A company has 125 personal computers. The probability that any one of them will require repair on a given day is 12.5%. What is the probability that exactly 20 of the computers will require repair on a given day?

Use the binomial formula to determine the desired probability. $$ P\left(X=x\right)=\left(_nC_x\right)\times p^x\times q^{n-x} $$ $$ P\left(X=20\right)=\left(_{125}C_{20}\right)\times\left(0.125\right)^{20}\times\left(1-0.125\right)^{125-20} $$ $$ P\left(X=20\right)\approx0.0505 $$