Four bad apples are mixed accidentally with 20 good apples. If 3 apples are drawn one by one with replacement then mean of number of good apples is 
 

Let X denote the no. of good apples. 

P (drawing good apple) $=\frac{20}{24}=\frac{5}{6}$

P (drawing bad apple) $=\frac{4}{24}=\frac{1}{6}$

Here X takes the values 0, 1, 2, 3.

Now $\mathrm{P}(\mathrm{X}=0)={}^3{\mathrm{C}}_0\left(\frac{1}{6}\right) \left(\frac{1}{6}\right) \left(\frac{1}{6}\right)={}^3{\mathrm{C}}_0{\left(\frac{1}{6}\right)}^3=\frac{1}{216}$

         $$\begin{gathered} \mathrm{P}(\mathrm{X}=1)={}^3{\mathrm{C}}_1\left(\frac{5}{6}\right) \left(\frac{1}{6}\right) \left(\frac{1}{6}\right) ={}^3{\mathrm{C}}_1 \left(\frac{5}{6^3}\right)=\frac{15}{216} \\ \mathrm{P}(\mathrm{X}=2)={}^3{\mathrm{C}}_2 \left(\frac{5}{6}\right) \left(\frac{5}{6}\right) \left(\frac{1}{6}\right) ={}^3{\mathrm{C}}_2 \left(\frac{5^2}{6^3}\right)=\frac{75}{216} \\ \mathrm{P}(\mathrm{X}=3)={}^3{\mathrm{C}}_3{\left(\frac{5}{6}\right)}^3=\frac{125}{216} \end{gathered}$$

Mean $=\mathrm{E}\left(\mathrm{x}\right)=\sum _{ }\mathrm{xP}\left(\mathrm{x}\right)$

                        $$\begin{gathered} =0+1\left(\frac{15}{216}\right)+2\left(\frac{75}{216}\right)+3\left(\frac{125}{216}\right) \\ =\frac{540}{216}=\frac{5}{2} \end{gathered}$$