Dear Anjani,

The problem you propose is an interesting classical one. I have used it or similar versions in order to introduce Bayesian statistics. It shows that Bayes' formula is a paradigm of information acquisition through an experiment.

In fact, before the experiment of picking balls from the urn, we have a prior knowledge: there are two urns and they are selected at random, i.e. P[A]=1/2, P[B]=1/2. The experiment gives us the result nb=2 (black's), nw=3 (white's). The key point is to establish the likelihood of experiments A and B given such a result. The likelihood is described by the two probabilities

P[ Nb=nb, Nw=nw | A ] , P[ Nb=nb, Nw=nw | B ] .

They are computed by using the binomial distribution (there is replacement of balls) assuming that nb+nw is the total number of extractions,

P[ Nb=nb, Nw=nw | A ] = Bin(nb+nw,nb) p_A^nb (1-p_A)^nw ,

p_A = 2/3 (prob. of black in A)

P[ Nb=nb, Nw=nw | B ] = Bin(nb+nw,nb) p_A^nb (1-p_A)^nw ,

p_B = 1/3 (prob. of black in B)

Now the Bayes' formula provides the solution after the result of the experiment (posterior)

P[A | Nb=nb, Nw=nw] = C P[ Nb=nb, Nw=nw | A ] P[A]

P[B | Nb=nb, Nw=nw] = C P[ Nb=nb, Nw=nw | B ] P[B]

where C is a constant so that the two a posterior probabilities add up to 1.

There are interesting cases depending on the extractions and the number of black and white balls in each urn. For instance, if the number of black and white balls in both urns are equal, then the experiment does not provide any information as the prior and posterior probabilities are equal, whatever is the observation of balls. Also, if the number of black and white observed balls are equal the result (not the experiment) is not informative.

I like this kind of problems! Thank you!

Hi Anjani. I would use Baye's theorem for this because you are asking for a conditional probability: what is the probability this sample was taken from urn A given we selected 2 black and 3 white balls. Let A denote the event urn A is chosen and C denote the event that 2 black and 3 white balls were selected. We would like to compute P(A|C)=P(C|A)P(A)/P(C). Because we're selecting urns at random then P(A)=0.5. Since there are twice as many black balls as white balls in urn A then P(C|A) = (5 choose 2)(2/3)^2(1/3)^3 by the binomial distribution. And I think P(C)=0.5*(P(C|A)+P(C|!A)) where P(C|!A) is the probability of selecting 2 black and 3 white balls from urn B given by P(C|!A)=(5 choose 2)(1/5)^2(4/5)^3. I get about about 45%. I hope this helps (and is right!).

Dear Anjani,

The problem you propose is an interesting classical one. I have used it or similar versions in order to introduce Bayesian statistics. It shows that Bayes' formula is a paradigm of information acquisition through an experiment.

In fact, before the experiment of picking balls from the urn, we have a prior knowledge: there are two urns and they are selected at random, i.e. P[A]=1/2, P[B]=1/2. The experiment gives us the result nb=2 (black's), nw=3 (white's). The key point is to establish the likelihood of experiments A and B given such a result. The likelihood is described by the two probabilities

P[ Nb=nb, Nw=nw | A ] , P[ Nb=nb, Nw=nw | B ] .

They are computed by using the binomial distribution (there is replacement of balls) assuming that nb+nw is the total number of extractions,

P[ Nb=nb, Nw=nw | A ] = Bin(nb+nw,nb) p_A^nb (1-p_A)^nw ,

p_A = 2/3 (prob. of black in A)

P[ Nb=nb, Nw=nw | B ] = Bin(nb+nw,nb) p_A^nb (1-p_A)^nw ,

p_B = 1/3 (prob. of black in B)

Now the Bayes' formula provides the solution after the result of the experiment (posterior)

P[A | Nb=nb, Nw=nw] = C P[ Nb=nb, Nw=nw | A ] P[A]

P[B | Nb=nb, Nw=nw] = C P[ Nb=nb, Nw=nw | B ] P[B]

where C is a constant so that the two a posterior probabilities add up to 1.

There are interesting cases depending on the extractions and the number of black and white balls in each urn. For instance, if the number of black and white balls in both urns are equal, then the experiment does not provide any information as the prior and posterior probabilities are equal, whatever is the observation of balls. Also, if the number of black and white observed balls are equal the result (not the experiment) is not informative.

I like this kind of problems! Thank you!