It is currently 18 Mar 2018, 06:29

### 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

# Bill has a set of 6 black cards and a set of 6 red cards. Each card

Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 13 Apr 2013
Posts: 347
Location: India
GMAT 1: 480 Q38 V22
GPA: 3.01
WE: Engineering (Consulting)
Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card [#permalink]

### Show Tags

21 Oct 2015, 04:21
gmatter0913 wrote:
Bill has a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card. Bill likes to play a game in which he shuffles all 12 cards, 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) $$\frac{8}{33}$$
(B) $$\frac{62}{65}$$
(C) $$\frac{17}{33}$$
(D) $$\frac{103}{165}$$
(E) $$\frac{25}{33}$$

My Solution:

We can solve it by find the opposite probability when bill finds no pair of cards that have the same value:

P = 1 -(none)

We have total 12 cards, then we have

12/12 * 10/11* 8/10* 6/9 = 16/33

Now subtract this from 1 we get

P = 1-16/33 = 17/33 Answer is Option C

_________________

"Success is not as glamorous as people tell you. It's a lot of hours spent in the darkness."

Director
Joined: 12 Nov 2016
Posts: 790
Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card [#permalink]

### Show Tags

19 Apr 2017, 17:37
gmatter0913 wrote:
Bill has a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card. Bill likes to play a game in which he shuffles all 12 cards, 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) $$\frac{8}{33}$$
(B) $$\frac{62}{65}$$
(C) $$\frac{17}{33}$$
(D) $$\frac{103}{165}$$
(E) $$\frac{25}{33}$$

This question can be solved using complementary probability or in other words 1-p. When translating the question into an equation, and translating the various probabilities for each card draw, we should use the 1-x technique which is a concept relating to pairs in probability.

So

We need to find the complementary probability of not drawing a pair on each draw. There are four draws, so when we find the probability of not drawing a pair for each draw we will multiply those four probabilities and subtract from one.

First Draw

The probability of not drawing a pair on the first draw is 12/12, or 1, because you cannot have a pair with two items of course

Second Draw

Here is where we start to apply the 1-x technique

10 ( we subtract 2 from the numerator ) / 11 (we subtract 1 from the numerator

Third Draw

8 (again subtract 2 from the numerator)/ 10 (again subtract 1 from the numerator)

Fourth Draw

6 (again subtract 2 from the numerator)/9 (again subtract 1 from the numerator)

1- ( [1] X [10/11] x [8/10] x [6/9] )
1-(16/33)= 17/33

Possibly, the most fundamental example of using the complementary probability is a dice roll problem- on a dice numbered 1-6 what is the chance of rolling five at least once. *note the words at least and exact are important in probability problems

1- (5^3/6^3)

Though a probability question that is more on par with this question is one regarding a pair of socks

https://gmatclub.com/forum/tony-owns-si ... 34352.html
Manager
Status: Profile 1
Joined: 20 Sep 2015
Posts: 75
GMAT 1: 690 Q48 V37
GPA: 3.2
WE: Information Technology (Investment Banking)
Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card [#permalink]

### Show Tags

01 May 2017, 02:02
mcwoodhill wrote:
Zarrolou wrote:
Probability at least one pair equal $$= 1-P(none)$$.

Now I am gonna find the probability of picking all different values.
You can chose the first card is 12 ways ( every card is OK ), the second in 10 ways (there are 11 cards remaining, and 1 has the same value as the first one), the third in 8 ways (there are 10 cards remaining, 2 with the same value as the first ones), and the fourth in 6 ways (same reasoning).

The first is chosen out of a pool of 12(all cards), the second out of a pool of 11, the third out of a pool of 10, the fourth out of a pool of 9.

$$P=1-\frac{12}{12}*\frac{10}{11}*\frac{8}{10}*\frac{6}{9}=1-\frac{16}{33}=\frac{17}{33}.$$

Hope it's clear

nice way. my way was a bit more complicated.

Hi All,

In the same line of above approach of finding probability (without permutation or combination), can we find solution without subtract by 1.
I mean in above approach we are finding no pair then to find at least one pair by subtracting by 1, so I am looking for a solution where directly we can find at least one pair(no subtraction and without finding no pair probability).

Thanks
Intern
Joined: 28 Jun 2011
Posts: 19
Location: India
Schools: ISB '19 (D)
GMAT 1: 730 Q50 V38
GPA: 3.17
WE: Project Management (Energy and Utilities)
Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card [#permalink]

### Show Tags

04 Jun 2017, 02:52
russ9 wrote:
Hi,

Maybe Bunuel can shed some light on this? I didn't use the 1-Not probable approach, rather I carried out the tedious calc:

Probability Method:

Theory: Prob of at least 1 pair + Prob of 2 pairs

$$\frac{12}{12}*\frac{1}{11}*\frac{10}{10}*\frac{9}{9}$$ * $$\frac{4!}{2!2!}$$

+

$$\frac{12}{12}*\frac{1}{11}*\frac{10}{10}*\frac{1}{9}$$ * $$\frac{4!}{2!2!}$$ = 20/33.

Where am I going wrong? EDIT: maybe in the permutation part?

Combinatorics Method

Theory: Prob of at least 1 pair + Prob of 2 pairs

(6c1)(6c1)/(12c4)

+

(6c2)(6c2)/(12c4) I get a completely wrong answer.

Any thoughts on what I'm doing wrong here?

Hi,
The way to solve with probablity method without using 1- probablity of(none) is
Using the full understanding of all cases:

Probablity of selecting one pair :
(12/12) * (1/12)* 10/10 *8/9= 8/99----assuming ist two cards are same.
12/12 * 10/11* 2/10 * 8/9 = 16/99 -------assuming 1st and third card are same .
12/12 * 10/11 * 8 /10 * 3/9 = 24/99-------assuming ist and fourth card are same.
Total probablity of only one card same is therefore : (8+16+24)/99 = 48/99 =16/33-----(1)
Probablity of two pair same in similar way is
12/12 *1/12* 10/10*1/9 = 1/99 ....ist two same and third and fourth same.
12/12*10/11*2/10*1/9 = 2/99----- This includes both 1-3 * 2-4 similar and 1-4 & 2-3 similar
Therefore total probablity of any two same is 3/99=1/33----(2)

therefore total probablity = (1) + (2) = 17/33.

Kudos if helped.
Intern
Joined: 09 Jun 2016
Posts: 11
Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card [#permalink]

### Show Tags

05 Mar 2018, 13:16
Hi Bunnel... need your intervention to explain on the logic of P (None) please

Sent from my SM-G935F using GMAT Club Forum mobile app
Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card   [#permalink] 05 Mar 2018, 13:16

Go to page   Previous    1   2   [ 25 posts ]

Display posts from previous: Sort by