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, for each value from 1 to 6, there are two cards in the deck with that 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?
8/33
62/165
17/33
103/165
25/33
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
Manhattan Math: Probability
This topic has expert replies
-
Ashishkapoor7
- Senior | Next Rank: 100 Posts
- Posts: 35
- Joined: Fri Jan 14, 2011 4:04 am
- Thanked: 1 times
-
brainturner
- Newbie | Next Rank: 10 Posts
- Posts: 5
- Joined: Mon Jun 25, 2012 4:55 am
Bill has 12 cards in his deck. After shuffling he chooses 4 cards out of 12 cards.
1st card - 12 ways
2nd card - 11 ways
3rd card - 10 ways
4th card - 9 ways
Total number of combinations = 12*11*10*9 / 4! = 495
The chance of Bill finding at least one pair of cards that have the same value
= 1 - chance of Bill finding no pair of cards that have same value
For a deck of 4 cards with all different cards
1st card - 12 ways
2nd card - 10 ways
3rd card - 8 ways
4th card - 6 ways
Number of combinations = 12*10*8*6/4! = 240
So the chance of Bill finding at least one pair of cards that have the same value = 1 - (240/495)
= [spoiler] 17/33 Answer C[/spoiler]
1st card - 12 ways
2nd card - 11 ways
3rd card - 10 ways
4th card - 9 ways
Total number of combinations = 12*11*10*9 / 4! = 495
The chance of Bill finding at least one pair of cards that have the same value
= 1 - chance of Bill finding no pair of cards that have same value
For a deck of 4 cards with all different cards
1st card - 12 ways
2nd card - 10 ways
3rd card - 8 ways
4th card - 6 ways
Number of combinations = 12*10*8*6/4! = 240
So the chance of Bill finding at least one pair of cards that have the same value = 1 - (240/495)
= [spoiler] 17/33 Answer C[/spoiler]
GMAT/MBA Expert
-
Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
I tackled this question by recognizing that P(at least one pair) = 1-P(no pairs)Ashishkapoor7 wrote: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, for each value from 1 to 6, there are two cards in the deck with that 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?
8/33
62/165
17/33
103/165
25/33
First, the number of possible outcomes:
We have 12 cards and we select 4.
This is accomplished in 12C4 ways = 495
Now we count the number of ways to select 4 different values with no pairs. In other words, we wand 4 different card values.
First look at how many different cards we can select (WITHOUT considering the suits)
There are 6 card values and we want to select 4.
This is accomplished in 6C4 = 15 ways.
For each of these 15 values (e.g., 1,2,4,5) each card can be of either suit.
So, for example, in how many ways can we have card values of 1,2,4,5?
For the 1-card we have 2 suits from which to choose.
For the 2-card we have 2 suits from which to choose.
For the 4-card we have 2 suits from which to choose.
For the 5-card we have 2 suits from which to choose.
So, for the selection 1,2,4,5 there are 2x2x2x2-16 possibilities.
So, the number of ways to select 4 cards such that there are no pairs is 15x16=240
So, the probability that there are no pairs = 240/(11x5x9) = 48/99 = 16/33
So, P(at least one pair) = 1- 16/33 =17/33
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
GMAT/MBA Expert
-
Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
We can also solve this question using probability rules.
First, recognize that P(at least one pair) = 1 - P(no pairs)
P(no pairs) = P(select any 1st card AND select any non-matching card 2nd AND select any non-matching card 3rd AND select any non-matching card 4th)
= P(select any 1st card) x P(select any non-matching card 2nd) x P(select any non-matching card 3rd) x P(select any non-matching card 4th)
= 1 x 10/11 x 8/10 x 6/9
= 16/33
So, P(at least one pair) = 1 - 16/33
= 17/33
= C
Cheers,
Brent
First, recognize that P(at least one pair) = 1 - P(no pairs)
P(no pairs) = P(select any 1st card AND select any non-matching card 2nd AND select any non-matching card 3rd AND select any non-matching card 4th)
= P(select any 1st card) x P(select any non-matching card 2nd) x P(select any non-matching card 3rd) x P(select any non-matching card 4th)
= 1 x 10/11 x 8/10 x 6/9
= 16/33
So, P(at least one pair) = 1 - 16/33
= 17/33
= C
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
-
willrc
- Junior | Next Rank: 30 Posts
- Posts: 13
- Joined: Tue Jul 17, 2012 12:43 am
- Location: London
- Thanked: 2 times
- GMAT Score:760
We just need to decide how to avoid getting a pair:
Pick any card first (probability 1)
x
Avoid 1 out of 11 remaining cards (P=10/11)
x
Avoid 2 out of the 10 remaining cards (P=8/10)
x
Avoid 3 out of the 9 remaining cards (P=6/9).
1 x 10/11 x 8/10 x 6/9 = 16/33
1-16/33 = 17/33
Pick any card first (probability 1)
x
Avoid 1 out of 11 remaining cards (P=10/11)
x
Avoid 2 out of the 10 remaining cards (P=8/10)
x
Avoid 3 out of the 9 remaining cards (P=6/9).
1 x 10/11 x 8/10 x 6/9 = 16/33
1-16/33 = 17/33
London-based private tutor
www.gmatprolondon.com
www.gmatprolondon.com
