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dumb probability question

  
 
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Numbr2intheWorld
Old 02-17-2008, 09:49 PM     Post subject: dumb probability question #1 (permalink)  
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 Help my friend out.

So there is a website, in which they generate a stat and find out that 40% of visitors are women and 60% are male. They are tracking a customer (who visited the page) who's last web page stop was dillards.com, a site where it is 3 times as likely that a women visits than a man.

what is the revised probabilty that the customer being tracked is a woman?

please explain the logic.
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mcatdog
Old 02-17-2008, 10:17 PM #2 (permalink)  
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Assuming that visiting the two websites are independent events it'd be

(.75)(.4) / ( (.75)(.4) + (.25)(.6) ) = 2/3

But you can't necessarily assume independence, for example if only 4 people have ever visited the Dillard's website (3 women and 1 man) and only one of those 4 people has ever visited your friend's website then the answer wouldn't be 2/3.
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swiggidy
Old 02-17-2008, 10:18 PM #3 (permalink)  
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 It seems to me you would need a joint probability that someone visits both sites.

I'm tired so I'm probably missing something (and I'll about it later).
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Ragnar4
Old 02-17-2008, 10:20 PM #4 (permalink)  
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And here I wanted to simply average the probability that a woman had hit both sites against eachother bringing a 55%.
The older I get, the more I start wondering; Just what in the hell is going on here?
 
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Numbr2intheWorld
Old 02-17-2008, 10:25 PM #5 (permalink)  
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Quote:
Originally Posted by mcatdog
Assuming that visiting the two websites are independent events it'd be

(.75)(.4) / ( (.75)(.4) + (.25)(.6) ) = 2/3

But you can't necessarily assume independence, for example if only 4 people have ever visited the Dillard's website (3 women and 1 man) and only one of those 4 people has ever visited your friend's website then the answer wouldn't be 2/3.
this is correct thank you.
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Numbr2intheWorld
Old 02-17-2008, 10:26 PM #6 (permalink)  
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 so then is the equation P(A and B)/ P( A and B)+ P (not A and B)?
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DrivingDog
Old 02-17-2008, 11:05 PM #7 (permalink)  
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I think you basically have it right. The facts are that someone visited both sites and their gender presumably did not change in between visits. Thus, the question is how likely is it that a woman visits both sites. To find this you use the probability that it was a woman visiting both sites relative to the probability that it was a man visiting both sites



pW = (pWD * pWO)/ (pWD * pWO + pMD * pMO)

= (.75 * .4) /(.75 * .4 + .25 * .6)

=.67 (what mcatdog said)


where:
p is probability
W is woman
M is man
D is dillard's
O is the other website

conversely,

pM = (pMD * pMO) / (pMD * pMO + pWD * pWO)
= (.25 * .6 ) / (.25 * .6 + .75 * .4)
= .33

since pM and pW are mutually exclusive, they should add up to 1 (and they do: .67 = .33 = 1)
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