A spaceship is controlled by three independent computers. The ship can function as long as at least two of the three computers are functioning. Suppose the lifetimes of the computers are i.i.d. exponential random variables with parameter $N>0$. What is the probability that the ship can function for the first $t$ units of time?
My answer: \begin{align*} P(\text{ship function for t units})&=P(\text{ two or three computers function})\\ &= P(\text{two function) + P(three function})\\ &=(3**C**2) P(X<t)P(X>t)^2 +(3**C**1)P(X>t)^3 \\ &=2(1-e^-Nt)e^-2Nt + 3e^-3Nt \end{align*}
Can someone just check my answer, and reassure me this is correct?
Let $Z_i$ denote the lifetime of computer $i$, then the random variables $Z_i$ are independent and for each nonnegative time $t$, $P(Z_i\geqslant t)=\mathrm e^{-Nt}$ for every $i$.
Fix some time $t$ and define $p=\mathrm e^{-Nt}$. The event $W$ that the spaceship functions at time $t$ occurs if and only if at most one computer lifetime is less than $t$. Thus, $$ W=U\cup V_1\cup V_2\cup V_3,$$ with $U=[Z_1\gt t,Z_2\gt t,Z_3\gt t]$ and, for every $i$ in $\{1,2,3\}$, $$V_i=[\forall j\ne i,Z_j\gt t\gt Z_i].$$ These four events are disjoint and by symmetry $P(V_i)$ does not depend on $i$ hence $$P(W)=P(U)+3P(V_1).$$ By independence of the random variables $Z_i$, $$ P(U)=p^3,\qquad P(V_1)=p^2(1-p), $$ hence $$ P(W)=p^3+3p^2(1-p)=3p^2-2p^3=3\mathrm e^{-2Nt}-2\mathrm e^{-3Nt}. $$ Thus, the lifetime $S$ of the spaceship has density $f_S$ defined by $$ f_S(t)=6\mathrm e^{-2Nt}(1-\mathrm e^{-Nt}). $$