I was watching this video on Khan Academy about condition probability where they demonstrated a problem using a tree. I tried to solve that problem using Bayes rule, but my answer doesn't match the one presented in the video.
The problem: There are $2$ fair coins and 1 biased coin. The biased coin has a $2/3$ probability of landing a heads. All three coins are put in a bag. One coin is picked out of the bag at random and flipped. The result is Heads. What is the probability that the coin picked out of the bag was biased?
The result reported on the video is $4/10$.
I tried to use Bayes rule on this, but this is what I have come with so far…
Let $A$ be the event that the coin picked out of the bag is biased
Let $B$ be the event that a coin flip results in heads
$P(A) = 1/3$ because there are three coins and only one biased coin
$P(B\mid A) = 2/3$ because the biased coin has a $2/3$-rd chance of landing a heads
$P(B) = ?$
$P(A\mid B) = ?$
They're asking for $P(A\mid B)$. I can't figure out $P(B)$. How do I proceed?
Best Answer
$$P(B\mid A)\times(P(A)=P(A\cap B)=P(A\mid B)\times P(B)\\ P(A\mid B)=\frac{2}{3}\frac{1}{3}\frac{1}{P(B)}=\frac{2}{9P(B)}\\ P(B)=P(B\mid A)\times P(A) + P(B\mid A^C)\times P(A^C) = \frac{1}{9}+\frac{1}{3}=\frac{4}{9}\\ P(A\mid B) = \frac{2}{9\times \frac{4}{9}}=\frac{1}{2} $$
Hope it's clear enough.