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A box contains $n$ number of coins, $m$ of which are fair and the rest are biased. The probability of getting a head when a fair coin is tossed is $1/2$, while it is $2/3$ when a biased coin is tossed. A coin is drawn from the box at random and is tossed twice. The first time it shows a head and the second time it shows tail. What is the probability that the coin drawn is fair.

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What have you done so far? Hint: Without the flips, what is the chance you draw a fair coin? If you draw a fair coin, what is the chance you get the flips specified? If you draw a biased coin, what is the chance?

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Let

  • $A$ = Even that coin selected is fair

  • $B$ = Coin selected is biased

  • $C$= Even that first results in a head, second results in a tail.

Now use Bayes Theorem to get the answer. Answer is $\frac{9m}{8n+m}$

You have:

  • $P(A)=\frac{m}{n}$ and $P(B)=\frac{n-m}{n}$

  • $P(C/A) = \frac{1}{4}$ and $P(C/B) = \frac{2}{3} \times \frac{1}{3} = \frac{1}{9}$.

  • By Bayes theorem $$P(A/C) = \frac{P(A) \times P(C/A)}{P(A) \times P(C/A) + P(B) \times P(C/B)}=\frac{9m}{8n+m}$$

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    $\begingroup$ it's Bayes (or Bayes') theorem, from Thomas Bayes; not Baye's $\endgroup$ – leonbloy Jan 19 '11 at 20:12
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I know this isn't really answering your question, but a coin toss, is rarely ever fair. And I'm not talking a few thousandths of a percent, but more like 52% vs 48%. Below makes for fascinating reading:

http://www.codingthewheel.com/archives/the-coin-flip-a-fundamentally-unfair-proposition

And remember, always be the tosser!

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