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) Write the sample space showing all possible outcomes
(b) What is the probability that two black balls 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) Write the sample space showing all possible outcomes
(b) What is the probability that two black balls are chosen?

Solution:

It is given that one urn contains two black balls and one white ball
and the second urn contains one black ball and two white balls.

Also given that if one of the two urns is chosen, then a ball is randomly chosen from the urn, then the second ball is chosen at random from the same urn without replacing the first ball.
(a) Sample Space \mathrm{S}=\left\{\mathrm{B}_{1} \mathrm{~B}_{2}, \mathrm{~B}_{1} \mathrm{~W}, \mathrm{~B}_{2} \mathrm{~W}, \mathrm{~B}_{2} \mathrm{~B}_{1}, \mathrm{WB}_{1}, \mathrm{WB}_{2}, \mathrm{~W}_{1} \mathrm{~W}_{2}, \mathrm{~W}_{1} \mathrm{~B}, \mathrm{~W}_{2} \mathrm{~B}, \mathrm{~W}_{2} \mathrm{~W}_{1}, \mathrm{BW}_{1}, \mathrm{BW}_{2}\right\}
The total no. of sample space =12
(b) If the two black balls are chosen
The total outcomes =12
The favourable outcomes are B_{1} B_{2}, B_{2} B_{1}
\therefore The total favourable outcomes =2
It is known that,
\begin{array}{l} \text { Probability }=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }} \\ \therefore \text { Required Probability }=\frac{2}{12}=\frac{1}{6} \end{array}