Author: Dan Newman
Date: 02:39:43 08/29/98
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On August 28, 1998 at 23:30:44, Serge Desmarais wrote: >On August 18, 1998 at 07:26:14, fca wrote: > >ut I didn't answer (c). > >>I already gave an example of a *bad question* (I repeat, with some >>simplification): >> >>Given: A child may be only a boy (B) or a girl (G). Twins and higher do not >>occur. Bs and Gs are equiprobable. The sex of any existing child does in no >>way affect the probability of the sex of its later-born sibling or half-sibling >>being B or G. >>A man has exactly two biological children. >>Question (1): "One of them is a B. What is the probability that both are Bs?" >>Question (2): "The older of them is a B. What is the probability that both are >>Bs?" >> >>Now the answer to (1) is 1/3, and to (2) is 1/2. > > > > I don't understand how you can say that the answer to 1 is 1/3? If the first >one is a boy, the other child has 50% chance of being a boy and 50% of being a >girl, so that should be 1/2, instead! > > >Serge Desmarais > The reasoning for 1/3 must go something like this: There are four possibilities for two children which we might specify as BB, BG, GB, and GG, each of which is equally probable. Now, when we are told that one is a boy, we must have one of BB, BG, GB (again each of which has the same probability)--hence the 1/3 for BB. Another way of looking at it is to think about the set of all sibling pairs in the world. Aproximately 1/4 will be BB, 1/2 BG or GB, and 1/4 GG. If we toss the GG's out of the set, then the BB's will be 1/3 of the remaining pairs. -Dan. > > > >> >>In computer-chess >> >>Kind regards >> >>fca >> >>PS: Yes I can count. The mathematician was a woman. >> >>;-)
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