Question

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that (i) both the balls are red. (ii) first ball is black and second is red. (iii) one of them is black and other is red.

Answer 1

$$\begin{gathered} \mathrm{Given}:\hspace{0.17em}\mathrm{Box}=\left(10B+8R\right)balls \\ \left(\mathrm{i}\right)P\left(\text{both red balls}\right)=\frac{8}{18}\times \frac{8}{18} \\ =\frac{64}{324}=\frac{16}{81} \\ \left(\mathrm{ii}\right)P\left(\text{first black and second red}\right) \\ =\frac{8}{18}\times \frac{10}{18} \\ =\frac{80}{324}=\frac{20}{81} \\ \left(\mathrm{iii}\right)P\left(\text{one is red and one is black}\right)=P\left(\text{first red and second black}\right)+P\left(\text{first red and}\mathrm{second}\text{black}\right) \\ =\frac{8}{\text{'}18}\times \frac{10}{18}+\frac{10}{18}\times \frac{8}{18} \\ =\frac{80}{324}+\frac{80}{324} \\ =\frac{40}{81} \end{gathered}$$