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Bill has a small deck of 12 playing cards made up of only 2 [#permalink]
21 Jan 2007, 16:16

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A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

69% (03:27) correct
31% (03:04) wrong based on 102 sessions

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, there are 2 cards in the deck that have the same 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?

counting quadruples with no pairs:
first card - no restrictions - 12 options
second card - cannot be same as first - 10 options (out of 11 cards)
third card - cannot be same as first or second - 8 options (out of 10 cards left)
forut card - cannot be same as first three - hence 6 options out of 9 remaining cards).

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, there are 2 cards in the deck that have the same 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?

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, there are 2 cards in the deck that have the same 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

I know the answer is C but i can't understand why i can't get the answer in the following way....
12C4 gives all the combinations = 11*9*5
6*10C2 gives all the possible combinations with at least one pair=6*9*5
divide giving 6/11 ?????
I don't see what i am doing wrong

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, there are 2 cards in the deck that have the same 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

I know the answer is C but i can't understand why i can't get the answer in the following way.... 12C4 gives all the combinations = 11*9*5 6*10C2 gives all the possible combinations with at least one pair=6*9*5 divide giving 6/11 ????? I don't see what i am doing wrong

I dunno, bout that approach, im really bad using the combinatorics approach.

Well try and find the probabilty that there are no pairs.
Lets say we pick the following cards: 1,2,3,4. In this order.

1 * 10/11 * 8/10 * 6/9

1: b/c the probability of picking any number is 1.

10/11: is the prob of not picking the next pair
we have 1,2

8/10: we have 1,2,3,3,4,4,5,5,6,6 left. 10 numbers and 2 unfavorables here. So its 8/10

6/9: we have 1,2,3,4,4,5,5,6,6 left. So 9 numbers and 3 unfavorables so 6/9.

Thanks GMATBLACKBELT but i know and understand your approach. I am just trying to standardise my thinking. If combinatorics approach doesn't apply here then i will be confused when to use and when not to use combinatorics.

Hmmmmm i think you have struck a cord
6*10C2
There are Six sets if (10C2).
Assume i have selected the first pair then i have i have 10 cards to choose 2 from. And there are 6 such scenarios because there are 6 pairs of cards.

BUT you have just reminded me that this way includes some duplicates so my method nearly correct.

so
[10C2+(10C2-1)+(10C2-2)+(10C2-3)+(10C2-4)+(10C2-5) ]/12C4=17/33

p=1-q, where q is the probability to turn over 4 cards with different values.

Same starting point. But aftwerwards applied the following logic

probability that the second card will not make a pair to the first withdrwan
times
probability that the third will not make a pair with any of the two previously witdrawn cards
times
probability that the fourth one will not make a pair with any of the three previously withdrawn cards

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, there are 2 cards in the deck that have the same 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

D103/165

E 25/33

Amar

OK here it is, I originally got the answer, but did it through some method that doesn't really make a whole lot of sense to me anymore. I dont think it was correct either, I just got really lucky and somehow things worked out =)

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, there are 2 cards in the deck that have the same 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

D103/165

E 25/33

Amar

OK here it is, I originally got the answer, but did it through some method that doesn't really make a whole lot of sense to me anymore. I dont think it was correct either, I just got really lucky and somehow things worked out =)

Probability of NOT picking any pairs.

1*10/11* 8/10*6/9

1/11*8*2/3 --> 16/33 --> 1-16/33 =17/33

Ok there we go.

can someone tell me where I am wrong????

all possible outcomes: 12c4=990

let's choose 1 pair of identic cards from the 2 sets: 6c1 possibilities=6 let's multiply this result * 10c2, combinations for the 2 free places=6*45=270

let's consider the possibility that we have 2 pairs of identic cards: 6c2=15

thus we have 285/990 = 19/33....please help! I add 15 but I am sure it needs to subtract it from 270.....

My last interview took place at the Johnson School of Management at Cornell University. Since it was my final interview, I had my answers to the general interview questions...