Question

An urn contains 7 red and 4 blue balls. Two balls are drawn at random with replacement. Find the probability of getting
(i) 2 red balls
(ii) 2 blue balls
(iii) One red and one blue ball.

Answer 1

$$\begin{gathered} \text{It is given that the urn contains 7 red and 4 black balls.} \\ \left(\mathrm{i}\right)P\left(\text{2 red balls}\right)=\frac{7}{11}\times \frac{7}{11} \\ =\frac{49}{121} \\ \left(\mathrm{ii}\right)P\left(\text{2 blue balls}\right)=\frac{4}{11}\times \frac{4}{11} \\ =\frac{16}{121} \\ \left(\mathrm{iii}\right)P\left(\text{one red ball and one blue ball}\right)=P\left(\text{blue ball followed by red ball}\right)+P\left(\text{red ball followed by blue ball}\right) \\ =\frac{4}{11}\times \frac{7}{11}+\frac{7}{11}\times \frac{4}{11} \\ =\frac{28}{121}+\frac{28}{121} \\ =\frac{56}{121} \end{gathered}$$