A pair of fair dice is thrown 4 times. If getting the same number on both dice is considered as a success, then the probability of two successes are

The probability of getting identical numbers on both dice in a single throw is $p = \frac{6}{36} = \frac{1}{6}$, with complementary probability $q = \frac{5}{6}$.

With $n = 4$ independent trials and $k = 2$ successes required, the binomial probability is $P(X=2) = C(4,2) \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^2$.

The binomial coefficient evaluates to $C(4,2) = \frac{4!}{2!2!} = 6$, so:

$P(X=2) = 6 \times \frac{1}{36} \times \frac{25}{36} = \frac{150}{1296} = \frac{25}{216}$

This corresponds to option A.