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From: leimkuhl@uiuccsb.UUCP
Newsgroups: net.math
Subject: Re: simple (?) statistics problem solved - (nf)
Message-ID: <2633@uiucdcs.UUCP>
Date: Fri, 26-Aug-83 05:29:15 EDT
Article-I.D.: uiucdcs.2633
Posted: Fri Aug 26 05:29:15 1983
Date-Received: Sat, 27-Aug-83 19:46:27 EDT
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#R:rabbit:-181400:uiuccsb:9700001:000:1650
uiuccsb!leimkuhl    Aug 25 12:50:00 1983

/***** uiuccsb:net.math / grkermit!larry / 11:48 am  Aug 19, 1983 */
>From Laura@utcsstat:
To begin with, you have an equal chance of picking a 2-gold, a 2-silver or a
1-of-each cabinet. You eliminate one of these when you make your choice. I
toss this one out of the window and say that you have a 50% chance of either
cabinet. The answer says that I should not have tossed the other cabinet
out of the window. Why is that silly cabinet relavant to the question?

~~~~~~~~~~~~~~~~

here's why it works.  First, lets define exactly what the probability
is.  IF we were to do this experiment 100 times, and every time we get
a gold coin, we check to see what cabinet we are in, the percentage of
the time that wee choose the 2 gold cabinet is the probability that
that is the cabinet on any given time.

So, there a three gold coins.  Since we have an equal chance of
choosing any one of them, if we do choose one it is equally likely that
it is the one in the silver gold cabinet, or that it is one of the two
in the gold gold cabinet.  Since the odd that it is any particular one
given that we have already chosen a gold one is 1/3, the odds that it
was in the gold gold cabinet is the sum of the probabilities of all the
coins in that cabinet, 1/3 + 1/3 = 2/3.

Think of it like this, since there are twice as many gold coins in one
cabinet as the other, you are going to choose gold coins out of that
cabinet twice as often.  Since there are 3 coins, 2 out of every three
will be chosen from the gold gold cabinet, thus 2/3.

-- 
Larry Kolodney (The Devil's Advocate)
{linus decvax}!genrad!grkermit!larry
(ARPA)  rms.g.lkk@mit-ai
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