Question

There are two bags. The first bag contains 4 white and 5 black balls and the second bag contains 5 white and 4 black balls. Two balls are drawn at random from the first bag and are put into the second bag without noticing their colours. Then two balls are drawn from the second bag. Find the probability that the balls are white and black.___

Answer 1

A white and a black ball can be drawn from the second bag in the following mutually exclusive ways:
(i) By transferring 2 white balls from the first bag to the second bag and then drawing a white and a black ball from it.
(ii) By transferring 2 black balls from the first bag to the second bag and then drawing a white and a black ball from it.
(iii) By transferring 1 white and 1 black ball from the first bag to the second bag and then drawing a white and a black ball from it.
Let A, B, C and D be the events as defined below:
A = Two white balls are drawn from the first bag
B = Two black balls are drawn from the first bag.
C = One white and one black ball are drawn from the first bag.
D = Two balls drawn from the second bag are white and black.
We, have $P\left(A\right),\frac{{}^4{C}_2}{{}^9{C}_2}=\frac{6}{36}=\frac{1}{6}P\left(B\right)=\frac{{}^5{C}_2}{{}^9{C}_2}=\frac{10}{36}=\frac{5}{18}P\left(C\right)=\frac{{}^4{C}_1{\times }^5{C}_1}{{}^9{C}_2}=\frac{20}{36}=\frac{5}{9}$
If A has already occurred, i.e., if two white balls have been transferred from the first bag to the second bag, then the second bag will contain 7 white and 4 black balls, therefore the probability of drawing a white and a black from the second bag is
$P\left(\frac{D}{A}\right)=\frac{{}^7{C}_1{\times }^4{C}_1}{{}^{11}{C}_2}=\frac{28}{55}Similarly,P\left(\frac{D}{B}\right)=\frac{{}^5{C}_1{\times }^6{C}_1}{{}^{11}{C}_2}=\frac{30}{55}=\frac{6}{11}andP\left(\frac{D}{C}\right)=\frac{{}^6{C}_1{\times }^5{C}_1}{{}^{11}{C}_2}=\frac{30}{55}=\frac{6}{11}$
$\therefore$ By the law of total probability, we have
$P\left(D\right)=P\left(A\right)P\left(\frac{D}{A}\right)+P\left(B\right)P\left(\frac{D}{B}\right)+P\left(C\right)P\left(\frac{D}{C}\right)=\frac{1}{6}\times \frac{28}{55}+\frac{5}{18}\times \frac{6}{11}+\frac{5}{9}\times \frac{6}{11}=\frac{14}{165}+\frac{5}{33}+\frac{10}{33}=\frac{14+25+50}{165}=\frac{89}{165}$