A die is tossed twice A success is getting an odd number on a toss. Find the variance of the number of successes.

Explanation

In a single toss,
P(success)=P(getting and odd number)`=(3)/(6)=(1)/(2)` and P(non-success)`=(1-(1)/(2))=(1)/(2).`
Let X be the number of successes. Then, X=0, 1 or 2.
P(X=0)P [(non-success in the 1st toss and non-success in the 2nd)]
`=((1)/(2)xx(1)/(2))=(1)/(4).`
P(X=1)P[(success in the 1st toss and non-success in the 2nd) or (non-success in 1st toss and success in the 2nd)]
`((1)/(2)xx(1)/(2))+((1)/(2)xx(1)/(2))=(1)/(2).`
P(X=2)=P[(success in the 1st toss and success in the 2nd)]
`=((1)/(2)xx(1)/(2))=(1)/(4).`