Four bad oranges are accidently mixed with sixteen good oranges. Find…

The correct answer is Option (1) → Mean = $\frac{2}{5}$, Variance = $\frac{144}{475}$, Standard Deviation = $\frac{12}{\sqrt[5]{19}}$

Let the random variable X be defined as the number of bad oranges in a draw of two oranges. Then X can take values 0, 1, 2.

Total number of oranges = 4 + 16 = 20.

Two oranges can be drawn (simultaneously) in ${^{20}C}_2$ ways.

$P(X = 0)$ = P(drawing no defective orange) = P(both good oranges)

$=\frac{{^{16}C}_2}{{^{20}C}_2}=\frac{16×15}{1×2}×\frac{1×2}{20×19}=\frac{12}{19}$,

$P(X = 1)$ = P(drawing one defective orange and one good orange)

$=\frac{{^4C}_1 × {^{16}C}_1}{{^{20}C}_2}=\frac{4}{1}×\frac{16}{1}×\frac{1×2}{20×19}=\frac{32}{95}$,

$P(X = 2)$ = P(drawing two defective oranges)

$=\frac{{^4C}_2}{{^{20}C}_2}=\frac{4×3}{1×2}×\frac{1×2}{20×19}=\frac{3}{95}$

∴ The probability distribution is $\begin{pmatrix}0&1&2\\\frac{12}{19}&\frac{32}{95}&\frac{3}{95}\end{pmatrix}$

To calculate the mean and variance of the distribution, we construct the following table:

$x_i$	$p_i$	$p_ix_i$	$p_i{x_i}^2$
0	$\frac{12}{19}$	0	0
1	$\frac{32}{95}$	$\frac{32}{95}$	$\frac{32}{95}$
2	$\frac{3}{95}$	$\frac{6}{95}$	$\frac{12}{95}$
Total		$\frac{38}{95}=\frac{2}{5}$	$\frac{44}{95}$

Mean, $μ = Σp_ix_i =\frac{2}{5}$

Variance, $σ^2 = 2p_i{x_i}^2 - μ^2 =\frac{44}{95}-(\frac{2}{5})^2=\frac{44}{95}-\frac{4}{25}=\frac{144}{475}$.

Standard deviation = $\sqrt{\frac{144}{475}}=\frac{12}{5\sqrt{19}}$