Your friend rolls either one, two, or three dice (
n=1, n=2 or n=3). Each die is a normal cube with six sides, displaying a number between 1 and 6. She doesn't tell you what
n is, but tells you that the sum of the numbers on the dice is 7.
For example, she could have rolled 4 and 3 with
n=2; or perhaps 5, 1, and 1 with
n=3 etc. Obviously,
n cannot be equal to one.
What are the odds of
n=2 v/s
n=3 given that the sum is 7?
Answer coming up in a week, but this is an example of simple Bayesian analysis.
Credits: picture from
http://www.pwcphoto.com/studio/studio-07.htm.
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