Let X be the random variable for a ‘success’ for getting an even number on a toss.

∴ X = 0, 1, 2

n = 2

Even number on dice = 2, 4, 6

∴ Total possibility of getting an even number = 3

Total number on dice = 6

p = probability of getting an even number on a toss

q = 1 – p

Now, the probability of x successes in n–Bernoulli trials is ⁿC_rp^rq^n-r

Thus, P( x successes) = ⁿC_rp^rq^n-r

=1

Also, Var(X) = E(X²) – [E(X)]²

= ΣX²P(X) – [E(X)]²

= 1.5 – 1

= 0.5