Probability with uniform random variables Let $X_{1}$,$X_{2}$... $X_{n+1}$ be independent uniform random variables on [0,1]. 
let Z be the 2nd highest value in $\{X_{1},X_{2}... X_{n}\}$ 
let $c$ be a constant s.t $c \in [0,1]$ 
let $b$ be a constant s.t $b \in [0,1]$ and $b<c$ 
What is the $prob(X_{n+1}<Z<c,b<Z)$?
My attempt using the method from Conditional expected value of a maximum of uniform random variables.
$$prob(X_{n+1}<Z<c,b<Z)={\int_{b}^{c} n(n-1)(1-z) z^{n - 1} dz}.$$
Is this correct? If not where am I going wrong?
 A: You're right:
$Z$ follows a Beta$(n-1,2)$ distribution as the second largest random variable in a sample of $n$ independent uniform random variables - check order statistics for this. The density of $Z$ is then:
$$f(z)= n(n-1) z^{n-2} (1-z) 1_{(0,1)}(z) $$
(This is intuitively clear : $z^{n-2}$ is the probability that $n-2$ independent uniform are below $z$, and $1-z$ that a further independent uniform random variable is above; once you have chosen one of the $n$ indices to be the one that achieves the second record, you have $n-1$ others to choose to be the one of the first record)
Since the event $X_{n+1}$ is independent of $\{X_1, \ldots, X_n\}$, the two r.v. in the pair $(X_{n+1},Z)$ are independent, with density the product of their respective maginals. Set $D=\{(x,z) \in [0,1]^2: x<z, b<z<c \}$ and compute :
\begin{align}
P(X_{n+1}<Z ,b<Z<c)
& = P((X_{n+1},Z) \in D ) \\
& =\int x f(z) 1_{(x,z) \in D} dx dz \\
&= \int_{b}^{c} z f(z) dz  \\
& =  n(n-1) \int_{b}^c z^{n-1} (1-z) 1_{(0,1)}(z)
\end{align}
A: Every arrangement is equally likely to occur
So z is second highest so n-2 things are less are and 1 is greater . This has probability k.z^(n-2)*(1-z)
To calculate k we will put integration from 0 to 1 = 1
Giving K=n.(n-1),  Kn+1 < z have probability z again. Yeah I think u r absolutely correct.
