Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP
Posting-Version: version B 2.10.1 6/24/83; site dciem.UUCP
Path: utzoo!dciem!ntt
From: ntt@dciem.UUCP (Mark Brader)
Newsgroups: net.math
Subject: Re: New Probability problems: answers
Message-ID: <414@dciem.UUCP>
Date: Wed, 28-Sep-83 17:24:15 EDT
Article-I.D.: dciem.414
Posted: Wed Sep 28 17:24:15 1983
Date-Received: Thu, 29-Sep-83 04:09:58 EDT
References: <516@houxz.UUCP>, <1210@stolaf.UUCP>
Organization: NTT Systems Inc., Toronto, Canada
Lines: 11

The answer previously posted to the "lost club?" problem is wrong;
stolaf!koch (Paul Koch) has the right result, but there is a much
easier way to obtain it.
	5 cards were drawn and are known to be clubs.  The remaining
47 cards including the lost card contain 8 clubs.  We have no information
about which of the 47 cards was lost.  Therefore the probability that the
lost card is a club is 8/47 (or .1702 for those who prefer approximations).
The fact that the card was lost before the 5 cards were drawn is irrelevant,
because we did not see it and therefore have no information about it.

Mark Brader, NTT Systems Inc., Toronto