# Probability of choosing a biased coin $C$ which has probability $3/15$ of getting heads, assuming we got head on the first toss

Full question: there are 3 biased coins $$A$$, $$B$$, and $$C$$ each with probability $$5/15, 3/15, 1/15$$ of getting heads respectively. Also, they have probability $$1/4$$ for $$A$$, $$1/4$$ for $$B$$, and $$1/2$$ for $$C$$ of getting picked. If a coin was picked and tossed and the result was heads, what is the probability that the coin was coin $$C$$?

My approach: Since the coin picked was $$C$$ and the result was head we merely multiply the probability of both those things happening concerning $$C$$: $$\frac1{15} \cdot \frac12 = \frac1{30}$$

An intuitive argument based on Bayes' Theorem, says that getting heads was possible in one of three different ways:

1. Draw $$A$$ with probability $$1/4$$ and flip heads with probability $$5/15$$, total chance of $$1/4 \times 5/15 = 1/12$$.
2. Draw $$B$$: $$1/4 \times 3/15 = 1/20$$.
3. Draw $$C$$: $$1/2 \times 1/15 = 1/30$$.

Thus, the chance that $$C$$ was drawn is $$\frac{1/30}{1/12 + 1/20 + 1/30} = \frac{1}{5/2+3/2 + 1} = \frac15.$$

• Thank you, this makes a lot of sense. – hussain sagar Dec 24 '18 at 4:57
• @hussainsagar you are welcome – gt6989b Dec 24 '18 at 4:58

What you computed is the probability that coin $$C$$ is chosen and you get a head $$H$$, that is $$P(H \cap C)$$. It is not equalty to $$P(C|H)$$.

Guide: Use Bayes rule, that is

$$P(C|H)= \frac{P(H|C)P(C)}{P(H)}=\frac{P(H|C)P(C)}{P(H\cap A)+P(H \cap B)+P(H \cap C)}$$