Correct Answer - C
probability of selecting ace, `p=4/(52)=1/(13)`
Probabllity of not selecting ace, `q=1-1/(13)=(12)/(13)`
`p(X=0)=""^2C_0xx(1/(13))^0xx((12)/(13))^2=(144)/(169)`
`P(X=1)=""^2C_1xx(1/(13))xx((12)/(13))=(24)/(169)`
`P(X=2)=""^2C_2(1/(13))^2.((12)/(13))^0=1/(169)`
Mean `=SigmaP_iX_i=0xx(144)/(169)+1xx(24)/(169)+2xx1/(169)`
`=(24)/(169)+2/(169)=2/(13)`