Question

There are three urns containing 2 white and 3 black balls, 3 white and 2 black balls, and 4 white and 1 black balls, respectively. There is an equal probability of each urn being chosen. A ball is drawn at random from the chosen urn and it is found to be white. Find the probability that the ball drawn was from the second urn.

Answer 1

We have 3 urns:

∴ Probabilities of choosing either of the urns are 

P(U₁) = P(U₂) = P(U₃) = `1/3`

Let H be the event of drawing white ball from the chosen urn.

∴ `"P"("H"/"U"_1) = 2/5`

`"P"("H"/"U"_2) = 3/5`

And `"P"("H"/"U"_3) = 4/5`

∴ `"P"("U"_2/"H") = ("P"("U"_2)*"P"("H"/"U"_2))/("P"("U"_1)*"P"("H"/"U"_1) + "P"("U"_2)*"P"("H"/"U"_2) + "P"("U"_3)*"P"("H"/"U"_3))`

= `(1/3*3/5)/(1/3*2/5 + 1/3*3/5 + 1/3*4/5)`

= `(3/5)/(2/5 + 3/5 + 4/5)`

= `3/9`

= `1/3`

Hence, the required probability is `1/3`.