# Conditional Probability Question given 3 Coins

I had this question as a bonus problem on a previous exam and thought it was interesting, but I had no idea how to tackle it.

There are three coins in a box. One is a two-headed coin, another is a fair coin, and the third is a biased coin that comes up heads 75% of the time. When one of the three coins is selected at random and flipped, it shows heads. What is the probability that it was the first coin?

• This is a problem for Bayes' theorem. No more no less. – Makina Dec 7 '18 at 0:46

One possibility to think about it "in a very naive way" is as follows. The coins are numbered as 1, 2, 3 - say. Instead of the given experiment, we can imagine an equivalent one. Consider a hat with $$12$$ tickets in it, labeled as $$1H,\ 1H,\ 1H,\ 1H;\ 2H,\ 2H,\ 2T,\ 2T;\ 3H,\ 3H,\ 3H,\ 3T.$$ Choosing a coin $$k$$ has the same probability as choosing a ticket $$k?$$ from the hat. For each fixed $$k$$, getting head (H) for the coin $$k$$ has the same probability as picking the ticket $$kH$$.

Now we pass to the conditional probability. We have only

$$\color{red}{1H,\ 1H,\ 1H,\ 1H};\ 2H,\ 2H;\ 3H,\ 3H,\ 3H$$ at our disposal. Picking $$1H$$ has then probability $$\frac 49\ .$$

Bayes' theorem approach:

$$P(A_1), P(A_2)$$ and $$P(A_3) = \frac{1}{3}$$ - choice of coins

$$P(B)$$ = get heads

You are asked a question to find $$P(A_1 | B)$$

$$P(A_1 | B) = \frac{P(A_1)*P(B | A_1)}{P(B)} = \frac{P(A_1)*P(B | A_1)}{P(B | A_1)*P(A_1) + P(B | A_2)*P(A_2) + P(B | A_3)*P(A_3)} = \frac{4}{9}$$