Mathematics - Probability Question with Solution | TestHub

Given apples are: $3$ bad, $15$ good

Let $X$ be no. of bad apples.

$\Rightarrow P\left(X=0\right)=\frac{{}^{15}{C}_2}{{}^{18}{C}_2}=\frac{{\displaystyle \frac{15!}{2!\times 13!}}}{{\displaystyle \frac{18!}{2!\times 16!}}}$

$\Rightarrow P\left(X=0\right)=\frac{15\times 14}{18\times 17}=\frac{5\times 7}{3\times 17}$

$\Rightarrow P\left(X=0\right)=\frac{35}{51}$

$\Rightarrow P\left(X=1\right)=\frac{{}^3{C}_1{\times }^{15}{C}_1}{{}^{18}{C}_2}=\frac{3\times 15}{9\times 17}$

$\Rightarrow P\left(X=1\right)=\frac{5}{17}$

$\Rightarrow P\left(X=2\right)=\frac{{}^3{C}_2}{{}^{18}{C}_2}=\frac{3}{9\times 17}$

$\Rightarrow P\left(X=2\right)=\frac{3}{153}$

$\Rightarrow E\left(X\right)=0\times \frac{35}{51}+1\times \frac{5}{17}+2\times \frac{3}{153}$

$\Rightarrow E\left(X\right)=\frac{1}{3}$

We know that, $\mathrm{Var}\left(X\right)=E\left(X^2\right)-{\left(E\left(X\right)\right)}^2$

$\Rightarrow \mathrm{Var}\left(X\right)=0\times \frac{35}{51}+1\times \frac{5}{17}+4\times \frac{1}{51}-{\left(\frac{1}{3}\right)}^2$

$\Rightarrow \mathrm{Var}\left(X\right)=\frac{19}{51}-\frac{1}{9}$

$\Rightarrow \mathrm{Var}\left(X\right)=\frac{40}{153}$