# Conditional Probability (Baye's Rule) question with urns

There are two urns A and B. Urn A contains 1 red ball and 2 white balls, whereas urn B contains 2 red balls and 1 white ball. Calculate the conditional probability that a randomly chosen ball belonged to urn A given that it is white.

I know, that the answer is 0.6666... But I can't figure the way how to apply the Baye's rule to the conditions.

• Have you tried anything? Do you have any thoughts on this problem? Dec 10 '17 at 20:09
• Are we to take that Urn A and Urn B may be chosen with equal probability, and one random ball extracted ? Dec 10 '17 at 20:12

Let A be the event that urn A chosen, B be the event that Urn B chosen,
and W be the event that a white ball is chosen.

Using Bayes' Rule and the law of total probability,

P(From Urn A | it is white) $= \dfrac{P(A)\cdot P(W|A)}{P(A)\cdot P(W|A) + P(B)\cdot(P(W|B)} =\dfrac{\dfrac12 \cdot\dfrac23}{\dfrac12\cdot\dfrac23+\dfrac12\cdot\dfrac13}=\dfrac23$

$$P(A|W) = \frac{P(A \cap W)}{P(W)}$$ Assuming either urn is equally probable ..

$P(W) = \frac 12$ ( 3 out of 6 balls are white )

$P(A \cap W)=\frac12 \times \frac 23$