Let X be the random variable. Then, X= number of kings obtained in two draws.
Clearly, X can assume the value 0, 1 or 2.
`P("drawing a king")=(1-(1)/(13))=(12)/(13)cdot`
P(X=0)=P(not a king in the 1st draw and not a king in the 2nd draw)
`=((12)/(13xx(12)/(13))=(144)/(169)cdot`
P(X=1)=P( a king in the 1st draw and not a king in the 2nd draw) or P(not a king in the 1st draw and a king in the 2nd draw)
`=((1)/(13)xx(12)/(13)+(12)/(13)xx(1)/(13))=(24)/(169)cdot`
P(X=2)=P(a king in the 1st draw and a king in the 2nd draw)
`=((1)/(13)xx(1)/(13))=(1)/(169)cdot`
Hence, the probability distribution is given by
`therefore" mean," mu=Sigmax_(i)p_(i)=(0xx(144)/(169))+(1xx(24)/(169))+(2xx(1)/(169))=(2)/(13)cdot`
Variance, `sigma^(2)=Sigmax_(i)^(2)p_(i)-mu^(2)`
`=[(0xx(144)/(169))+(1xx(24)/(169))+(4xx(1)/(169))-(4)/(169))]=(24)/(169)cdot`