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?

Home » One urn contains two black balls (labelled and ) and one white ball. A second urn contains one black ball and two white balls (labelled and ). 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?

CBSE | Class 11 | Maths | NCERT Exemplar | Probability

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 $=\frac{8}{12}=\frac{2}{3}$