Three bags contain a number of red and white balls as follows Bag I : 3 red balls, Bag II : 2 red balls and 1 white ball and Bag III : 3 white balls. The probability that bag i will be chosen and a ball is selected from it is $\frac{i}{6}$, where $i = 1,2,3.$ What is the probability that

(i) a red ball will be selected?

(ii) a white ball is selected?

Bag I : 3 red balls and 0 white ball.
Bag II : 2 red balls and 1 white ball.
Beg III : 0 red ball and 3 white balls.

Let ${E_1},{E_2}$ and ${E_3}$ be the events that bag 1 ,

bag II and bag III is selected and a ball is chosen from it.

$P\left( {{E_1}} \right) = \frac{1}{6},P\left( {{E_2}} \right) = \frac{2}{6}$

and $P\left( {{E_3}} \right) = \frac{3}{6}$

(i) Let $E$ be the event that a red ball is selected. Then, probability that red ball will be selected

$P(E) = P\left( {{E_1}} \right) \cdot P\left( {E/{E_1}} \right) + P\left( {{E_2}} \right) \cdot P\left( {E/{E_2}} \right) + P\left( {{E_3}} \right) \cdot P\left( {E/{E_3}} \right)$

$= \frac{1}{6} \cdot \frac{3}{3} + \frac{2}{6} \cdot \frac{2}{3} + \frac{3}{6} \cdot 0$

$= \frac{1}{6} + \frac{2}{9} + 0$

$= \frac{{3 + 4}}{{18}} = \frac{7}{{18}}$

(ii) Let $F$ be the event that a white ball is selected.

$\therefore$ $P(F) = P\left( {{E_1}} \right) \cdot P\left( {F/{E_1}} \right) + P\left( {{E_2}} \right) \cdot P\left( {F/{E_2}} \right) + P\left( {{E_3}} \right) \cdot P\left( {F/{E_3}} \right)$

$= \frac{1}{6} \cdot 0 + \frac{2}{6} \cdot \frac{1}{3} + \frac{3}{6} \cdot 1 = \frac{1}{9} + \frac{3}{6} = \frac{{11}}{{18}}$

Note $P(F) = 1 - P(E) = 1 - \frac{7}{{18}} = \frac{{11}}{{18}}$

[since, we know that $P(E) + P(F) = 1$]