I don't understan the following question (There is no similarity with the other exponential distribution problems thats why I got confused) so,
Assume there is a computer lab with two computers. The times until the failures of computers 1 and 2 are independent and exponentially distributed with means of 2 and 3 years respectively. (a) What is the probability that the failure time of computer 1 is more than twice of the failure time of computer 2? (b) Find the probability distribution of the time until the first failure.
(c) Using part (b), nd the expected time until the first failure. (d) Find the probability that the computer which fails later works more that 20 months?
Since $\{2T_2\leqslant t\} = \left\{ T_2\leqslant \frac12 t\right\}$ for $t\geqslant0$, it follows that $$F_{2T_2}(t) = F_{T_2}\left(\frac12 t\right) = 1 - e^{-\frac13\left(\frac 12 t\right)}= 1-e^{-\frac16 t}, $$ so $2T_2\sim\mathrm{Exp}\left(\frac16 \right)$. We then compute $$\mathbb P\left(T_1>2T_2\right) = \frac{1/2}{1/2 + 1/6} = \frac34. $$ The time until the first failure is the minimum of $T_1$ and $T_2$, and so has exponential distribution with rate $\frac12+\frac13 = \frac56$. The expected time until the first failure is then $\left(\frac56\right)^{-1}=\frac65$.
Let $W = T_1\vee T_2$ denote the maximum of $T_1$ and $T_2$. For $t\geqslant 0$, we have $\{W\leqslant t\} = \{T_1\leqslant t, T_2\leqslant t\}$, so from independence it follows that $$ \mathbb P(T_1\leqslant t, T_2\leqslant t) = \mathbb P(T_1\leqslant t)\mathbb P(T_2\leqslant t) $$ and hence $$F_W(t) = F_{T_1}(t)F_{T_2}(t) = \left(1-e^{-\frac12 t}\right)\left(1-e^{-\frac13 t}\right). $$ Therefore the probability that $W$ is greater than $20 $ months ( which is $20/12$ years) is $$1-F_W\left(\frac53\right) = 1 - \left(1-e^{-\frac12\cdot\frac53}\right)\left(1-e^{-\frac13\cdot\frac53}\right) = 1-\left(1-e^{-\frac56}\right)\left(1-e^{-\frac59}\right) .$$