This kind of situation is modelled by Bernoulli's formula. It applies everytime there is an experiments with two possible outcomes repeated n times. Each repetition is independent on the others, and we know the probability of the two outcomes p and 1-p. If we want the outcome with probability p to appear k times, the probability is
[tex] \binom{n}{k}p^k(1-p){n-k} [/tex]
In your case, you run the "experiment" 4 times (you choose 4 printers) and want that all of them to be non-defective. A printer is non-defective with probability 1/2, since there are 5 defective and 5 non-defective printers.
So, our model is built with n = k = 4, p = 1-p = 1/2. The probability is
[tex] P=\binom{4}{4}\left(\cfrac{1}{2}\right)^4\left(\cfrac{1}{2}\right)^0 = 1 \cdot \left(\cfrac{1}{2}\right)^4 \cdot 1 = \cfrac{1}{16} [/tex]
