Initially $P$ (working computer) =$\dfrac{4}{10},$ $P$ (non-working computer) = $\dfrac{6}{10}.$
Case 1 : three computers are functional : There are $4$ sub-cases $WWWN, WWNW, WNWW, NWWW,$ where $W$ means working, $N$ means non-working, but $P(WWWN) = P(WWNW) = P(WNWW) = P(NWWW),$
because for example $P(WWWN)=\dfrac{4}{10}\times \dfrac{3}{9}\times \dfrac{2}{8}\times \dfrac{6}{7}=\dfrac{144}{5040}$
In all other $3$ sub-cases, we get same numerators and denominators (in different order), so total prob in this case is
$\dfrac{4\times 144}{5040} = \dfrac{576}{5040}$.
Case 2 : all $4$ are working
$P(WWWW)=\dfrac{4}{10}\times \dfrac{3}{9}\times \dfrac{2}{8}\times \dfrac{1}{7}=\dfrac{24}{5040}$
$P$(atleast $3$ are working) =$\dfrac{600}{5040}$
So $100\times p =11.90$