Let $X$ denotes the number of successes in $3$ trials. Then, $X$ is a binomial variate with $n=3$, 

$p=\frac{3}{6}=\frac{1}{2}$.

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

Therefore, the required probability of getting at least two successes is

$P(X\ge 2)=P(X=2)+P(X=3)$

$={}^{3}C_2{\left(\frac{1}{2}\right)}^3+{}^{3}C_3{\left(\frac{1}{2}\right)}^3$

$=3\times {\left(\frac{1}{2}\right)}^3+{\left(\frac{1}{2}\right)}^3$

$={\left(\frac{1}{2}\right)}^3\times 4$

$=\frac{1}{2}$