Probability a coin is heads given 5 heads and distribution 
A machine produces a weighted coin that lands on heads with an unkown
probability $p$, and we know that $P(p\le x) = x^4$.  You flip the
coin $5$ times, and every time it lands on heads.  What is the
probability that the next flip will also land on heads?

I am not exactly sure how to tackle this problem. I attempted going a conditional probability route, but was not sure what to put as the values for the numerator and denominator. It looks like the probability the coin is less than x is a cumulative distribution function, but I don't know if that intuition is correct or helpful. How should I go about solving this?
 A: $F(x) = x^4, f(x) = 4x^3$, $P(N = 5 \mid p = x) =  x^5$.
Define $H(x) = P(p\leq x\mid N = 5) = \int_0^x h(t)\,dt$. Since
$
P(N = 5) = \int_0^1 P(N = 5 \mid p = t)f(t)\,dt
=\int_0^1 t^5 f(t)\,dt
= \frac{4}{9},
$ the density is
$$
h(x) = P(p=x\mid N = 5) = \frac{P(N=5\mid p = x) f(x)}{P(N=5)} = x^5 4 x^3\frac{9}{4} = 9 x^8.
$$
Therefore $
E[p \mid N = 5] = \int_0^1 x h(x) dx
= 9 \int_0^1 x^9 dx = \frac{9}{10}.
$
Perhaps its easier to calculate
$P(N^+ = 6) =\int_0^1 t^6 f(t)\,dt= \frac{2}{5}$
and
$$
P(N^+ = 6 \mid N = 5) = \frac{P(N^+ = 6 \cap N = 5)}{P(N=5)} = \frac{2/5}{4/9} =\frac{9}{10}
$$
but this does not give the distribution of $p$.
A: 
It looks like the probability the coin is less than x is a cumulative distribution function,

Indeed.
We have the CDF for $p$, which gives us the pdf.
$$\begin{align}
 F_p(x) &= x^4\,\mathbf 1_{0\leq x\lt 1}~+~\mathbf 1_{1\leq x}
\\[2ex]
 f_p(x) &= 4x^3\,\mathbf 1_{0\leq x\leq 1}
\end{align}$$

I attempted going a conditional probability route, but was not sure what to put as the values for the numerator and denominator.

You have the evidence, and the prediction.
Let $E$ be the event of the evidence ($5$ flips that result in $5$ heads).   Now the conditional distribution for this given a $p$ is binomial: $(E\mid p\,{=}\,x)\sim\mathcal{Bin}(5,x)$ .
$$\mathsf P(E\mid p=x) = \binom 55 x^5 \,\mathbf 1_{x\in[0,1]}$$
  Thus the marginal probability for $E$ is the exponential:
$$\mathsf P(E) = \int_0^1 \binom 55\, x^5\cdot f_p(x)\,\mathrm d x$$
Let $N$ be the event that the next flip is also heads.   Likewise we find:
$$\begin{align}\mathsf P(N\cap E\mid p=x) ~&=~ \binom 66 x^6 \,\mathbf 1_{x\in[0,1]}
\\
\mathsf P(N\cap E) ~&=~ \int_0^1 \binom 66\, x^6\cdot f_p(x)\,\mathrm d x\end{align}$$
So...
We seek the probability for the next coin toss being heads given that the prior five were heads, which is to say $\mathsf P(N\mid E)$.   This is, by definition:
$$\mathsf P(N\mid E)=\dfrac{\mathsf P(N\cap E)}{\mathsf P(E)}$$
A: The probability that the next flip will land on heads is $p$, the same as the probability of any single flip.
$P(\text{next flip is heads}|\text{5 flips were heads}) = \frac{P(\text{next flip is heads and 5 flips were heads})}{P(\text{5 flips were heads})} = \frac{P(\text{next flip is heads})\cdot P(\text{5 flips were heads})}{P(\text{5 flips were heads})} = P(\text{next flip is heads})$
$P(\text{next flip is heads}) = p$
