Let $X_{i}$ denote life time of i's component, and $X$ denote life time of the system as your notations. $X$ will work if at least two components of it work. In other words It fails to work if two components will fail to work. This means at time the life of system finished second component is failed to work (pay attention I'm not pointing to $X_{2}$, the mean is second component which is stopped). So if life time of $X$ be equal to $x$ means before time $x$ one of components is stopped, at time $x$ one different component is stopped, and the last component is stopped after time $x$. Therefore $X=x$ will accurse if $(x_{1},x_{2},x_{3})\in \{(x_{1},x_{2},x_{3})\in [0,\infty)\times[0,\infty)\times[0,\infty)\;|\; x_{1}=x,x_{2}\leq x,x_{3}\geq x\}\cup \{(x_{1},x_{2},x_{3})\in [0,\infty)\times[0,\infty)\times[0,\infty)\;|\; x_{1}=x,x_{2}\geq x,x_{3}\leq x\}\cup\{(x_{1},x_{2},x_{3})\in [0,\infty)\times[0,\infty)\times[0,\infty)\;|\; x_{1}\leq x,x_{2}=x,x_{3}\geq x\}\cup\{(x_{1},x_{2},x_{3})\in [0,\infty)\times[0,\infty)\times[0,\infty)\;|\; x_{1}\geq x,x_{2}=x,x_{3}\leq x\}\cup\{(x_{1},x_{2},x_{3})\in [0,\infty)\times[0,\infty)\times[0,\infty)\;|\; x_{1}\leq x,x_{2}\geq x,x_{3}=x\}\cup\{(x_{1},x_{2},x_{3})\in [0,\infty)\times[0,\infty)\times[0,\infty)\;|\; x_{1}\geq x,x_{2}\leq x,x_{3}=x\}$ Because Life time of components are i.i.d. exponential random variables with mean 1, so $f_{X}(x)=e^{-x}\int_{0}^{x}e^{-x_{2}}dx_{2}\int_{x}^{\infty}e^{-x_{3}}dx_{3}+e^{-x}\int_{x}^{\infty}e^{-x_{2}}dx_{2}\int_{0}^{x}e^{-x_{3}}dx_{3}+(\int_{0}^{x}e^{-x_{1}}dx_{1})e^{-x}(\int_{x}^{\infty}e^{-x_{3}}dx_{3})+(\int_{x}^{\infty}e^{-x_{1}}dx_{1})e^{-x}(\int_{0}^{x}e^{-x_{3}}dx_{3})+(\int_{0}^{x}e^{-x_{1}}dx_{1}\int_{x}^{\infty}e^{-x_{2}}dx_{2})e^{-x}+(\int_{x}^{\infty}e^{-x_{1}}dx_{1}\int_{0}^{x}e^{-x_{2}}dx_{2})e^{-x}$ By calculating above integrals we have $\forall x\in[0,\infty)\; :\; f_{X}(x)=6(e^{-2x}-e^{-3x})$ So $E(X)=\int_{0}^{\infty}6(e^{-2x}-e^{-3x})xdx=6\int_{0}^{\infty}xe^{-2x}dx-6\int_{0}^{\infty}xe^{-3x}dx=6(\frac{1}{4})-6(\frac{1}{9})=\frac{3}{2}-\frac{2}{3}=\frac{5}{6}$ So the choice "c" is correct.