probability - expected lifetime The distribution of lifetime $X$ of a component is continuous, and the probability that $X$ exceeds $x>0$ (in months) is   $1/(x+1)^3$ . What is the expected lifetime of that component? What is the probability that $2$ out of $5$ such components, working independently, do not fail in $2$ first months?
 A: For nonnegative random variables, the expected value can be calculated using $$E[X]=\int_0^\infty P\{X > x\}\,\mathrm dx.$$ Can you figure out what the integrand would be for your question? And if so, can you calculate the value of the integral?
A: So this question is tricky because you are given the cdf (cumulative distribution function) which I will call $F(x)$, but in order to calculate expectation you need the pdf (probability distribution function) which I will call $f(x)$.  We can find the pdf by noting that
$$
F(x)=P\{X\leq x\}=1-P\{X>x\}=1-\frac{1}{(x+1)^3}=\int_0^xf(x)\,dx
$$
Now, we take a closer look at that last bit of the equality:
$$
1-\frac{1}{(x+1)^3}=\int_0^xf(x)\,dx
$$
By the fundamental theorem of calculus, we can differentiate both sides to find
$$
f(x) = \frac3{(x+1)^4}
$$
In order to evaluate the expected lifetime, we find
$$
E(x)=\int_0^\infty x\,f(x)\,dx=\int_0^\infty \frac{3x}{(x+1)^4}\,dx
$$
As you stated, the integral (and thus the expectation) should come out to $\frac12$.
That the other answerer's method should work equally well is explained here
http://en.wikipedia.org/wiki/Expected_value#Continuous_distribution_taking_non-negative_values
As for the second part, the probability of a single component not failing in the first 2 months is
$$
p=P(X\geq 2)=\frac1{(2+1)^3}=\frac1{27}
$$
The probability that this happens exactly 2 out of 5 times is simply
$$
P=\binom{5}{2}p^2(1-p)^3
$$
