One urn contains two black balls (labelled B_1 and B_2) and one white ball. A second urn contains one black ball and two white balls (labelled W_1 and W_2). Suppose the following experiment is performed. One of the two urns is chosen at random. Next a ball is randomly chosen from the urn. Then a second ball is chosen at random from the same urn without replacing the first ball.
(a) What is the probability that two balls of opposite colour are chosen?
One urn contains two black balls (labelled B_1 and B_2) and one white ball. A second urn contains one black ball and two white balls (labelled W_1 and W_2). Suppose the following experiment is performed. One of the two urns is chosen at random. Next a ball is randomly chosen from the urn. Then a second ball is chosen at random from the same urn without replacing the first ball.
(a) What is the probability that two balls of opposite colour are chosen?

Solution:
(a) If two balls of opposite colours are chosen then
The favourable outcomes are \mathrm{B}_{1} \mathrm{~W}, \mathrm{~B}_{2} \mathrm{~W}, \mathrm{WB}_{1}, \mathrm{WB}_{2}, \mathrm{~W}_{1} \mathrm{~B}, \mathrm{~W}_{2} \mathrm{~B}, \mathrm{BW}_{1}, \mathrm{BW}_{2}
\therefore The total no. favourable outcomes =8 and total no. outcomes =12
It is known that,
\text { Probability }=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}
\therefore The required Probability =\frac{8}{12}=\frac{2}{3}