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A bag contains 1000 coins. 999 are normal, fair coins; 1 coin has 2 heads.

You pick a coin at random from the bag. You flip it 10 times. You get 10 heads.

What is the probability that the coin is the 2-headed coin?

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closed as off-topic by user223391, user284331, Leucippus, Shailesh, samjoe Apr 21 '18 at 7:39

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  • $\begingroup$ Please do not use codeblock for anything that is not actually code. Quoteblock using > or emphasis or strong emphasis using * should be applied where necessary. $\endgroup$ – Nij Apr 20 '18 at 23:10
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Option 1: you picked a fair coin and it came up heads 10 times in a row.

$P = 0.999\cdot \frac {1}{2^{10}}$

Or you picked the fake coin

$P = 0.001$

$\frac {0.001}{0.999\cdot \frac {1}{2^{10}} + 0.001} = \frac {1.024}{0.999 + 1.024} = \frac {1.024}{2.023}$

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You are right about using Bayes' Theorem. Let's systematically assemble the ingredients you need to apply that theorem.

Let $T$ be the event that you pick a two-headed coin; $P(T) = 1/1000.$ Let $F$ be the event that you pick a fair coin; $P(F) = 999/1000.$ Let $E$ be the event of getting ten heads in a row; $P(E|T) = 1$ and $P(E|F) = (1/2)^{10}.$

Bayes' Theorem: $$P(T|E) = \frac{P(T\cap E)}{P(E)} = \frac{P(T\cap E)}{P(T\cap E)+P(F\cap E)}.$$

Now, $P(T \cap E) = P(T)P(E|T) = ?$ And $P(F \cap E) = P(F)P(E|F) = ?$

I'll let you put the pieces together.

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