expectation calculation problem I got the answers for this and i know its 1.05 but the way it explains is very difficult to understand so im seeking for some help here.
A system made up of 7 components with independent, identically distributed lifetimes will operate until any of 1 of the system's components fails. If the life time X of each component has density function
$f(x) =
\begin{cases}
3/x^4,  & \text{for 1<x}\\
0, & \text{otherwise}
\end{cases}$
what is the expected lifetime until failure of the system?
I tried to find the intersect of 7 components by integrating and power it by 7 but it doesnt give me anything useful...

Since the system fails as soon as the first component fails, we are looking for the minimum of the lifetimes of the seven components, call it $Y=\min(X_1,X_2\ldots,X_7)$.  The cumulative distribution $G$ for $Y$ is $$G(y)=\Pr(Y\le y)$$$$=1-\Pr(Y>y)$$$$=1-\Pr(X_1>y,X_2>y,\ldots,X_7>y)$$
$$=1-[\Pr(X_1>y)\Pr(X_2>y)\ldots\Pr(X_7>y)]$$
(Because of independence, we could split the probabilities)
$$=1-[(1-\Pr(X_1\le y))(1-\Pr(X_2\le y))\ldots(1-\Pr(X_7\le y))]$$
$$=1-[(1-F(y))(1-F(y))\ldots(1-F(y))]$$
$$=1-[1-F(y)]^7,$$
where $F$ is your cumulative distribution function for any $X$.  Thus your density function $g$ will be the derivative,
$$g(y) = 7[1-F(y)]^6f(y)$$
Can you go from there?
