Bayes Theorem for coins problem

[Adyssa]

Homework Statement

I have a bag of n coins, and 1 is fake - it has 2 heads.

a) Determine the probability that if I flip a coin and it comes up heads, the coin is fake.
b) If a coin is flipped k times and comes up heads k times, what is the probability that the coin is fake?

Homework Equations

Bayes Theorem: P(A|B) = [P(B|A) P(A)] / P(B)

The Attempt at a Solution

For part a) I think I have this correctly.

P(fake) = 1/n
P(!fake) = (n-1) / n
P(heads) = P(heads|fake) P(fake) + P(heads|!fake) P(!fake)
= 1 * 1/n + 1/2 * (n-1) / n
= 1/n + (n-1) / 2n
= 2/2n + (n-1) / 2n
= (n+1) / 2n

P(fake|heads) = [P(heads|fake) P(fake)] / P(heads)
= 1 * 1/n / (n+1) / 2n
= 1/n / (n+1) / 2n
= 1/n * 2n / (n+1)
= 2n / n(n+1)
= 2 / (n+1)

For part b), I'm a little stuck. I think the probability of the coin coming up heads k times = P(heads)^k or [(n+1) / 2n]^k. My reasoning here is that ignoring the fake coin the probability of getting 5 heads in a row would be 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = (1/2)^k.

So P(fake|(k heads)) = [P((k heads) | fake) * P(fake)] / P(k heads)
and P((k heads) | fake) = 1, so

P(fake|(k heads)) = [1 * 1/n] / [(n+1) / 2n]^k
= [1/n] / [(n+1) / 2n]^k
= 1/n * [2n / (n+1)]^k
= 1/n * [2n^k/(n+1)^k]
= 2n^k / n(n+1)^k

But I'm unsure of my reasoning.

 

[clamtrox]

I think the point is that you flip the same coin k times and it comes up heads k times. So the probability of getting that from a real coin is P(k heads|!fake) = 1/2^k, and P(k heads|fake)=1 . Otherwise the calculation is exactly the same.