Question

A box contains 16 bulbs, out of which 4 bulbs are defective. Three bulbs are drawn at random from the box. Let X be the number of defective bulbs drawn. Find the mean and variance of X.

Answer 1

Correct Answer - `"Mean "=(3)/(4), "variance "=(39)/(80)`
Three bulbs drawn one by one without replacement is the same as drawing 3 bulbs simultaneously.
Let X = number of defective bulbs in a lot of 3 bulbs drawn.
Then, X=0,1,2 or 3.
P(X=0)=P(one of the bulbs is defective)
`=(""^(12)C_(3))/(""^(16)C_(3))=((12xx11xx10)/(3xx2xx1)xx(3xx2xx1)/(16xx15xx14))=(11)/(28).`
P(X=1)=P(1 defective bulb and 2 nondefective bulbs)
`(""^(4)C_(1)xx""^(12)C_(2))/(""^(16)C_(3))=((4xx12xx11)/(2xx1)xx(3xx2xx1)/(16xx15xx14))=(33)/(70).`
P(X=2)=P(2 defective bulbs and 1 nondefective bulb)
`=((""^(4)C_(2)xx""^(12)C_(1)))/(""^(16)C_(3))=((4xx3)/(2xx1)xx12xx(3xx2xx1)/(16xx15xx14))=(9)/(70).`
P(X=3)=P(3 defective bulbs)
`=(""^(4)C_(3))/(""^(16)C_(3))=((4xx3xx2)/(3xx2xx1)xx(3xx2xx1)/(16xx15xx14))=(1)/(140).`