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A deck of playing cards - probability

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A deck of playing cards - probability

by nikhilsrl » Sat Mar 19, 2011 7:09 am
Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, for each value from 1 to 6, there are two cards in the deck with that value. Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?

a) 8/33

b) 62/165

c) 17/33

d) 103/165

e) 25/33

OA is C.

I want to know how to solve this.

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by force5 » Sat Mar 19, 2011 8:43 am
IMO

17/33

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by AIM GMAT » Sat Mar 19, 2011 9:01 am
The below is the probability of getting all 4 diffrent cards

(12/12) X (10/11) X (8/10) X (6/9) = 16 / 33

Prob of atleast one couple = 1 - prob of no couple
= 1 - (16/33)
= 17/33

IMO C.
Thanks & Regards,
AIM GMAT

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by gmat7202011 » Sat Mar 19, 2011 10:03 am
AIM GMAT wrote:The below is the probability of getting all 4 diffrent cards

(12/12) X (10/11) X (8/10) X (6/9) = 16 / 33

Prob of atleast one couple = 1 - prob of no couple
= 1 - (16/33)
= 17/33

IMO C.
AIM GMAT, Can you please explain how you went about this ?

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by AIM GMAT » Sat Mar 19, 2011 10:22 am
First of all let us find out prob. of not getting any couple :

__ X __ X __ X __

Now we need to fill each blank with the prob .

1st blank : There are 12 cards so we can select 12 out of 12 cards ,hence the blank is filled with (12/12). Let us assume we chose "A" for 1st blank .

2nd blank : There are 11 cards left , but to not get a couple we also cannot chose the number that we chose for 1st blank i.e. A so we need to exclude that from out count , so we can chose from 10 cards , prob wud be (10/11). Let us assume we chose "B" for 2nd blank .

3rd blank : There are 10 cards left , out of which we need to exclude A and B , so left with 8 cards , prob wub become (8/10).Let us assume we chose "C" here .

4th blank : 9 cards left out of which we can select excluding A , B and C , so 6 cards , prob wub be (6/9)

Put the respective values calculated ans we get (16/33).

NOw to calculate atleast one pair of cards => [1 pair + 2 pair] OR [1 - NO PAIR]

I have followed the second approach = 1 - prob.(NO PAIR)
= 1 - (16/33)
= 17/33


Hope that helps and i havnt made it confusing :) .

Let me know if u dnt understand any part
Thanks & Regards,
AIM GMAT

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by gmat7202011 » Sat Mar 19, 2011 11:52 am
Thank You AIM GMAT. The explanation provided helps greatly

Thanks again,