Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 28 Apr 2015, 06:57

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:
Manager
Joined: 07 May 2007
Posts: 181
Followers: 2

Kudos [?]: 18 [0], given: 0

Peter has a small deck of 12 playing cards made up of only 2 [#permalink]  16 Jul 2007, 16:40
1
This post was
BOOKMARKED
00:00

Difficulty:

(N/A)

Question Stats:

100% (00:00) correct 0% (00:00) wrong based on 1 sessions
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
Joined: 08 Jun 2005
Posts: 1147
Followers: 6

Kudos [?]: 127 [0], given: 0

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

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

1 - 16/33 = 17/33

Senior Manager
Joined: 04 Mar 2007
Posts: 442
Followers: 1

Kudos [?]: 16 [0], given: 0

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?
VP
Joined: 08 Jun 2005
Posts: 1147
Followers: 6

Kudos [?]: 127 [0], given: 0

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 ).

Director
Joined: 14 Jan 2007
Posts: 780
Followers: 2

Kudos [?]: 71 [0], given: 0

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.
Senior Manager
Joined: 04 Mar 2007
Posts: 442
Followers: 1

Kudos [?]: 16 [0], given: 0

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?
Similar topics Replies Last post
Similar
Topics:
10 Randolph has a deck of 12 playing cards made up of only 2 22 29 Jul 2012, 09:11
46 Bill has a small deck of 12 playing cards 29 18 Jun 2010, 21:47
10 Bill has a small deck of 12 playing cards made up of only 2 50 21 Jan 2007, 16:16
Bill has a small deck of 12 playing cards made up of only 2 7 20 Oct 2007, 12:16
Bill has a small deck of 12 playing cards made up of only 2 3 05 Oct 2007, 22:39
Display posts from previous: Sort by