I have two coins, one of which is biased P(heads) = p and one of which is fair.
I pick one of the coins at random and use it for the rest of the game.
I flip the coin 100 times and want to find the PMF for number of heads
I know how to get the PMF of a biased coin where x = # heads
$$ P(X=x) = {100 \choose x} p^x(1-p)^{10-x} $$
and of the fair coin
$$ P(X=x) = {100 \choose x} .5^{10} $$
My problem comes with putting them together.
There's a 50/50 chance I get either coin so I want to just multiply each PMF by $\frac{1}{2}$ and add them but this seems too simple.
I haven't been able to find examples of this sort of problem anywhere so I'm not sure I'm not the right track.
Thanks!
Best Answer
You have, conditioned on picking the fair coin, a pmf denoted by $p_f$ that you know how to compute.
You have, conditioned on picking the bias coin, a pmf denoted by $p_b$ that you know how to compute.
So the pmf overall is $p^\ast$ given by $$\begin{align} p^\ast(n) &= \Pr[X=n\mid \text{fair}]\cdot\Pr[\text{fair}]+\Pr[X=n\mid \text{biased}]\cdot\Pr[\text{biased}] \\ &= \frac{1}{2}\left(\Pr[X=n\mid \text{fair}]+\Pr[X=n\mid \text{biased}]\right) \\ &= \frac{1}{2}\left(p_f(n) + p_b(n)\right) \end{align}$$ as you suggested.
Edit: this follows from the Law of Total Probability. If $A_1,\dots, A_m$ are events that partition the space of outcomes, then for any event $E$, $$ \Pr[E] = \sum_{k=1}^m \Pr[E\cap A_k] = \sum_{k=1}^m \Pr[E\mid A_k]\cdot\Pr[A_k]. $$