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A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not).

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    $\begingroup$ What do you think? Where are you stuck? $\endgroup$ – cardinal Aug 7 '11 at 20:02
  • $\begingroup$ should I consider this problem as independent of previous toss or not? $\endgroup$ – rohit Aug 7 '11 at 20:17
  • $\begingroup$ Hint: Consider how you'd answer if at least one of the previous tosses had been a tail. $\endgroup$ – cardinal Aug 7 '11 at 20:19
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Alternately, you could fully build the tree. What is the probability that you picked a fair coin? What is the probability that it shows heads five times in a row? The unfair coin? Five times in a row?

And don't forget conditional probability.

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Hint: what is the total probability that you will get 5 heads in a row starting from scratch? How much of that comes from the double head coin?

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The question can be easily answered if you can determine the probability that the coin you chose is the double-headed coin. For this, you want to use Bayes' theorem for inverting conditional probabilities:

$$P(B|A)=\frac{P(A|B)P(B)}{P(A)}=\frac{P(A|B)P(B)}{P(A\cap B)+P(A\cap B^c)}=\frac{P(A|B)P(B)}{P(A|B)P(B)+P(A|B^c)P(B^c)}$$

Here, we want $A$ to be the event that we get $5$ heads in a row, and $B$ the event that we have the double-headed coin. All of the quantities in the final formula can be determined.

If you don't have experience applying Bayes' theorem (or even if you do), it might be worthwhile to followed mixedmath's suggestion of building a probability tree and seeing how the terms in the formula come from the nodes of the tree.

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There's a 1/5 chance that you chose the two-headed coin. If you did, then the probability that you'd get five heads in a row is 1.

There's a 4/5 chance that you chose the normal coin. If you did, then the probability that you'd get heads five times in a row would be just $(\frac{1}{2}) ^ 5$.

Using Bayes' Theorem, the probability that you chose the two-headed coin is $\frac{(\frac{1}{5}) \times 1}{(\frac{1}{5}) \times 1 + (\frac{4}{5})(\frac{1}{2}) ^ 5}$, or about 0.889. The probability that you chose a normal coin is $1 - 0.889 = .111$ (there are only two options).

Since there's a 0.889 probability that you have a two-headed coin, you start of with a 0.889 probability that the next flip is heads. If the coins is normal, there's still a .5 probability that it will be heads ($0.111 \times .5 = 0.0555$). Your final probability of getting heads on the next flip is $0.889 + 0.0555 = 0.944$.

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