Two bad eggs are accidently mixed up with ten good ones. Three eggs are drawn at random with replacement from this lot. compute the mean for the number of bad eggs drawn.

Answer

`"Mean"=1/2`

Explanation

Let `X` denote the number of bad eggs


Total number of eggs`=` bad eggs+good eggs`=2+10=12`


Now the probability of drawing bad eggs are:


` P(X=0)=P("no bad egg or all three good eggs")`


`=(10C_3)/(12C_3)=(frac{10!}{3!7!})/(frac{12!}{3!9!})`


`=((10 times 9 times 8 times 7!)/(3 times 2 times 7!))/((12 times 11 times 10 times 9!)/(3 times 2 times 9!))=12/22`


`P(X=1)=P("one bad egg and two good eggs")`


`=(2C_1 times 10C_2)/(12C_3)=(frac{2!}{1!1!} times frac{10!}{2!8!})/(frac{12!}{3!9!})`


...