We have \(n = 5\) days, and probability of rain on any day is \(p = 0.5\), probability of no rain is \(q = 0.5\).
We need the probability of exactly \(k = 3\) rainy days. This follows a binomial distribution:
\[
P(X = k) = \binomnk p^k q^n-k
\]
Substitute values:
\[
P(X = 3) = \binom53 (0.5)^3 (0.5)^2
\]
\(\binom53 = 10\), and \((0.5)^5 = \frac132\).
Thus:
\[
P(X = 3) = 10 \times \frac132 = \frac1032 = \frac516
\]
Hence, the probability is \(\frac516\).