A die is tossed thrice. If event of getting an even number is a success, then the probability of getting at least two successes is

$\frac{1}{2}$

Let X denote the number of successes in 3 trials.

The, X is a binomial variate with $n = 3, p =\frac{3}{6}=\frac{1}{2}$ such that

$P(X=r)= {^3C}_r \left(\frac{1}{2}\right)^3, r= 0, 1, 2, 3$

∴ Required probability $= P(X≥2) - P(X=2) +P(X=3)$

$={^3C}_2 \left(\frac{1}{2}\right)^3+{^3C}_3 \left(\frac{1}{2}\right)^3=\frac{1}{2}$