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# Queens and Diamond

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Manager
Joined: 29 May 2008
Posts: 113
Queens and Diamond [#permalink]

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16 Jun 2009, 06:25
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What is the probability of drawing a queen or a diamond from a normal shuffled deck?

a) 2/3
b) 4/13
c) 17/52
d) 16/51
e) 1/4

I can't understand this questions would you please help me ?
Current Student
Joined: 13 May 2008
Posts: 141
Schools: LBS
Re: Queens and Diamond [#permalink]

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16 Jun 2009, 07:05
1
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KUDOS
b) 4/13...

there are 4 Queens and 13 diamonds IN A 52 deck card.. so:

P(queen or diamond) = P(queen) + P(diamond) − P(queen and diamond)
P(queen or diamond) = (4/52) + (13/52) − (1/52)
P(queen or diamond) = 16/52
P(queen or diamond) = 4/13..

hope that helps you understand it!
Manager
Joined: 29 May 2008
Posts: 113
Re: Queens and Diamond [#permalink]

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16 Jun 2009, 08:18
Absolutely now I got it
Current Student
Joined: 13 May 2008
Posts: 141
Schools: LBS
Re: Queens and Diamond [#permalink]

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16 Jun 2009, 08:39
1
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KUDOS
i don't get no kudos!!!
Manager
Joined: 16 Apr 2009
Posts: 233
Schools: Ross
Re: Queens and Diamond [#permalink]

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19 Jun 2009, 21:27
One kudo for MJ2009...
Your post helped me a lot...
_________________

Keep trying no matter how hard it seems, it will get easier.

Manager
Joined: 07 Jun 2009
Posts: 207
Re: Queens and Diamond [#permalink]

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19 Jun 2009, 22:37
MJ2009 wrote:
b) 4/13...

there are 4 Queens and 13 diamonds IN A 52 deck card.. so:

P(queen or diamond) = P(queen) + P(diamond) − P(queen and diamond)
P(queen or diamond) = (4/52) + (13/52) − (1/52)
P(queen or diamond) = 16/52
P(queen or diamond) = 4/13..

hope that helps you understand it!

Good explanation
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"Always....Read between the lines"

Director
Joined: 04 Jan 2008
Posts: 898
Re: Queens and Diamond [#permalink]

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19 Jun 2009, 23:15
Simply awesome
more because of the datailing part
+1 from me as well

MJ2009 wrote:
b) 4/13...

there are 4 Queens and 13 diamonds IN A 52 deck card.. so:

P(queen or diamond) = P(queen) + P(diamond) − P(queen and diamond)
P(queen or diamond) = (4/52) + (13/52) − (1/52)
P(queen or diamond) = 16/52
P(queen or diamond) = 4/13..

hope that helps you understand it!

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http://gmatclub.com/forum/math-polygons-87336.html
http://gmatclub.com/forum/competition-for-the-best-gmat-error-log-template-86232.html

Re: Queens and Diamond   [#permalink] 19 Jun 2009, 23:15
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# Queens and Diamond

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