Question

A bag contains 4 balls.Two balls drawn at random without replacement and are found to be white. What is the probability that all balls are white?

Answer 1

Let A be the event of drawing 2 white balls.

from the bag containing 4 balls.

The remaining 2 balls of the bag has three options.

Let $E_1$ be the event that the remaining 2 balls of the bag are not white

Let $E_2$ be the event that are remaining 2 balls of the bag are one white and one not white.

Let $E_3$ be the event that rae remaining 2 balls of the bag are white.

$\therefore P\left(E_1\right)=P\left(E_2\right)=P\left(E_3\right)=\frac{1}{3}$

$P\left(A/E_1\right)=P$ (drawing 2 white balls from the bag contains 2 white and 2 not white)

$=\frac{{}^2{C}_2}{{}^4{C}_2}=\frac{1}{6}$

$P\left(A/E_2\right)=P$ (drawing 2 white balls from the bag containing 2 white and 1 non-white)

${=}^3.C_2{/}^4.C_2=\frac{3}{6}=\frac{1}{2}$

$P\left(A/E_3\right)=P$ (drawing 2 white ball from bag containing 4 whire balls)

= 1.

$P\left(E_2/A\right)=\frac{P\left(E_2\right)\left(PA/E_2\right)}{P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)+P\left(E_3\right)P\left(A/E_3\right)}$

$=\frac{1}{3}\times \frac{1}{6}+\frac{1}{3}\times \frac{1}{2}+\frac{1}{3}.\frac{1}{1}$

$=\frac{1}{3}(\frac{1}{6}+\frac{1}{2}+1)$

$=\frac{1}{3}$

$P\left(E_2/A\right)=\frac{\frac{1}{3}.\frac{1}{2}}{\frac{1}{3}.\frac{5}{3}}=\frac{\frac{1}{6}}{\frac{5}{9}}$

$=\frac{1}{6}\times \frac{3}{5}=\frac{3}{10}$