What is wrong with my method of finding the probability 
One way to solve this and my book has done it is by : 


This is a well known way, but I have a different method, and it seems logical to me (but I don't know what the mistake is). And yes it's wrong, but I don't understand what's wrong with my following method :
For 1 component the mean is $E(X)=2.5$ so for 5 components it's : 
$$E(5X)=5E(X)=5(2.5)=12.5$$
So for 5 items we can say : 
$$\lambda= 1/E(5X)= 1/12.5$$
$$X ~ Expo (1/12.5)$$
$$P(T \geq 3)=1-e^{-3/12.5}$$
$$P(T \geq3)=0.21$$
Which is not the same as in the book. Please help me, what is wrong with my method.
 A: Let $X_1,X_2,\dots,X_5$ be the lifetimes of the components. Let $Y=\min(X_1,X_2,\dots,X_5)$. Then $Y$ is the lifetime of the system.
We have $Y\gt y$ precisely if $X_i\gt y$ for $i=1,2,\dots,5$.  Thus by independence
$$\Pr(Y\gt y)=\prod_1^5\Pr(X_i\gt y)=e^{-5\lambda y}.$$
It follows that the lifetime of the system has exponential distribution with parameter $5\lambda$. This means that the lifetime of the system is, in our case, exponentially distributed with mean $\frac{2.5}{5}$.
A: Let's make the problem simpler by considering only two component lifetimes $X_1$ and $X_2$, both of which are exponentially distributed. The expectation of the sum $\mathbb{E}[X_1 + X_2]$ is not the expected lifetime of a system having both components concurrently. Instead, it is the expected value of a random variable $Y = X_1 + X_2$, which is the sum of the lifetimes of both components. In fact, $Y$ represents the lifetime of a system with the two components running sequentially (with one component replacing the other after the first "died"). What the problem is asking is for a system in which all five components are concurrently operating at the same time. 
