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Example There are red balls and white balls in the three bags as shown below. Bag 1: 4 red balls 1 white ball, Bag 2: 3 red balls 3 white balls, Bag 3: 2 red balls 4 white balls When the balls were taken out from one of the bags, they were white balls. Find the probability that this ball was taken out of bag 2

Answer A = [The ball taken out is white],$E_1$ = [Event to take out from bag 1] $E_2$ = [Event to take out from bag 2] $E_3$ = [Event to take out from bag 3]

We find $P(E_2\vert A)$．Note that $A = A \\ cap \\ Omega = A \\ cap (E_1 \\ cup E_2 \\ cup E_3)$

$P(E_2\vert A)$ is P(bag 2 $\vert$ ball is white)．Note that $P(E_2\vert A) = \frac{P(A \cap E_2)}{P(A)}$. Thus, we can write $P(A \cap E_2) = P(A \vert E_2)P(E_2)$. In other words, P(ball is white $\vert$ bag 2)P(bag 2) Here, P(ball is white $\vert$ bag 2) $= \frac{3}{6}$, P(bag 2) $= \frac{1}{3}$. Also, $P(A) = P(A \cap \Omega ) = P(A \cap E_1) + P(A \cap E_2) + P(A \cap E_3)$implies that

$\displaystyle P(A) = \frac{1}{5}\times \frac{1}{3} + \frac{3}{6}\times \frac{1}{3} + \frac{4}{6}\times \frac{1}{3} = \frac{41}{90}$

Therefore,

$\displaystyle P(E_2\vert A) = \frac{1/6}{41/90} = \frac{15}{41}$