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]
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:
70% (03:22) correct
30% (03:01) wrong based on 112 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.....
The “3 golden nuggets” of MBA admission process With ten years of experience helping prospective students with MBA admissions and career progression, I will be writing this blog through...
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...