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Peter has a small deck of 12 playing cards made up of only 2

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Peter has a small deck of 12 playing cards made up of only 2 [#permalink]

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16 Jul 2007, 16:40
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Peter 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, there are 2 cards in the deck that have the same value.

Peter 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 he 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
VP
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16 Jul 2007, 20:09
2/12*10/11*8/10*6/9 = 8/99

6*8/99 = 48/99 = 16/33

1 - 16/33 = 17/33

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17 Jul 2007, 06:01
Hi Killer!
I get the same answer, but using a bit different method
P (1st card) = 1 I think as you can chose any card
then Prob (getting all the cards of different values)=1 * 10/11 * 8/10 * 6/9 = 16/33
1 - 16/33 =17/33

I am not sure that we have to multiply by 6..

Any other opinions/methods?
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17 Jul 2007, 07:07
Caas wrote:
Hi Killer!
I get the same answer, but using a bit different method
P (1st card) = 1 I think as you can chose any card
then Prob (getting all the cards of different values)=1 * 10/11 * 8/10 * 6/9 = 16/33
1 - 16/33 =17/33

I am not sure that we have to multiply by 6..

Any other opinions/methods?

I Agree - using 1 is better.

2/12*10/11*8/10*6/9 = 8/99

(choosing 1 from 12 cards)*(not choosing 1 from 11 cards - lets assume we choose 2)*(not choosing 1 or 2 from 10 cards - lets assume we choose 3)*(not choosing 1,2 or 3 from 9 cards) = 8/99

but we can start this with 1,2,3,4,5,6 hence I multiply by 6.

we need the completing event so I subtracted from 1.

caas way is better since it start with 1 (assuming any card for the first pick and continuing from there so - 12/12*10/11*8/10*6/9 = 48/99 ).

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17 Jul 2007, 08:24
Caas wrote:
Hi Killer!
I get the same answer, but using a bit different method
P (1st card) = 1 I think as you can chose any card
then Prob (getting all the cards of different values)=1 * 10/11 * 8/10 * 6/9 = 16/33
1 - 16/33 =17/33

I am not sure that we have to multiply by 6..

Any other opinions/methods?

This method is sweet.

But I solved it different way.

Total ways = 12C4 = 495

ways of getting at least a pair:

1 1 ----10C2
2 2 ----- 10C2 - 1 ( as 1 1 2 2 is covered above)
3 3 -----10C2 - 2
4 4 -----10C2 - 3
5 5 -----10C2 - 4
6 6-----10C2 - 5

So such ways are = 6 * 10C2 - 15 = 255

So prob = 255/495 = 17/33

Though this is lengthy, I am comfortable with it.
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18 Jul 2007, 03:58
I see. Here, though in different ways, but we all come to the same answer.

Try the following one! When you solve it using combinations and not, you would get different answers.

A black sack contains three green balls, five yellow balls, and four white ones. Three balls are taken at random without repetition. What is the probability of having all the the three balls of different colors?

A 3/22
B 1/22
C 3/11
D 1/121
E 1/33

How would u solve it?
18 Jul 2007, 03:58
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