This kind of problems is modeled by Bernoulli's formula, which states that, if you have probability [tex] p [/tex] of success and you want [tex] k [/tex] successes over [tex] n [/tex] trials, the probability is
[tex] \binom{n}{k}p^k(1-p)^{n-k},\quad \mbox{where}\ \binom{n}{k} = \frac{n!}{k!(n-k)!} [/tex]
In your case, [tex] p = 0.25 [/tex], [tex] k=3 [/tex] and [tex] n=5 [/tex]
So, the formula becomes
[tex] \binom{5}{3} 0.25^30.75^2 = \frac{5!}{3!2!}\cdot 0.015625\cdot 0.5625 = 10\cdot 0.015625\cdot0.5625 = 0.087890625 [/tex]
So the probability is around [tex] 9\% [/tex]
