A die is tossed twice . If a success is getting an even number on a toss , then find the variance of the number of successes .

Explanation

Let X be the random variable for a success for getting an even number on a toss .
`therefore " " X = 0,1,2 , n = 2 , p = (3)/(2) = (1)/(2) ` and q = `(1)/(2)`
At X = 0 , `" " P(X = 0) = "^(2)C_(0)((1)/(2))^(0) ((1)/(2))^(2-0) = (1)/(4)`
At X = t , `" " P(X = 1) = "^(2)C_(0)((1)/(2))^(1) ((1)/(2))^(2-1) = 2 * (1)/(2) * (1)/(2) = (1)/(2)`
At X = 2 , `" " P(X = 2) = "^(2)C_(2)((1)/(2))^(2) ((1)/(2))^(2-2) = (1)/(4)`

`therefore " " sumXP(X) = 0 + (1)/(2) + (1)/(2) = 1 " " .....(i) `
and `" " sumX^(2) P(X) = 0 + (1)/(2) + 1 = (3)/(2) " " .....(ii)`
`because " "` Var(X) = E`(X^(2)) - [E(X)]^(2)`
`= sumX^(2)P(X) -[sumXP(X)]^(2) = (3)/(2) - (1)^(2) = (1)/(2)` [using Eqs. (i) and (ii)]