Answer to Question #254061 in Statistics and Probability for Olim Vicis

Question

Two Urns - one has 4 white balls and one has 3 white & 1 black ball. An urn is chose at random and ball is taken out. This ball is white. What is the probability that the urn was the one that contained the 4 white balls (Show working giving reasons) Another ball is taken out. It is white. What now is the probability that the urn was the one that had the 4 white balls? A third ball iis taken out It is white. Again, what is the probability that the urn was the one that had 4 white balls in it?

Answer 1

To solve all of those problems it is appropriate to use conditional probability formula

$P(A|B) = {\frac {P(B|A)*P(A)} {P(B)}}$

Let urn1 - urn with 4 white balls

in our case: A - ''it was taken from urn 1", B - "the white ball was taken"

1 ball taken: $P(A|B) = {\frac {P(B|A)*P(A)} {P(B)}}= {\frac {1*0.5} {1*0.5+0.75*0.5}} = 0.57$

1 urn now contains 3 white balls

2 ball taken: $P(A|B) = {\frac {P(B|A)*P(A)} {P(B)}}= {\frac {1*0.5} {1*0.5+0.75*0.5}} = 0.57$

1 urn now contains 2 white balls

3 ball taken: $P(A|B) = {\frac {P(B|A)*P(A)} {P(B)}}= {\frac {1*0.5} {1*0.5+0.75*0.5}} = 0.57$

We received 3 equivalent probability values for all cases. It was expected, cause we didn't change the probability of chosen any urn, didn't even once took the ball from the urn 2, and in the urn 1 were 4 white balls, so, if we pick a ball from urn 1 it will definately be white as long as there are any balls in the urn 1