Question

An urn contains four white and three red balls. Find the probability distribution of the number of red balls in three draws with replacement from the urn.

Answer 1

As three balls are drawn with replacement, the number of white balls, say X, follows binomial distribution with n =3
$$\begin{gathered} p=\frac{3}{7}\text{and}q=\frac{4}{7} \\ P(X=\mathrm{r})={}^{3}C_r{\left(\frac{3}{7}\right)}^r{\left(\frac{4}{7}\right)}^{3-r},\mathit{}\mathit{}r=0,1,2,3 \\ P(X=0)={}^{3}C_0{\left(\frac{3}{7}\right)}^0{\left(\frac{4}{7}\right)}^{3-0} \\ P(X=1)={}^{3}C_1{\left(\frac{3}{7}\right)}^1{\left(\frac{4}{7}\right)}^{3-1} \\ P(X=2)={}^{\mathit{3}}C_{\mathit{2}}{\left(\frac{\mathit{3}}{\mathit{7}}\right)}^2{\left(\frac{4}{7}\right)}^{3-2}\mathit{}\mathit{} \\ P\mathit{(}X=3)\mathit{}\mathit{=}{}^3\mathrm{C}_3{\left(\frac{3}{7}\right)}^{\mathit{3}}{\left(\frac{4}{7}\right)}^{3-3} \\ X\mathit{}0123 \\ P\left(\mathrm{X}\right)\frac{64}{343}\frac{144}{343}\frac{108}{343}\frac{27}{343} \end{gathered}$$