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# Queens and Diamond

Author Message
TAGS:
Manager
Joined: 29 May 2008
Posts: 113
Followers: 1

Kudos [?]: 33 [0], given: 0

Queens and Diamond [#permalink]  16 Jun 2009, 05:25
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
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

Current Student
Joined: 13 May 2008
Posts: 141
Schools: LBS
Followers: 4

Kudos [?]: 24 [1] , given: 6

Re: Queens and Diamond [#permalink]  16 Jun 2009, 06:05
1
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
Followers: 1

Kudos [?]: 33 [0], given: 0

Re: Queens and Diamond [#permalink]  16 Jun 2009, 07:18
Absolutely now I got it
Current Student
Joined: 13 May 2008
Posts: 141
Schools: LBS
Followers: 4

Kudos [?]: 24 [1] , given: 6

Re: Queens and Diamond [#permalink]  16 Jun 2009, 07:39
1
KUDOS
i don't get no kudos!!!
Manager
Joined: 16 Apr 2009
Posts: 246
Schools: Ross
Followers: 2

Kudos [?]: 38 [0], given: 10

Re: Queens and Diamond [#permalink]  19 Jun 2009, 20:27
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: 212
Followers: 1

Kudos [?]: 25 [0], given: 9

Re: Queens and Diamond [#permalink]  19 Jun 2009, 21:37
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
_________________

Director
Joined: 04 Jan 2008
Posts: 919
Followers: 57

Kudos [?]: 244 [0], given: 17

Re: Queens and Diamond [#permalink]  19 Jun 2009, 22:15
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!

_________________
Re: Queens and Diamond   [#permalink] 19 Jun 2009, 22:15
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# Queens and Diamond

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