Question

An urn contains $10$ white and $3$ black balls while another urn contains $3$ white and $5$ black balls. Two balls are drawn from the first urn and put into the second urn then a ball is drawn from the latter. Find the probability that it is a white ball.

Answer 1

There are three mutually exclusive and exhaustive ways in which $2$ balls can be transferred from first bag to second bag.
First way: Two white balls are transferred from first bag to second bag, probability of which is $\frac{{{}^{10}C}_2}{{{}^{13}C}_2}$
In the second bag we now have $5$ white and $5$ black balls and probability of getting a white ball is $\frac{5}{10}$,
$\therefore$ Required probability$=\frac{{{}^{10}C}_2}{{{}^{13}C}_2}\times \frac{5}{10}=\frac{45}{78}\times \frac{5}{10}=\frac{225}{780}$
Second way: two black balls are transferred from first bag to the second bag
probability of which is $\frac{{{}^{3}C}_2}{{{}^{13}C}_2}$
Now we have $3$ white and $7$ black balls in second bag and probability of getting a white ball is $\frac{3}{10}$
$\therefore$ Required probability$\frac{{{}^{3}C}_2}{{{}^{13}C}_2}\times \frac{3}{10}=\frac{3}{78}\times \frac{3}{10}=\frac{9}{780}$
Third way: One black and one white ball are transferred from first bag to second bag, the probability of which is $\frac{{{}^{10}C}_1\times {{}^{3}C}_1}{{{}^{13}C}_2}$
In second bag, there are now $4$ white and $6$ black balls and the probability of drawing a white ball is $\frac{4}{10}$
$\therefore$ Required probability$\frac{{{}^{10}C}_1\times {{}^{3}C}_1}{{{}^{13}C}_2}\times \frac{4}{10}=\frac{30}{78}\times \frac{4}{10}=\frac{120}{780}$

$\therefore$ Total Probability $=\frac{225+9+120}{780}=\frac{354}{780}$