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

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From an ordinary deck of cards, only 12 picture cards are [#permalink] New post 13 Oct 2004, 16:45
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(i) From an ordinary deck of cards, only 12 picture cards are retained. They are shuffled (mixed) and a man draws 2 cards at random. He then announce one of these cards is the King or Spades. What is the probability that he holds two Kings? (hint: list the possible sample space carefully).

(ii) The 2 cards are replaced and the cards re-shuffled. He then draws 2 cards, and announces that he holds at least one King. Find the probability that he holds two Kings in his hand.


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 [#permalink] New post 13 Oct 2004, 17:54
1. There are 4 kings in a pack of cards. If we need to select 2 out of the 4, we can do it in 4C2 ways. The prob is then 4C2/12C2 => 1/11.

2. Atleast one king means the no. of ways would be: (4C1*8C1+4C2) / 12C2 =>19/33.
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 [#permalink] New post 13 Oct 2004, 20:20
Hi,

you did not consider that he has announced that he got a king.

[quote="Anonymous"]1. There are 4 kings in a pack of cards. If we need to select 2 out of the 4, we can do it in 4C2 ways. The prob is then 4C2/12C2 => 1/11.

2. Atleast one king means the no. of ways would be: (4C1*8C1+4C2) [b]/ [/b]12C2 =>19/33.[/quote]
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 [#permalink] New post 14 Oct 2004, 02:04
I think the answer to the first Q is not 4C2/2C12 as we already know one of the 2 cards in the king of spades. Thus it is 4C2 / 2 and final result is 1/22
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 [#permalink] New post 14 Oct 2004, 02:07
For the 2nd question, if it is the prob to hold 2 kings answer is 1/11 as developed by "guest". If the question is at least 1 king, (4C2+4C1.8C1)/12C2 is correct
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 [#permalink] New post 14 Oct 2004, 14:13
Weird OA... How is it possible to have a 19 at denominator...

I think we need a guru here to blast it all.
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 [#permalink] New post 15 Oct 2004, 21:25
What I understood is that there are total 12 cards, selected from a deck of 52 cards. These 12 cards are picture cards - ie - K, Q and J

applying - P(A+B) = P(B). P(A/B)

P(A+B) = (prob of two king) = PA.PB = (4/12)(3/11) = 1/11

P(B) = Prob of a King or a Spade = (4/12) + (3/12) - (4/12)(3/12) = 1/2

So, 1/11 = 1/2 . P(A/B) or P(A/B) = 2/11

So Prob of two kings given that he has one king or one spade = 2/11 ( However it does not match the answer)

2. Applying the same conditional probability

P(A+B) = P(B) . P(A/B)
prob of two kings = prob of atleast one king . P(a/b)

=> (4/12)(3/11) = (1-8/12). P(A/B)

P(A/B) = 3/11
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 [#permalink] New post 15 Oct 2004, 21:29
That was me..... :)
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PS Probability [#permalink] New post 16 Oct 2004, 10:39
By PracticeMore's method, I am able to get the second answer right(matching the OA).

P(K)=Probability of 2 Kings.

P(A)=Probability of atleast 1 King.

We need to find P(K/A) ie Probability of getting 2 Kings given that he holds atleast 1 King.

Remember this formula for conditional probability:

P(A)*P(K/A)= P(K)

P(A) = Probability of atleast 1 King=1-Probability of NO King = 1- 8C2/12C2 = 19/33

PK) = Probability of 2 Kings = 4C2/12C2 = 1/11

so P(K/A) = P(K)/P(A) = (1/11)/(19/33) = 3/19


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 [#permalink] New post 16 Oct 2004, 10:55
Great Anish, I see my mistake. I should have considered 1-8C2/12C2 as we are selecting two out of the card set. 1-8/12 would be the case if it was for selecting just one card.

Anyway, do we have answer for the first part ?
  [#permalink] 16 Oct 2004, 10:55
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