An urn contains 10 white balls and 3 black balls. Another urn contains
3 white balls and 5 black balls. Two balls are drawn at random from
the first urn and placed in the second urn. A ball is then taken from
the second urn. What is the probability that it is a white ball?
Here is my approach:
1. Find probability of getting two whites.
2. Find probability of getting two blacks.
3. Find probability of getting a white and a black.
Depending on what happens above, the number of white and black balls in urn 2 will change.
I get that I have to apply Baye's theorem but cannot wrap my head around it. Help will be appreciated. Simple words please.
Best Answer
There are $3$ possibilities for the balls we drew from the first urn: (i) both white; (ii) both black; (iii) mixed.
Let $w$ be the probability of both white, $b$ the probability of both black, and $m$ the probability of mixed.
(i) If both balls drawn from the first urn are white, the second urn will have $5$ white balls and $5$ black, and the probability of drawing a white ball is $\frac{5}{10}$.
(ii) If both drawn balls are black, then the probability of drawing a white from the second urn is $\frac{3}{10}$.
(iii) And if they are mixed, the probability of drawing a white from the second urn is $\frac{4}{10}$.
Our desired probability is $w\cdot\frac{5}{10}+b\cdot\frac{3}{10}+m\cdot\frac{4}{10}$.
I assume you know how to calculate $w$, $b$, and $m$.