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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}|A)． Note that A = A cap Omega = A cap(E_{1} cupE_{2} cupE_{3})

P (E_{2}|A) is P(bag 2 | ball is white)． Note that P (E_{2}|A) = \frac{P (A∩E_{2})}{P (A)}. Thus, we can write P (A ∩ E_{2}) = P (A|E_{2})P (E_{2}). In other words, P(ball is white | bag 2)P(bag 2) Here, P(ball is white | bag 2) = \frac{3}{6}, P(bag 2)= \frac{1}{3}. Also, P (A) = P (A ∩ Ω) = P (A ∩ E_{1}) + P (A ∩ E_{2}) + P (A ∩ E_{3})implies that

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

Therefore,

P (E_{2}|A) = \frac{1/6}{41/90} = \frac{15}{41}