RE: [Math4u] Re: probability problem

Alternate calculation, same conclusion:

 

1st card can be anything,

2nd card has 1/11 chance of being a pair => 1/11

Assume not, 10/11: 3rd card has 2/10 chance of being a pair => (10/11) * (2/10) = 2/11

Assume not, (10/11)*(8/10): 4th card has 3/9 chance of being a pair => (10/11)*(8/10)*(3/9) => 8/33

 

Add:

(1/11)+(2/11)+(8/33) = (3+6+8)/33 = 17/33

 

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From: Math4u@yahoogroups.com [mailto:Math4u@yahoogroups.com] On Behalf Of w7anf
Sent: 29 September 2007 16:15
To: Math4u@yahoogroups.com
Subject: [Math4u] Re: probability problem

495 ways to select 4 cards from 12.
255 ways to select a pair.
255/495=17/33
None of the answers are correct.
Jim FitzSimons

--- In Math4u@yahoogroups.com, "Alex Chandra" <mbs@...> wrote:
>
> Bill has a small deck of 12 playing cards made up of only 2 suits
of 6
> cards each. Each of the 6 cards within a suit has a different
value
> from 1 to 6; thus, there are 2 cards in the deck that have the
same value.
>
> Bill likes to play a game in which he shuffles the deck, turns
over 4
> cards, and looks for pairs of cards that have the same value. What
is
> the chance that Bill finds at least one pair of cards that have the
> same value?
>
> a. 8/33
> b. 62/165
> c. 14/33
> d. 103/165
> e. 25/33
>

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