Stuck at defining the parameter λ for an exponential distribution So I'm given this assignment in probability that states to the following:
A computer is made out of m subsystems, by which each of them has the same exponential distribution for the time of failure. The subsystems are independent from each other and the system will fail if one of the subsystems fails.
-Find the distribution function for the time of failure of the system.
-If the average time of failure for each subsystem is 800 hours and there are 5 subsystems find the probability that the time of failure is greater than 150 hours.
Now for the first one I assume that since each subsystem has an exponential distribution with the same parameter the if $Mi$ is an event that a system fails then the probability for the whole system failing P(M) is:
P(M) = P(M1) + P(M2) + P(M3) + .... + P(Mm)
or the probability density function for the time of failure of the system is:
$ p(x) = mλe^{-λx}$
and we just integrate the function.
Is that all because this seems pretty simple, or am I missing something?
And for the second one I'm having a hard time deciding. Since the average time is 800 hours is that the expected value and λ = $\frac{1}{800}$ or is that just λ and λ = 800?
 A: Let $X_1, \ldots, X_M$ be the time to failure of each of the $M$ components. Then let $X$ be the time to failure of the entire system. Now if one component fails, the entire system fails. Hence, we have that the event:
$$
\{X\leq x\} \ \ \text{occurs if and only if} \ \  \{X_1\leq x\} \ \ \text{or} \ \ \ldots \ \ \text{or} \ \ \{X_M\leq x\} \ \ \text{occurs}
$$
These are rather nasty to compute so instead we can look at the complement, or $\{X\geq x\}$. We have that:
$$
\{X\geq x\} \ \ \text{occurs if and only if} \ \  \{X_1\geq x\} \ \ \text{and} \ \ \ldots \ \ \text{and} \ \ \{X_M\geq x\} \ \ \text{occurs}
$$
Each $X_i$ is an independent exponential so:
$$
P(X \ge x)=\prod_{i=1}^M P(X_i \ge x) = \prod_{i=1}^M e^{- \lambda x} = e^{- \lambda M x}
$$
so that:
$$
P(X \le x) = 1- P(X \ge x) = 1-e^{- \lambda M x}
$$
which is the CDF of an exponential distribution with rate parameter $\lambda M$. 
The parameter $\lambda$ is defined as the number of failures per unit time. Hence, we want $\lambda = \frac{1}{800}$. Therefore the probability is:
$$
P(X \ge 150) = 1-P(X \le 150) = 1- e^{-\frac{1}{800} \cdot 5 \cdot 150}
$$
One helpful way to always ensure you have it correct is to make sure that the time units in the exponent cancel out.
