A box contains 8 red and 2 white balls. If three balls are drawn , one by one, at random without replacement, then the variance of the number of white balls drawn is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{28}{75} \)
Let X be the number of white balls drawn; X = 0, 1 or 2. Without replacement, we count selections of 3 balls out of 10, \( ^{10}C_{3}=120 \):
\( \displaystyle P(X=0)=\frac{^{8}C_{3}}{120}=\frac{56}{120}=\frac{7}{15} \)
\( \displaystyle P(X=1)=\frac{^{2}C_{1}\cdot{}^{8}C_{2}}{120}=\frac{56}{120}=\frac{7}{15} \)
\( \displaystyle P(X=2)=\frac{^{2}C_{2}\cdot{}^{8}C_{1}}{120}=\frac{8}{120}=\frac{1}{15} \)
Mean and \( E(X^2) \):
\( \displaystyle E(X)=0+\frac{7}{15}+\frac{2}{15}=\frac{9}{15}=\frac{3}{5} \)
\( \displaystyle E(X^2)=0+\frac{7}{15}+\frac{4}{15}=\frac{11}{15} \)
Variance:
\( \displaystyle \mathrm{Var}(X)=E(X^2)-[E(X)]^2 \)
\( \displaystyle =\frac{11}{15}-\frac{9}{25} \)
\( \displaystyle =\frac{55-27}{75} \)
\( \displaystyle =\frac{28}{75} \)
Shortcut: X is hypergeometric (N = 10, 2 white, n = 3), so
\( \displaystyle \mathrm{Var}(X)=n\cdot\frac{2}{10}\cdot\frac{8}{10}\cdot\frac{N-n}{N-1} \)
\( \displaystyle =3\cdot\frac{16}{100}\cdot\frac{7}{9} \)
\( \displaystyle =\frac{28}{75} \)