Finding combined time to repair two machines where time is exponentially distributed I am trying to solve the following problem.
The time $T$ required to repair a machine is an exponentially distributed random variable with mean 10 hours.
a) What is the probability that a repair takes at least 15 hours given that its duration
exceeds 12 hours?
b) What is the probability that the combined time to repair two machines is at least
20 hours?
Solution Attempt
Since mean is given to be 10 hours hence $\lambda = \dfrac {1}{10}$ and the probability distribution of the time is given as $e^{-\lambda t} = e^{-\dfrac {1}{10} t} $ 
a) $P(T>15 |T>12) = P(0 $ repairs in $ (12, 15]) = e^{-\dfrac {1}{10} 3}$
b) let $T_1$ be the r.v representing time to repair the first machine and $T_2$ be the r.v representing time to repair the second machine. So we seek to evaluate
$P(T_1 + T_2 > 20)$ we know both of these time should be independent as the exponential distribution process to memory less but i am not sure how to proceed from here. 
Any help would be much appreciated. 
 A: While the numerical answer you get for part (a) is correct, I think that your work
indicates some misinterpretation of the concepts.  $T$ is the time required to complete
a repair, and its complementary cumulative distribution function is $\exp(-t/10)$,
that is,
$$P\{T > t\} = 1 - F_T(t) = e^{-t/10}.$$
The question in part (a) asks for a conditional probability $P\{T > 15 \mid T > 12\}$
which is by definition given by
$$P\{T > 15 \mid T > 12\} = \frac{P(\{T > 15 \}\cap\{T > 12\})}{P\{T > 12\}}
= \frac{P\{T > 15 \}}{P\{T > 12\}} = \frac{e^{-15/10}}{e^{-12/10}}= e^{-3/10}$$
which is the same answer as you obtained, but it is not the probability
of $0$ repairs in $(12,15]$.  The repairing began at time $t =0$ and 
the question asks: if the repair is still ongoing at time $t = 12$, what is
the conditional probability that it is still ongoing at $t = 15$, and
thus completes at some time $T$ larger than $15$. Your use of the phrase

$$0 ~ \text{repairs in} ~(12,15]$$ 

almost makes it sound like the repairs are a Poisson arrivals
process.
A: $$\mathrm P(T_1+T_2\gt20)=\mathrm P(T_1\gt20)+\int_0^{20}\mathrm P(T_2\gt20-t)\cdot\lambda\mathrm e^{-\lambda t}\cdot\mathrm dt
$$
$$
\mathrm P(T_1+T_2\gt20)=\mathrm e^{-20\lambda}+\int_0^{20}\mathrm e^{-\lambda (20-t)}\cdot\lambda\mathrm e^{-\lambda t}\cdot\mathrm dt=\ ...$$
