Probability of one biased and two fair coins

Question

You have three coins. Two of them are ‘fair’ while one of them is biased in that, for the biased coin, P{Head} = 2/3 and P{Tail} = 1/3. All three coins look alike, so that you don’t know a priori which coin is biased and which is not. You flip three coins one by one and observe {Head, Head, Tail}.Based on the observation {Head, Head, Tail}, find the probability that the first coin is biased.

I know the solution uses Bayes Theorem but i am not aware of how to find probability that the first one is biased. Any help is appreciated.

Answer 1

Let $F_i$ be the event that $i$th coin was biased, and $Q$ be the event that we observe $(H,H,T)$. Then, by Bayes' rule,

$$ \begin{split} P(F_1|Q) &= \frac{P(Q|F_1)P(F_1)}{\sum_{i=1}^3 P(Q|F_i)P(F_i)}\\ &=\frac{P(Q|F_1)}{\sum_{i=1}^3 P(Q|F_i)}\\ &=\frac{\frac{2}{3}\cdot\frac{1}{2}\cdot\frac{1}{2}}{2\cdot\frac{2}{3}\cdot\frac{1}{2}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{3}}\\ &=\frac{2}{5} \end{split} $$

Answer 2

A coin that lands Head is twice more likely than a coin that lands Tail to be the biased one.

Let $R_i$ denote the result of flipping coin #$i$, and $P_i$ denote the probability that it is biased:

So the answer to your question is $\dfrac{2}{5}$.