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Five cards lie face down on a table; exactly 3 of them have

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Manager
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Five cards lie face down on a table; exactly 3 of them have [#permalink] New post 23 May 2007, 05:01
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Five cards lie face down on a table; exactly 3 of them have "winner" written on the underside. If Rene randomly selects 3 of the cards without replacement, what is the probability that all 3 have "winner" written on the underside?
A. 1/15
B. 1/10
C. 1/5
D. 1/3
E. 3/5
Manager
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Re: probability question [#permalink] New post 23 May 2007, 05:06
Jamesk486 wrote:
Five cards lie face down on a table; exactly 3 of them have "winner" written on the underside. If Rene randomly selects 3 of the cards without replacement, what is the probability that all 3 have "winner" written on the underside?
A. 1/15
B. 1/10
C. 1/5
D. 1/3
E. 3/5


(B) it is

(3/5)*(2/4)*(1/3) = 1/10
VP
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 [#permalink] New post 23 May 2007, 05:15
kirakira is right !

the probability for choosing the first card out of 5 = 3/5

the probability for choosing the second card out of 4 = 2/4

the probability for choosing the last card out of 3 = 1/3

since all those events are dependent then:

3/5*2/4*1/3 = 6/60 = 1/10

:-D
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 [#permalink] New post 23 May 2007, 06:03
Probability = 3C3/5C3 = 1/10
  [#permalink] 23 May 2007, 06:03
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Five cards lie face down on a table; exactly 3 of them have

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