Probability conditional on inequality Let $X_1,X_2,....,X_{n+1}$ independent uniform variables on the unit interval $[0,1]$, $V:=\max(X_1,X_2,...,X_n)$  and   $c \in [0,1].$ 
What is $\textrm{Prob}(X_{n+1}<V|V<c)$?
Initially I thought this would be equivalent to 
$$\textrm{Prob}(X_{n+1}<V<c)=\int_0^cnz^ndz=\frac{n}{n+1}z^{n+1},$$
but I am concerned that equivalence may not hold.
 A: The formal meaning of $P(X_{n+1}<V|V<c)$ is $E(1_{X_{n+1}<V}|1_{V<c})$. The conditional expectation writes as $f(1_{V<c}) = f(1)1_{V<c} + f(0)1_{V\geq c}$ for some measurable $f$.
The $\sigma$-algebra generated by $1_{V<c}$ is  $\{\emptyset, (V<c), (V\geq c), \Omega  \}$ thus the defining equations of the conditional expectation write as $$E(1_{V<c} 1_{X_{n+1}<V}) = E( 1_{V<c} (f(1)1_{V<c} + f(0)1_{V\geq c})) =  f(1)E( 1_{V<c} ) = f(1)P(V<c)$$ 
$$E(1_{V\geq c} 1_{X_{n+1}<V}) = E( 1_{V\geq c} (f(1)1_{V<c} + f(0)1_{V\geq c})) =  f(0)E( 1_{V\geq c} ) = f(0)P(V\geq c)$$ 
Thus $\displaystyle f(1)= \frac{P(X_{n+1}<V<c)}{P(V<c)}$ and $\displaystyle f(0)= \frac{P((X_{n+1}<V) \cap (V\geq c))}{P(V\geq c)}$.
Both quantities are easily computable by integral means.
A: Since $V < c$ means $X_1, \ldots, X_n < c$, the probability of $X_i$ conditioned on $V < c$ is uniform
$$
\Pr(X_i \mid V<c) = 1 / c,
$$
and
$$
X_i \mid V < c \sim U(0,c),
$$
for $i = 1\ldots n$.
Consider $V := \max(X_1, \ldots, X_n) = X_j$, for $1 \leq j \leq n$, which follows that
$$
\begin{align}
 &\Pr(V < v \mid V < c) \\
=& \prod_j \Pr(X_j < v \mid V <c) \\
=& (v / c)^n.
\end{align}
$$
and
$$
\Pr(V=v \mid V<c) = \frac{d}{dv}Pr(V < v \mid V < c) =nv^{n-1}/c^n.
$$ 
Given that $X_1, \ldots, X_n, X_{n+1}$ are independent, it follows that
$$
\Pr(X_{n+1} < V \mid V <c)  = \Pr(A < B),
$$
where $j = 1\ldots n$ , $A \sim U(0,1)$, and $B = V\mid V<c$.
From Probability and uniform distribution, we know that
$$
\begin{align}
 &\Pr(B>A) \\
=&  \int_{0}^1 P(B>A | A=x) f_{A}(x) dx \\
=&  \int_{0}^c P(B>A | A=x) f_{A}(x) dx \qquad \text{($A$ must be lesser than $B$)}\\
=&  \int_{0}^c (1-(x / c)^n)  \cdot 1 dx \\
=&  \left. c - \frac{x^{n+1}}{(n+1)c^n} \right\vert_0^c \\
=&  \frac{n}{n+1}c
\end{align}
$$
Therefore, 
$$
\Pr(X_{n+1} < V \mid V <c) = \frac{n}{n+1}c. \tag{1}
$$
Moreover, eq. $(1)$ can be validated by the conditional probability formula
$$
\begin{align}
 &\Pr(X_{n+1} < V \mid V < c) \Pr(V < c) \\
=&\frac{n}{n+1}c \cdot c^n \\
=&\Pr(X_{n+1} < V < c).
\end{align}
$$
