Question

Bag A contains 3 red and 5 black balls. While bag B contains 4 red and 4 black balls. Two balls are transferred at random from hag A to bag B and then a ball is drawn from bag B at random. If the ball drawn from bag B is found to be red. Find the probability that two red balls were transferred from A to B.

Answer 1

The two balls drawn from the bag A could be

$1$ both red, Probability $=\frac{{}^3{C}_2}{{}^8{C}_2}$

$2$. One red and one black. Probability $=\frac{3\times 5}{{}^8{C}_2}$

$3$. both black. Probability$=\frac{{}^5{C}_2}{{}^8{C}_2}$

The number of balls in bag B in each case would be

$1$. $6$ red and $4$ black. Probability of picking red $=\frac{6}{10}$

$2$. $5$ red and $5$ black. Probability of picking red$=\frac{5}{10}$

$3$. $4$ red and $6$ black. Probability of picking red$=\frac{4}{10}$

Probability of two red ball transferred under the condition that red ball is found in bag B is Using Baye's theorem.

$\frac{\frac{{}^3{C}_2}{{}^8{C}_2}\times \frac{6}{10}}{\frac{{}^3{C}_2}{{}^8{C}_2}\times \frac{6}{10}+\frac{3\times 5}{{}^8{C}_2}\times \frac{5}{10}+\frac{{}^3{C}_2}{{}^8{C}_2}\times \frac{4}{10}}$

$=\frac{18}{133}$.