A jar containing 2011 fair coins also contains one tricked coin (it has heads on both sides). A coin is randomly selected from this jar and flipped 2012 times. It comes up heads each time. What is the probability that this was one of the fair coins?

Question

rsadhvika · Answer

$(2010/2011)*(1/2^{2012})$

?

anonymous · Answer

It's supposed to be that, but how do you get that?

rsadhvika · Answer

(Probability of picking fair coin, while selecting randomly) * (probability of head) * 2012

rsadhvika · Answer

(Probability of picking fair coin, while selecting randomly) * (probability of head) ^ 2012

anonymous · Answer

Oh ok. So:
Probability of picking fair coin = $\frac{2011}{2012}$
Probability of heads = $\frac 12$
That's how it works?

rsadhvika · Answer

yes

anonymous · Answer

Alright. Why is it to the 2012 power though? I get that's the max number of coins, but why would it be raised to it?

rsadhvika · Answer

you are doing the experiment 2012 times

P(first time heads) = 1/2
P(first two times heads) = 1/2 * 1/2
P(first n times heads) = 1/2 * 1/2 *...... n times

rsadhvika · Answer

tossing a coin has two possible outcomes
each has a probability of 1/2

rsadhvika · Answer

if you perform the experiment n times probability is 1/2^n

anonymous · Answer

Oh. I see. Thank you :)

rsadhvika · Answer

welcome :)