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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 $C$: $$\frac1{15} \cdot \frac12 = \frac1{30}$$

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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 $$ \frac{1/30}{1/12 + 1/20 + 1/30} = \frac{1}{5/2+3/2 + 1} = \frac15. $$

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

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

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