Question

An urn contains 7 red and 4 blue balls. Two balls are drawn at random with replacement. Find the probability of getting (a) 2 red balls (b) 2 blue balls (C) one red and one blue ball.

Answer 1

Number of red balls = 7
Number of blue balls = 4
Therefore total number of balls = 7 + 4 = 11
Let p be the probability getting a red ball and q be the probability getting a blue ball
$$\begin{gathered} \mathrm{then},p=\frac{7}{11}\mathrm{and}q=\frac{4}{11} \\ \mathrm{Number}\mathrm{of}\mathrm{balls}\mathrm{drawn}=2 \\ \mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{.}\mathrm{P}(\mathrm{getting}2\mathrm{red}\mathrm{balls}\mathrm{with}\mathrm{replacement})=p.p=\frac{7}{11}\times \frac{7}{11}=\frac{49}{121} \\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{.}\mathrm{P}(\mathrm{getting}2\mathrm{blue}\mathrm{balls}\mathrm{with}\mathrm{replacement})=q.q=\frac{4}{11}\times \frac{4}{11}=\frac{16}{121} \\ \mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{.}\mathrm{P}(\mathrm{getting}1\mathrm{red}\mathrm{ball}\mathrm{and}1\mathrm{blue}\mathrm{ball}\mathrm{withh}\mathrm{replacement})=p.q+q.p=\frac{7}{11}\times \frac{4}{11}+\frac{4}{11}\times \frac{7}{11} \\ =\frac{28}{121}+\frac{28}{121}=\frac{56}{121} \end{gathered}$$