An urn contians 4 white and 3 red balls. Let X be the number of red balls in a random draw of 3 balls. Find the mean and variance of X.

Explanation

When 3 balls are drawn at random, there may be no red ball, 1 red ball, 2 red balls or 3 red balls.
Let X denote the random variable showing the no. of red balls in a draw of 3 balls.
Then, X can take value 0, 1, 2 or 3
P(X=0)= P(getting no red balls)
= P(getting 3 white balls)
`=(""^4C_3)/(""^7C_3)=4/35`
P(X=1) = P(getting 1 red and 2 white balls)
`=(""^3C_1xx""^4C_2)/(""^7C_3)=18/35`
P(X=2) = P(getting 2 red and 1 white ball)
`=(""^3C_2xx""^4C_1)/(""^7C_3)=12/35`
P(X=3) = P(getting 3 red balls)
`=(""^3C_3)/(""^7C_3)=1/35`
Thus, the probability distribution of X is given below

`therefore " "mu=sumx_ip_i`
`=(0xx4/35)+(1xx18/35)+(2xx12/35)+(3xx1/35)`
`=9/7`
Variance, `sigma^2=sumx_i^2p_i-mu^2`
`=(0xx4/35+1xx18/35+4xx12/35+9xx1/35)-81/49`
`=15/7-81/49`
`=24/49`
Standard deviation, `sigma=sqrt(Variance)`
`=sqrt(24/49)=(2sqrt6)/(7)`