A box contains 4 red and 8 white balls. If 4 balls are drawn one by one with replacement then the probability of  getting 3 white balls is $\_\_\_\_\_$

Given no. of red balls    =4

           no. of white balls=8

P (getting white ball) $=\frac{{}^8{\mathrm{c}}_1}{{}^{12}{\mathrm{c}}_1}=\frac{8}{12}=\frac{2}{3}$

P (getting red ball)     $=\frac{{}^4{\mathrm{c}}_1}{{}^{12}{\mathrm{c}}_1}=\frac{4}{12}=\frac{1}{3}$

There are 4 Possibilities

$\begin{array}{c} \\ \mathrm{I}\\ \mathrm{II}\\ \mathrm{III}\\ \mathrm{IV}\end{array} \begin{array}{c}1\\ \mathrm{W}\\ \mathrm{R}\\ \mathrm{W}\\ \mathrm{W}\end{array} \begin{array}{c}2\\ \mathrm{W}\\ \mathrm{W}\\ \mathrm{R}\\ \mathrm{W}\end{array} \begin{array}{c}3\\ \mathrm{W}\\ \mathrm{W}\\ \mathrm{W}\\ \mathrm{R}\end{array} \begin{array}{c}4\\ \mathrm{R}\\ \mathrm{W}\\ \mathrm{W}\\ \mathrm{W}\end{array}$

$\therefore$ Required Probability 

                  $$\begin{gathered} =\mathrm{P}\left(\mathrm{I}\right)+\mathrm{P}\left(\mathrm{II}\right)+\mathrm{P}\left(\mathrm{III}\right)+\mathrm{P}\left(\mathrm{IV}\right) \\ =(\frac{2}{3}\times \frac{2}{3}\times \frac{2}{3}\times \frac{1}{3})+(\frac{1}{3}\times \frac{2}{3}\times \frac{2}{3}\times \frac{2}{3})+(\frac{2}{3}\times \frac{1}{3}\times \frac{2}{3}\times \frac{2}{3})+(\frac{2}{3}\times \frac{2}{3}\times \frac{1}{3}\times \frac{2}{3}) \\ =4\left(\frac{2}{3}\times \frac{2}{3}\times \frac{2}{3}\times \frac{1}{3}\right) \\ =\frac{32}{81} \\ =0.39 \end{gathered}$$