A machine has 12 identical components which function independently. The probability that a component will fail is 0.2. The machine will stop working if more than three components fail. Find the probability that the machine will stop.
\(\text{The number of components that fail has a binomial distribution, $n=12,~p=0.2$}\\ P[k>3]=1-P[0]-P[1]-P[2]-P[3] = \\ 1 - \sum \limits_{k=0}^3 \dbinom{12}{k}(0.2)^k(0.8)^{12-k} = 0.205431\\ \)
