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...
 A: You want the expected time until the earliest component failure, of seven i.i.d. components.   This is the seventh least order statistic.
$$\begin{align}
\mathsf E[X_{(7)}] 
 & = \binom{7}{1}\int_1^\infty x\cdot f_{X}(x)\cdot (1-F_X(x))^6 \operatorname d x
\\ & = 7\int_1^\infty x \cdot\frac {3}{x^4}\cdot \left(\int_x^\infty \frac {3}{y^4}\operatorname d y\right)^6\operatorname d x
\end{align}$$
A: 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?
