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# From an ordinary deck of cards, only 12 picture cards are

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Manager
Joined: 14 Mar 2007
Posts: 235

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From an ordinary deck of cards, only 12 picture cards are [#permalink]

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11 Jun 2007, 10:16
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From an ordinary deck of cards, only 12 picture cards are retained. They are shuffled (mixed) and a man draws 2 cards at random and announces that he holds at least one king. Find the probability that he holds two kings in his hand.

(A) 3/11 (B)4/11 (C)2/9 (D)6/13 (E)3/4

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

CEO
Joined: 17 May 2007
Posts: 2950

Kudos [?]: 654 [0], given: 210

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11 Jun 2007, 10:22
Once the guy has pulled out 1 king, there are 11 cards left. Out of which 3 are kings. So the probability of 2 kings is 3/11.

Option A.

Kudos [?]: 654 [0], given: 210

Intern
Joined: 02 Jun 2007
Posts: 27

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11 Jun 2007, 13:36

If he says he has atleast one king, then he has either 1 king, or 2 kings.

So, probability of having 2 kings given he has atleast 1 is:

4c2*8c0 / ( 4c2*8c0 + 4c1*8c1) = 3/19

This isn't even an option

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Manager
Joined: 23 Dec 2006
Posts: 134

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12 Jun 2007, 21:08
bsd_lover wrote:
Once the guy has pulled out 1 king, there are 11 cards left. Out of which 3 are kings. So the probability of 2 kings is 3/11.

Option A.
Huh?

4/12 * 3/11 = 1/11. How'd you figure it?

Kudos [?]: 29 [0], given: 0

CEO
Joined: 17 May 2007
Posts: 2950

Kudos [?]: 654 [0], given: 210

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12 Jun 2007, 21:19
Yes, that is correct. But the question is phrased in a way that shows that the event of first card being a king has already occured. So you dont need to multiply 3/11 by 4/12. 3/11 should be sufficient.

What is the OA ?

Kudos [?]: 654 [0], given: 210

Senior Manager
Joined: 27 Jul 2006
Posts: 294

Kudos [?]: 18 [0], given: 0

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12 Jun 2007, 22:32
Yeah, the caveat here is him stating that he has one king. If we do not have to calculate the probability of him bluffing into the situation, we can assume that there is a 100 percent chance that he holds a king.

All we need to do is calculate the probability that the next card is a king. In this case 3/11.

Kudos [?]: 18 [0], given: 0

Director
Joined: 30 Nov 2006
Posts: 591

Kudos [?]: 307 [0], given: 0

Location: Kuwait

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13 Jun 2007, 07:57
We have 12 cards out of which 4 are kings.

Since we know that one king has been drawn, only three are left.
So the probability of having two kings is the probability of drawing a king out of group of 11 cards that includes 3 kings.

So P(2 kings)= 3/11

Kudos [?]: 307 [0], given: 0

13 Jun 2007, 07:57
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