Question

$4$ bad apples accidentally got mixed up with $20$ good apples. In a draw of $2$ apples at random, expected number of bad apples is

• A. $1$
• B. $2/3$
• C. $1/3$
• D. $1/6$

Answer 1

The correct option is C $1/3$

Total apples are: $4$ Bad apples $\left(B\right)$ + $20$ Apples $\left(A\right)$$=$ $24$
Selection of $2$ apples from $24$ apples ${=}^{24}{C}_2$
Case $A$: $1B$ and $1A$ $=\frac{{}^4{C}_1{\times }^{20}{C}_1}{{}^{24}{C}_2}=\frac{80}{276}$
Case $B$: $2B$ and $0A$ $=\frac{{}^4{C}_2{\times }^{20}{C}_0}{{}^{24}{C}_2}=\frac{6}{120}$
Expected number of Bad apples drawn is $=1\times Case\left(A\right)+2\times Case\left(B\right)$
$=\frac{80}{276}+2\times \frac{6}{120}$
$=\frac{1}{3}$