
   Events $A_1$, $A_2$ and $A_3$ are mutually exclusive and exhaustive events.
   
   $\therefore \;$ $P \left(B\right) = P \left(A_1\right) \times P \left(B | A_1\right) + P \left(A_2\right) \times P \left(B | A_2\right) + P \left(A_3\right) \times P \left(B | A_3\right)$ $\;\;\; \cdots \; (1)$
   
   Probability of selecting urn $I$ $= P \left(A_1\right) = \dfrac{1}{3}$
   
   Probability of selecting urn $II$ $= P \left(A_2\right) = \dfrac{1}{3}$
   
   Probability of selecting urn $III$ $= P \left(A_3\right) = \dfrac{1}{3}$
   
   P(selecting a white chip when urn $I$ is selected) $= P \left(B | A_1\right)$
   
   $P \left(B | A_1\right) = \dfrac{{^{4}}{C}_{1}}{{^{10}}{C}_{1}} = \dfrac{4}{10} = \dfrac{2}{5}$
   
   P(selecting a white chip when urn $II$ is selected) $= P \left(B | A_2\right)$
   
   $P \left(B | A_2\right) = \dfrac{{^{5}}{C}_{1}}{{^{8}}{C}_{1}} = \dfrac{5}{8}$
   
   P(selecting a white chip when urn $III$ is selected) $= P \left(B | A_3\right)$
   
   $P \left(B | A_3\right) = \dfrac{{^{6}}{C}_{1}}{{^{10}}{C}_{1}} = \dfrac{6}{10} = \dfrac{3}{5}$
   
   Substituting the values of $P \left(A_1\right)$, $P \left(A_2\right)$, $P \left(A_3\right)$, $P \left(B | A_1\right)$, $P \left(B | A_2\right)$ and $P \left(B | A_3\right)$ in equation $(1)$ we have,
   
   $P \left(B\right) = \left(\dfrac{1}{3} \times \dfrac{2}{5}\right) + \left(\dfrac{1}{3} \times \dfrac{5}{8}\right) + \left(\dfrac{1}{3} \times \dfrac{3}{5}\right) = \dfrac{13}{24}$
   
   
2. To find: $\;$ P(of selecting urn $II$ when a white chip is selected) $= P \left(A_2 | B\right)$
   
   Since events $A_1$, $A_2$ and $A_3$ are mutually exclusive and exhaustive events, therefore we have by Bayes' theorem,
   
   $\begin{aligned} P \left(A_2 | B\right) & = \dfrac{P \left(A_2\right) \times P \left(B | A_2\right)}{P \left(A_1\right) \times P \left(B | A_1\right) + P \left(A_2\right) \times P \left(B | A_2\right) + P \left(A_3\right) \times P \left(B | A_3\right)} \\\\ & = \dfrac{\dfrac{1}{3} \times \dfrac{5}{8}}{\dfrac{1}{3} \times \dfrac{2}{5} + \dfrac{1}{3} \times \dfrac{5}{8} + \dfrac{1}{3} \times \dfrac{3}{5}} = \dfrac{5/8}{13/8} = \dfrac{5}{13} \end{aligned}$