This problem can be modeled using a binomial distribution where the probability of success (Martin playing games when his friend phones) is 0.25, and the probability of failure is 0.75. We are looking for the probability of exactly 2 successes in 8 trials.

The probability mass function for a binomial distribution is given by:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

where \(n\) is the number of trials, \(k\) is the number of successes, and \(p\) is the probability of success.

Substituting the given values:

\(P(X = 2) = \binom{8}{2} (0.25)^2 (0.75)^6\)

Calculating each part:

\(\binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28\)

\((0.25)^2 = 0.0625\)

\((0.75)^6 = 0.17803125\)

Therefore,

\(P(X = 2) = 28 \times 0.0625 \times 0.17803125 = 0.311\)