An urn contains 6 green balls and an unknown number, which is ≤ 6, of blue balls. Three balls are drawn successively at random, and not replaced and are all found to be blue.
Determine the probability that a green ball will be drawn at the next draw.
I concur with Charlie's analysis.
Just to be clear,
(a) Bayesian analysis can be applied with any a priori distribution, not just uniform
(b) Nothing better than a uniform priori probability distribution is available. In other words, a priori
blue probability
0 1/7
1 1/7
2 1/7
3 1/7
4 1/7
5 1/7
6 1/7
(c) Then the posterior probability distribution is P(T)/Total of all P(T)'s, which calculates to approximately:
3 6.05%
4 16.94%
5 30.80%
6 46.20%