Question

Three defective bulbs are mixd with 7 good ones. Let X be the number of defective bulbs when 3 bulbs are drawn at random. Find the mean and variance of X.

Answer 1

Let X denote the random variable showing the number of defective bulbs.
Then, X can take the value 0, 1, 2 or 3.
`therefore" "P(X=0)=P("none of the bulbs is defective")`
=P(all the 3 bulbs are good ones)
`=(""^(7)C_(3))/(""^(10)C_(3))=((7xx6xx5)/(3xx2xx1)xx(3xx2xx1)/(10xx9xx8))=(7)/(24)cdot`
P(X=1)=P(1 defective and non-defective bulbs)
`=(""^(3)C_(1)xx""^(7)C_(2))/(""^(10)C_(3))=(3xx(7xx6)/(2xx1)xx(3xx2xx1)/(10xx9xx8))=(21)/(40)cdot`
P(X=2)=P(2defective and 1 good one)
`=(""^(3)C_(2)xx""^(7)C_(1))/(""^(10)C_(3))=((3xx2)/(2xx1)xx7xx(3xx2xx1)/(10xx9xx8))=(7)/(40)cdot`
P(X=3)=P(3 defective bulbs)
`=(""^(3)C_(3))/(""^(10)C_(3))=(1xx(3xx2xx1)/(10xx9xx8))=(1)/(120)cdot`
Thus, the probability distribution is given by

`therefore" mean,"mu=Sigmax_(i)p_(i)=(0xx(7)/(24))+(1xx(21)/(40))+(2xx(7)/(40))+(3xx(1)/(120))=(9)/(10)cdot`
Variance, `sigma(2)=Sigmax_(i)^(2)p_(i)-mu^(2)`
`=(0xx(7)/(24))+(1xx(21)/(40))+(4xx(7)/(40))+(9xx(1)/(120))-(81)/(100)=((13)/(10)-(81)/(100))=(49)/(100)cdot`