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Multiplication Rule

A bag contain...

Question

A bag contains $4$ balls. Two balls are drawn at random and are found to be white. what is the probability that all the balls in the bag are white?

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Solution

$E_{1} : 2$ balls are white, $2 -$ are non white
$E_{2} : 3$ balls are white, $1$ is not white
$E_{3} : 4$ balls are white
$A :$ Two balls are white
$P (E_{1}) = \frac{1}{3}, P (E_{2}) = \frac{1}{3}, P (E_{3}) = \frac{1}{3}$
$P (A | E_{1}) = \frac{2 C_{2}}{4 C_{2}} = \frac{1}{6}$
$P (A | E_{2}) = \frac{3 C_{2}}{4 C_{2}} = \frac{3}{6} = \frac{1}{2}, P (A | E_{3}) = \frac{4 C_{2}}{4 C_{2}} = 1$
$∴ P (E_{3} | A) = \frac{P (A | E_{3}) P (E_{3})}{P (A | E_{1}) P (E_{1}) + P (A | E_{2}) P (E_{2}) + P (A | E_{3}) P (E_{3})}$
$= \frac{1 \times \frac{1}{3}}{\frac{1}{6} \times \frac{1}{3} + \frac{1}{2} \times \frac{1}{3} + \frac{1}{3} \times 1} = \frac{1}{\frac{1}{6} + \frac{1}{2} + 1} = \frac{6}{10} = \frac{3}{5}$

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