# Bayes Rule in 2 Fair and 1 Biased coin

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?

• Hint: $P(B)=P (B|A)P (A)+P (B|A^c)P (A^c)$, where $A^c$ denotes the complement of $A$. May 15 '15 at 10:19
• Perfect! I got P(B) = 5/9, and P(A|B) = 2/5, and this matches the result on the video. Very cool! I've never come across the formula you've posted though. What is it? Is it a derivative of Bayes' rule?
– user240753
May 15 '15 at 10:36
• Bayes rule is often stated in alternative form with $P (B)$ written out as such. I feel like the formula is pretty intuitive. What happens is that we partition the probability space into "$A$" and "not $A$" to get that $P (B)=P (B\cap A)+P (B\cap A^c)$. Now use the definition of conditional probability: $$P(X|Y)=\frac{P(X\cap Y)}{P (Y)}$$ to get $P (B)=P (B|A)P (A)+P (B|A^c)P (A^c)$. May 15 '15 at 10:51
• This is awesome. Thank you so much. I cannot find a like/upvote button here, but please know I really appreciate your input. Thanks again.
– user240753
May 15 '15 at 10:54
• @DouglasSangHue $\mathsf P(B)=\mathsf P(B\mid A)\,\mathsf P(A)+\mathsf P(B\mid A^c)\,\mathsf P(A^c)$ is an application of the Law of Total Probability. May 15 '15 at 12:15

$$P(B\mid A)\times(P(A)=P(A\cap B)=P(A\mid B)\times P(B)\\ P(A\mid B)=\frac{2}{3}\frac{1}{3}\frac{1}{P(B)}=\frac{2}{9P(B)}\\ P(B)=P(B\mid A)\times P(A) + P(B\mid A^C)\times P(A^C) = \frac{1}{9}+\frac{1}{3}=\frac{4}{9}\\ P(A\mid B) = \frac{2}{9\times \frac{4}{9}}=\frac{1}{2}$$