Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

21 Jan 2007, 16:16

1

This post received KUDOS

5

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

68% (03:05) correct
32% (02:39) wrong based on 143 sessions

HideShow timer Statistics

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.....

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...

Marketing is one of those functions, that if done successfully, requires a little bit of everything. In other words, it is highly cross-functional and requires a lot of different...