|Nov7-05, 09:21 AM||#1|
I have a short question from a homework. given a family with 3 children, the sexes of which are unknown: if the eldest child is a boy and there is at least one other boy, then what is the probability one child is a girl?
To me it seems obvious that there are at least two boys from this info, then either there are 3 boys or 2 boys and 1 girl, so the probability one child is a girl is 0.5. Is this method correct?
|Nov7-05, 11:29 AM||#2|
A family with three children and the eldest is a boy. Writing B for boy, G for girl, from oldest to youngest, the possibilities are:
Assuming "boy" or "girl" are equally likely at each birth, these are equally likely.
But we are told that at least one child is a boy: that throws out BGG leaving
BGB and they are still equally likely.
Since 2 of those 3 correspond to "one child is a girl", the probability that one of the children is a girl is 2/3, not 1/2. Yes, it is true that there must be "either 3 boys or 2 boys and a girl", but those are NOT "equally likely".
|Nov19-05, 03:09 PM||#3|
As far as I'm concerned you should better explain the underlying reason: probability of each sex child birth is fifty-fifty. In a formal proof is often request.
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