Repeat Exercise 11 if the system works if and only if the fourth component works and at least two of the other components work.

The total number of states remains the same as in Exercise 11. The number of states where the whole system works does change to \(\binom{3}{2} + \binom{3}{3} = 4\) so the final probability is \(\frac{\binom{3}{2} + \binom{3}{3}}{2^4} = \frac{1}{4}\)