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Topic: probability question
Replies: 9   Last Post: Mar 13, 2008 4:08 PM

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quasi

Posts: 12,057
Registered: 7/15/05
Re: probability question
Posted: Mar 12, 2008 6:25 AM
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On Wed, 12 Mar 2008 00:33:19 -0700, The World Wide Wade
<aderamey.addw@comcast.net> wrote:

>In article <b1set3l134nn2kt3buh893q8jjfg3kau6s@4ax.com>,
> quasi <quasi@null.set> wrote:
>

>> On Tue, 11 Mar 2008 21:56:27 EDT, Steven <sgottlieb60@hotmail.com>
>> wrote:
>>

>> >Suppose you meet me on a street corner and I introduce you to my son who is
>> >with me. I inform you that I have another child at home. What is the
>> >probability that my other child is a girl.

>>
>> The problem is not adequately specified.
>>
>> It depends on how the child accompanying the father is selected.
>>
>> If the child that accompanies the father is selected at random by a
>> flip of a fair coin, then the probability that the other child is a
>> girl is 1/3.

>
>The sample space for the children is (b b), (b g), (g b), (g g) where
>the first slot is the youngest child, the second slot is the oldest.
>These oredered pairs all have probability 1/4. Now we select a child
>at random for a walk. We get a new sample space: (b b b), (b g g), (b
>g b), (g b g), (g b b), (g g g), with the probabilities being 1/4 for
>the first and last triples, and 1/8 for the others. The probability
>the other child is a girl given the randomly selected child out with
>daddy is a boy is thus
>
>p((b g b) (g b b))/p((b b b) (b g b) (g b b))
>
> = (1/8 + 1/8)/(1/4 + 1/8 + 1/8) = 1/2.


Even without calculation, I should have realized my error based on the
following intuitive idea ...

If there is no gender bias in the method by which the child who went
with the father is selected, then there can be no gender bias for the
child who wasn't selected.

quasi



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