what is the probability of flipping three heads in a row?
I started by doing binomial,
so to calculate the probability of three consecutive head would be (.5)^3 which yields .125 but that assumes the coin is fair. How would you account for a biased coin?
On
You do not know the parameter $p$ of your binomial distribution (i.e., coin flip) exactly but only as a distribution, i.e., in this case the uniform distribution on $[0,1]$. You know that you get heads twice in a row so you can update this distribution via $$f(p|2 \; heads) = \dfrac{P_{Binom}(2 \; heads|p)f_{unif}(p)}{P(2\;heads)}=\dfrac{P_{Binom}(2 \; heads|p)f_{unif}(p)}{\int_{\mathbb{R}}P(2\;heads|p)f_{unif}(p)dp}.$$
Now, we can derive $\int_{\mathbb{R}}P(2\;heads|p)f_{unif}(p)dp = \int_0^1p^2 dp = \dfrac{1}{3}$.
So, $f(p|2\;heads) = 3p^2\;1_{[0,1]}(p)$. Hence, $$P(3\;heads|2\;heads) = \int_{\mathbb{R}}P(3\;heads|2\;heads,\;p)f(p|2\;heads)dp \\ = \int_0^1 p\cdot 3p^2 dp = \dfrac{3}{4}.$$
Denote by $X$ the random variable which produces $p$ (the probability that the coin gets heads) - it's distribution is $U[0,1]$.
$$P(2 \space heads)=\int_{x=0}^ {1} {f_X(x)\bullet x^2dx} =\int_{x=0}^ {1} {1 x^2dx}= \frac{1}{3}$$
$$P(3 \space heads)=\int_{x=0}^{1} {f_X(x)\bullet x^3} = \frac{1}{4}$$
$$P(3\space heads\space |\space 2\space heads) = \frac{P(3 \space heads)}{P(2\space heads)} = \frac{\frac{1}{4}}{\frac{1}{3}} = \frac{3}{4}$$
And that's our answer.