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 number of red balls in a draw of 3 balls.
Then, x can take the value 0, 1, 2 or 3.
P(X=0)=P(getting no red ball)
=P(getting 3 white balls)
`=(""^(4)C_(3))/(""^(7)C_(3))=((4xx3xx2)/(3xx2xx1)xx(3xx2xx1)/(7xx6xx5))=(4)/(5).`
P(X=1)=P(getting 1 red and 2 white balls)
`=(""^(3)C_(1)xx""^(4)C_(2))/(""^(7)C_(3))=((3xx4xx3)/(2)xx(3xx2xx1)/(7xx6xx5))=(18)/(35).`
P(X=2)=P(getting 2 red and 1 white ball)
`=(""^(3)C_(2)xx""^(4)C_(1))/(""^(7)C_(3))=((3xx2)/(2xx1)xx4xx(3xx2xx1)/(7xx6xx5))=(12)/(35).`
`P(X=3)=P("getting 3 red balls")=(""^(3)C_(3))/(""^(7)C_(3))=(1xx3xx2xx1)/(7xx6xx5))=(1)/(35).`
Thus, the probability distribution of X is given below.
`therefore" mean,"mu=Sigmax_(i)p_(i)=(0xx(4)/(35))+(1xx(18)/(35))+(2xx(12)/(35))+(3xx(1)/(35))=(9)/(7).`
Variance, `sigma^(2)=Sigmax_(i)^(2)p_(i)-mu^(2)`
`=[(0xx(4)/(35))+(1xx(18)/(35))+(4xx(12)/(35))+(9xx(1)/(35))-(81)/(49)]`
`=((15)/(7)-(81)/(49))=(24)/(49).`
