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Queens and Diamond

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Manager
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Queens and Diamond [#permalink] New post 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 ?
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Re: Queens and Diamond [#permalink] New post 16 Jun 2009, 07:05
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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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Re: Queens and Diamond [#permalink] New post 16 Jun 2009, 08:18
Absolutely now I got it
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Re: Queens and Diamond [#permalink] New post 16 Jun 2009, 08:39
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i don't get no kudos!!! :cry: :cry: :cry:
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Re: Queens and Diamond [#permalink] New post 19 Jun 2009, 21:27
One kudo for MJ2009...
Your post helped me a lot...
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Manager
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Re: Queens and Diamond [#permalink] New post 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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Re: Queens and Diamond [#permalink] New post 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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Re: Queens and Diamond   [#permalink] 19 Jun 2009, 23:15
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