Let       C = Event of 2 boys

            D = Event told at least 1 is a boy

Q:       

            P[C|D] = P[CD]/P[D] = P[C] P[D|C] / P[D]

Note:

            P[C|D] = P[2 boys | told at least 1 boy]

             P[D|C] = P[told at least 1 boy | 2 boys] = 1

So:

            P[CD] = P[C] = P[of 2 boys]

           

Now:    P[D] = P[D and 2 boys]

                        + P[D and first child is a boy and 2nd is a girl]

                        + P[D and first girl, 2nd boy]

But

            P[D and 2 boys]=P[2 boys]=1/4

 and

             P[D and first child is boy and 2nd is girl]=P[D and bg]=P[bg]P[D|bg] =(1/4)P[D|bg]

and

             P[D and first child is girl and 2nd is boy]=P[gb]P[D|gb]=(1/4)P[D|gb].

 Then

            P[D] totally hinges on P[D|bg] and P[D|gb].

 

Therefore: the answer hinges on:

 

P[told at least 1 is a boy | bg] 

and 

P[told at least 1 is a boy | gb]

 

If these values are 1, i.e. always say it is a boy, the answer is 1/3. (call this extreme prejudice) If this probability is 1/2 (call this no prejudice i.e. what you assume if not told different) then the answer is 1/2.

 

Notice that getting the 1/3 answer for our woman is a function of whether or not there is a prejudice; it is not a function of whether or not we have the “at least one is” statement.

This math was done by Dr. H.L. Gray, Professor of Mathematical Statistics at SMU. Dr. Gray suggested the term, "extreme prejudice". I had erroneously referred to this as, "a Bayes' solution. Dr Terry Moore pointed out that this is not Bayes'.

 

 

 

 

 

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