A bag contains 4 white and 6 black balls.

Question

A bag contains 4 white and 6 black balls. Three balls drawn at random; $X$=number of white balls. If $\sigma^2$ is the variance, find $100\sigma^2$.

Answer: 56

Accepted Answer

$$P(0)=\frac{{}^6C_3}{{}^{10}C_3}=\frac16,\ P(1)=\frac{{}^6C_2{}^4C_1}{{}^{10}C_3}=\frac12,\ P(2)=\frac{{}^6C_1{}^4C_2}{{}^{10}C_3}=\frac{3}{10},\ P(3)=\frac{{}^4C_3}{{}^{10}C_3}=\frac{1}{30}$$

$$\sigma^2 = \sum x^2P(x)-\left(\sum xP(x)\right)^2 = \left(\frac12+\frac{12}{10}+\frac{3}{10}\right)-\left(\frac12+\frac{6}{10}+\frac{1}{10}\right)^2 = \frac{56}{100}$$

$$100\sigma^2 = 56$$

Solutions · 1