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- Path: sparky!uunet!utcsri!torn!news.ccs.queensu.ca!qucdn!boydj
- Organization: Queen's University at Kingston
- Date: Tue, 15 Sep 1992 18:24:02 EDT
- From: Jeff Boyd <BOYDJ@QUCDN.QueensU.CA>
- Message-ID: <92259.182402BOYDJ@QUCDN.QueensU.CA>
- Newsgroups: sci.math.stat
- Subject: Re: two pairs
- Lines: 35
-
-
- Does anyone else here feel that counting is akin to witchcraft?
-
- I took a slightly different approach to this one, treating each pair
- -AND- the odd card as selections from sets of 4. The result is the
- same, but I think it's a bit more rigid in its logic.
-
-
- 1) select 3 of 13 sets of 4 (two pair slots, one singleton slot)
-
- 13c3
-
- 2) select 1 pair from one of the 3 sets of 4
-
- 4c2
-
- 3) select 1 pair from another of the 3 sets of 4
-
- 4c2
-
- 4) select 1 singleton from the last of the 3 sets of 4
-
- 4c1
-
- 5) select 2 of the 3 sets of 4 for the pair slots, -OR-
- select 1 of the 3 sets of 4 for the singleton slot
-
- 3c2 or 3c1
-
- Then
-
- P[2 pair] = (13c3)(4c2)(4c2)(4c1)(3c1) / (52c5)
-
- = 0.047539
-
-