Q.316.Probability - Application of Bayes' Theorem

Question 316.
Three urns contain colored balls at follows:
Urn One: 3 red, 4 white, 1 blue
Urn Two: 1 red, 2 white, 3 blue
Urn Three: 4 red, 3 white, 2 blue
One urn is chosen at random and a ball is withdrawn. It turns out to be red. What is the probability that is came from Urn Two?

Answer 316.
Let Ui = the event that urn i is selected
and R = the event that the red ball is selected from the selected urn

As the urns are selected randomly, P(U1) = P(U2) = P(U3) = 1/3

=> P(R/U1) = No. of red balls in urn 1 / total no. of balls in urn 1
                           = 3/8
Similarly, P(R/U2) = 1/6 and P(R/U3) = 4/9

By Bayes' Rule,
P(U2/R)
= P(U2) * P(R/U2) / [P(U1) * P(R/U1) + P(U2) * P(R/U2) + P(U3) * P(R/U3)]
= (1/3) * (1/6) / [(1/3) * (3/8) + (1/3) * (1/6) + (1/3) * (4/9)]
= (1/6) / (3/8 + 1/6 + 4/9)
= (1/6) * (72/71)
= 12/71.

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