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%