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A box contains n number of coins, m of which are fair and 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 head and second time it shows tail. What is the probability that the coin drawn is fair.

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3 Answers 3

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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it's Bayes (or Bayes') theorem, from Thomas Bayes; not Baye's –  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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