Given two coins with $P_1(H) = \frac{1}{2}$ and $P_2(H) = \frac{1}{3}$, one is chosen at random and tossed $5$ times, what's the probability of tossing at least $3$ heads?
If $A$ is the probability of getting at least three heads, then $P(A) = P(3) + P(4) + P(5)$ where $P(i)$ is the probability of getting $3, 4,$ and $5$ heads exactly respectively.
I know see that $P(3) = C_3^5P(H)^3(1-P(H))^{5-3}$, but this depends on the coin being tossed.
I also know that $P(3) = P(3 | B_1)P(B_1) + P(3 | B_2)P(B_2)$ for some partition ${B_1, B_2}$. But what would be an appropriate partition here?
Best Answer
Rather than determining the probability of getting $n$ heads as a function of the probability of heads, and then conditioning the probability of heads on which coin you chose, swap the order.
In other words, assume the probability of choosing coin $1$ or coin $2$ is $1/2$ either way. Then, given you choose coin $1$, the probability of getting at least $3$ heads is trivially $1/2$ (do you see why?). Given you choose coin $2$, the probability of getting at least $3$ heads is smaller, but (as you point out) can be obtained as
$$ P(\text{at least $3$ heads $\mid$ coin $2$}) = \sum_{k=3}^5 \binom{5}{k} \left(\frac{1}{3}\right)^k \left(\frac{2}{3}\right)^{5-k} $$
This should be straightforward to compute. Then we can compute
\begin{align} P(\text{at least $3$ heads}) & = P(\text{at least $3$ heads $\mid$ coin $1$}) P(\text{coin $1$}) \\ & + \, P(\text{at least $3$ heads $\mid$ coin $2$}) P(\text{coin $2$}) \end{align}