How do I get the probability? (Balls picked from 2 urns)

Question

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.

Answer 1

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 $m$.

Answer 2

You have defined three cases that are exhaustive and mutually exclusive. Figure out the probability of obtaining a white ball from the second urn using the law of total probability and these three cases as conditionals. To this end, say $W_2$ and $B_2$ are the events that you draw a white ball and a black ball from the second urn, respectively. Also, let $W_{1}$, $B_{1}$, and $BW_1$ be the events that you draw both white balls, both black balls, and a white and a black ball from the first urn, respectively. You have $$P(W_2)=P(W_2\,|\,W_1)P(W_1)+P(W_2 \, | \, B_1)P(B_1)+P(W_2 \,|\,BW_1)P(BW_1)\\=\frac{5}{10}\left(\frac{10}{13} \cdot\frac{9}{12}\right) + \frac{3}{10}\left(\frac{3}{13} \cdot\frac{2}{12}\right)+\frac{4}{10}\left(\frac{10}{13} \cdot\frac{3}{12} + \frac{3}{13} \cdot \frac{10}{12}\right)\, \\=\frac{59}{130},$$ where $P(BW_1)$ was calculated as a sum of two probabilities since you can choose the white ball first or the black ball first from the first urn (and then, the other colored ball in the second draw). In a similar way, you could also compute the probability that the ball that you draw from the second urn is black. Here, you would have $$P(B_2)=P(B_2\,|\,W_1)P(W_1)+P(B_2 \, | \, B_1)P(B_1)+P(B_2 \,|\,BW_1)P(BW_1)\\=1-P(W_2)=\frac{71}{130}\,,$$ where the last equality comes from the fact that $W_2$ is the complement of $B_2$.

Answer 3

Please go through the diagram to understand clearly. According to the problem, 3 cases appear => WW,BB,RB. Ultimately, the required Probabilistic Equation will be P(W) = $\frac{\binom{10}{2}}{\binom{13}{2}}\ast \frac{5}{10} + \frac{\binom{3}{2}}{\binom{13}{2}}\ast \frac{3}{10} + \frac{\binom{10}{1}\binom{3}{1}}{\binom{13}{2}}\ast \frac{4}{10}$