A machine has four identical components, whose duration times in years are independent and identically distributed random variables, with exponential distribution of the parameter $1/4$. Determine the life of the machine, assuming it fails
$1.$ When all components have failed.
$2.$ When one of the components fails.
$3.$ When only one component remains in operation.
In the three cases above, find the probability that the machine will last at least $5$ years.
My approach: Let $Y_{1},Y_{2},Y_{3},Y_{4}$ the indentical components in the machine. Now, by hypothesis we know that $Y_{i} \overset{i.i.d}{\sim}\textbf{Exp}(1/4)$.
We need to know when all components have failed, so taking $U_{4}=\max\{Y_{1},Y_{2},Y_{3},Y_{4}\}$ and since $$f_{U_n}(y)=nf(y)[F(y)]^{n-1}$$ because $Y_{i}\overset{i.i.d}{\sim}\textbf{Exp}(1/4)$ and since that here $n=4$, we have that $$f_{U_n}(y)=4\left(\frac{1}{4}e^{-y/4} \right)(1-e^{-y/4})=e^{-y/4}(1-e^{-y/4})^{3}$$ Note that $F(y)=\int_{-\infty}^{y} \frac{1}{4}e^{-t/4}dt=\int_{0}^{y}\frac{1}{4}e^{-t/4}dt=1-e^{-y/4}$ Also, $$\mathbb{P}(U_{n}\geq 5)=\int_{5}^{\infty}e^{-y/4}(1-e^{-y/4})^{3}dy\approx 0.7408$$.
We need to know when one of the components fails, so taking $U_{1}=\min\{Y_{1},Y_{2},Y_{3},Y_{4}\}$ and since the $Y_{i}$ are i.i.d, so we have that $$f_{U_1}(y)=nf(y)[1-F(y)]^{n-1}$$ and here $n=4$, so we have that $$f_{U_{1}}(y)=e^{-y}, \quad y\geq 0$$ also, $$\mathbb{P}(U_{1}\geq 5)=\int_{5}^{\infty}e^{-y}dy\approx 0.006738$$.
3.But, I don't know how can I solve this part. Can you help me in this part?
Hint: Writing the serving times of components in a variational series
$$ \min\{Y_1, Y_2, Y_3, Y_4\} := Y_{(1)} \le Y_{(2)} \le Y_{(3)} \le Y_{(4)} := \max\{Y_1, Y_2, Y_3, Y_4\} $$
one notices that $U_3 = Y_{(3)}$ is the quantity you are asked in the question. The latter is the $3$-rd order statistics of $Y_1, \dots, Y_4$.