Four identical urns each contain 3 balls.
In urn one, all three balls are black; urn two, 2 black 1 white; urn three, 1 black and 2 are white; urn four, all balls are whites.
One of the urn is picked at random, and a ball is chosen from the urn, which turns out to be white. What is the probability that all 3 balls in the selected urn are white?
This might be a silly question, but I am very confused. I know that there are 4 urns in total and only one urn contains three white balls. The part which I don't understand is if the probability of getting the urn with all white balls is 1/4 and it asks that all 3 balls in urn are white? So $1/4 * 1$?
Please explain, how you derived your answer from the question.
Best Answer
Let $U_i$ be the $i$th urn just like you described, and let $A =\{\text{Picked a white ball}\},B_i = \{\text{Pick $U_i$}\}.$ Then the event "all 3 balls in the selected urn are white" is the same as event $B_4$. Thus, if we use conditioning, the probability we are interested in is \begin{align*} P(B_4|A)&=\frac{P(B_4A)}{P(A)}\\ &=\frac{P(A|B_4)P(B_4)}{P(A)}\\ &=\frac{P(A|B_4)P(B_4)}{P(A|B_1)P(B_1)+P(A|B_2)P(B_2)+P(A|B_3)P(B_3)+P(A|B_4)P(B_4)}\\ &=\frac{1\cdot \frac{1}{4}}{0(1/4)+(1/3)(1/4)+(2/3)(1/4)+1(1/4)}\\ &=\frac{1}{2}. \end{align*}