Re: Loads of Doubts.

Posted by NM on Jun 12, 2014; 9:20pm
URL: http://discussion-forum.276.s1.nabble.com/Loads-of-Doubts-tp7592546p7592564.html

I am not using Bayes Theorem. This is a simplified procedure.

We need to calculate Pr(Coin being the fair one given the occurrence of 2 heads) = Pr(F/H) = (Pr(F intersection H)/Pr(H)  

The probability of  choosing the fair coin = 1/2
The probability of obtaining 2 heads in a row = 1/4
Hence, Pr(F intersection H) = 1/2 * 1/4 = 1/8

Now, Pr(H) = 1/2 * (1/2 * 1/2) + 1/2 * 1 = 5/8

Hence, Pr(F/H) = (1/8)/(5/8) = 1/5.