1. Joined
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    25 Jun '04 21:57
    1) In a two-child family, one child is a boy. What is the probability that the other child is a girl?


    2) In a two-child family, the older child is a boy. What is the probability that the other child is a girl?

    .
  2. Standard memberCribs
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    25 Jun '04 22:251 edit
    Originally posted by THUDandBLUNDER
    1) In a two-child family, one child is a boy. What is the probability that the other child is a girl?


    2) In a two-child family, the older child is a boy. What is the probability that the other child is a girl?

    .
    In case 1, 2/3.
    In case 2, 1/2.

    Cribs
  3. Standard memberopsoccergurl11
    rockin soccer kid
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    25 Jun '04 23:06
    1/2 for both
  4. Standard memberNemesio
    Ursulakantor
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    25 Jun '04 23:58
    Originally posted by Cribs
    In case 1, 2/3.
    In case 2, 1/2.

    Cribs
    Ow ow ow...

    Wait.

    The combinations are (caps for older, lower case for younger):

    Bb, Bg
    Gb, Gg

    In case one, our choices could be:

    Bb, Bg, or Gb, thus 2/3.

    In case two, our choice could be:

    Bb, or Bg, thus 1/2.

    Ahhhhh.

    My head hurts less...

    But still a little.

    Nemesio
  5. Standard memberopsoccergurl11
    rockin soccer kid
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    26 Jun '04 01:33
    but no matter what the propbability of boy or girl is 1/2, no matter what the other child is...am i missing something?
  6. Standard memberNemesio
    Ursulakantor
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    26 Jun '04 04:20
    A friend of mine sent me this.

    http://mathforum.org/dr.math/faq/faq.boy.girl.html
  7. Joined
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    27 Jun '04 08:332 edits
    Being a backgammon player, I prefer the puzzle in the following format:

    The probability of rolling a 7 with a pair of ordinary dice is 1/6 because, of the 36 possible (and equally likely) combinations, six of them sum to 7. Now, suppose the dice are rolled out of sight, and some honest person who can see the results tells us that at least one of the dice came up 6. This restricts the total number of possible combinations to the following eleven pairs:

    (1,6) (2,6) (3,6) (4,6) (5,6) (6,6) (6,1) (6,2) (6,3) (6,4) (6,5)

    of which exactly two sum to 7. So we would now assess the probability of a 7 as 2/11, just slightly better than 1/6. Of course, the same analysis would give a probability of 2/11 if our honest friend had reported "at least one 5" instead of "at least one 6". Or if he had reported "at least one 4". In fact, we would arrive at the same probability if he reported "at least one n" for ANY value of n. Obviously we have "at least one n" for SOME value of n, so why not just assume this without waiting for our friend to tell us? It would seem that this magically improves our a priori odds of rolling a seven from 1/6 to 2/11.

    .


  8. Standard memberopsoccergurl11
    rockin soccer kid
    over there
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    27 Jun '04 15:41
    i dont play backgammon, but i understand what u are getting at. so in a sense, this is also like the prisoner problem - where knowing precisely what instead of one or the other seems to change the probability of the same thing
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