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by Anurag@Gurome » Sun Jun 03, 2012 1:33 am
gmatter2012 wrote:As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the likelihood Exactly 2 people choose same number and other 2 choose different numbers ?
Total number of ways to select numbers without any restriction = 4^4 = 256

The selected set of numbers will be like {A, A, B, B}
We can view the situation as: two numbers are selected from four. Then two of each of the number, i.e. a total of four numbers are distributed among four persons. To clarify more, say we are selecting to numbers A and B from {A, B, C, D}. Then we are distributing two A's and two B's among four persons.

Hence, number of ways to select as mentioned = (Number of ways to select two different integers out of four)*(Number of ways to distribute four objects out of which two are of one kind and other two are of another kind among four persons)

Number of ways to select two different integers out of four = 4C2 = 6
Number of ways to distribute four objects out of which two are of one kind and other two are of another kind among four persons = 4!/[(2!)(2!)] = 6

Hence, required probability = 6*6/256 = 36/256 = 9/64

9/64 = (9/64)*100% = (9*25/16)% = (225/16)% = 14.(something)%

C is the closest option.
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by coolhabhi » Sun Jun 03, 2012 1:38 am
Since each person can choose any of the four numbers the total number of cases will be 4^4 = 256

Now exactly 2 people have to choose same number and other 2 have to choose different numbers - {a,a,b,c}:

The number of ways to choose which 2 persons will have the same number is 4C2 = 6

The number of ways to choose which number it will be = 4

The number of ways to choose 2 different numbers out of 3 numbers for 2 other persons is 3P2 = 6 (Here order matters).

So the likelihood of exactly 2 people choose same number and other 2 choose different numbers = (6*4*6)/256
56.25%
Answer id D

What is the OA? and what is the source?
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by lathmanu » Sun Jun 03, 2012 4:11 am
Anurag@Gurome wrote:
gmatter2012 wrote:As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the likelihood Exactly 2 people choose same number and other 2 choose different numbers ?
Total number of ways to select numbers without any restriction = 4^4 = 256

The selected set of numbers will be like {A, A, B, B}
We can view the situation as: two numbers are selected from four. Then two of each of the number, i.e. a total of four numbers are distributed among four persons. To clarify more, say we are selecting to numbers A and B from {A, B, C, D}. Then we are distributing two A's and two B's among four persons.

Hence, number of ways to select as mentioned = (Number of ways to select two different integers out of four)*(Number of ways to distribute four objects out of which two are of one kind and other two are of another kind among four persons)

Number of ways to select two different integers out of four = 4C2 = 6
Number of ways to distribute four objects out of which two are of one kind and other two are of another kind among four persons = 4!/[(2!)(2!)] = 6

Hence, required probability = 6*6/256 = 36/256 = 9/64

9/64 = (9/64)*100% = (9*25/16)% = (225/16)% = 14.(something)%

C is the closest option.
My reasoning

Probability A will choose a number = 4/4
Probability for B to choose same as A = 4/4 * 1/4
Probability for C to choose diff from A,B = 4/4 * 1/4 * 3/4
Probability for C to choose diff from A,B,C = 4/4 * 1/4 * 3/4 * 2/4
Total ways to choose 2 persons from a set of 4 = 4C2 = 6

Probability = 4C2 * 4/4 * 1/4 * 3/4 * 2/4

Ans = Approx 56% = D
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by gmatter2012 » Sun Jun 03, 2012 6:25 am
coolhabhi wrote:Since each person can choose any of the four numbers the total number of cases will be 4^4 = 256

Now exactly 2 people have to choose same number and other 2 have to choose different numbers - {a,a,b,c}:

The number of ways to choose which 2 persons will have the same number is 4C2 = 6

The number of ways to choose which number it will be = 4

The number of ways to choose 2 different numbers out of 3 numbers for 2 other persons is 3P2 = 6 (Here order matters).

So the likelihood of exactly 2 people choose same number and other 2 choose different numbers = (6*4*6)/256
56.25%
Answer id D

What is the OA? and what is the source?
OA is E= 56 %
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by gmatter2012 » Sun Jun 03, 2012 6:28 am
Anurag@Gurome wrote:
gmatter2012 wrote:As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the likelihood Exactly 2 people choose same number and other 2 choose different numbers ?
Total number of ways to select numbers without any restriction = 4^4 = 256

The selected set of numbers will be like {A, A, B, B}
We can view the situation as: two numbers are selected from four. Then two of each of the number, i.e. a total of four numbers are distributed among four persons. To clarify more, say we are selecting to numbers A and B from {A, B, C, D}. Then we are distributing two A's and two B's among four persons.

Hence, number of ways to select as mentioned = (Number of ways to select two different integers out of four)*(Number of ways to distribute four objects out of which two are of one kind and other two are of another kind among four persons)

Number of ways to select two different integers out of four = 4C2 = 6
Number of ways to distribute four objects out of which two are of one kind and other two are of another kind among four persons = 4!/[(2!)(2!)] = 6

Hence, required probability = 6*6/256 = 36/256 = 9/64

9/64 = (9/64)*100% = (9*25/16)% = (225/16)% = 14.(something)%

C is the closest option.
Anurag I think You have provided answer to the following question

"2 people choose same number and other 2 also choose same number - {a,a,b,b}:"
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by myselfhari » Sun Jun 03, 2012 10:33 pm
The 1st person can chose 1 number from a pool of 4 nos. =>
P(selecting a no.) = 1/4

The 2nd person, who should select the same no. as the first person, can chose only 1 no. =>
Probability(selecting the same no) = 1

The 3rd person, has to chose a no. that has not been selected by the 1st (and 2nd, obviously). So he has to select 1 number from a pool of 3 nos now (as per the ques) =>
Probability(selecting a no.) = 1/3

Similarly the fourth person can select 1 different no from a pool of 2 nos (as already 2 nos. are consumed) =>
Probability(selecting a no.) = 1/2

Therefore total probability is the product of all the 4 mentioned above => (1/4)*1*(1/3)*(1/2) = 1/24

Now remember that you still have to select 2 people from 4 people who will have the exactly the same number => 4C2

FINAL ANSWER = 4C2 * 1/24 = 6*1/24 = 1/4 = 25%

I somehow feel this is how the problem should be attacked.
Plz. correct me if i am wrong.
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by bubbliiiiiiii » Tue Jun 05, 2012 3:36 am
Looking forward for a consistent answer and approach.
Regards,

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by eagleeye » Tue Jun 05, 2012 5:14 am
[spoiler]The correct answer is 9/16 (which is ~56%)[/spoiler]. Let me explain:

First way: Combinations:

Overall combinations: Four people can select one from four numbers each. No. of ways of doing this = (4C1)^4= 4^4.

Now, We know that we have to set up the numbers in a way that two people get the same number while others get different numbers. So we have three numbers overall, one repeated and two distinct. Let's go selecting.
First, we are using only three out of the four numbers. Let's select those 3. No. of ways of doing this = 4C3 = 4.
Out of these three that we just selected, let's choose the one that is going to repeat. No of ways of doing this = 3C1 = 3 (Note that, if we select the one repeating, its the same as selecting two non-repeating ones).
So now we have 3 numbers, and we know which one repeats and which one doesn't.
Now we will go selecting from the other group.

We have four people. Let's select two people of the four who will get the repeating number. No. of ways of doing this = 4C2 = (4*3)/(2*1) = 6.
From the other two people, who get the distinct numbers, we have to choose the one who will who will receive one of the distinct numbers(The other number goes to the one remaining). No of ways. of selecting one out of 2 = 2C1 = 2
.
So overall probability = 4*3*6*2/(4^4) = 9/16 = 56.25%. Hence E

Second way: Combination and then permutation.

First, we are using only three out of the four numbers. Let's select those 3. No. of ways of doing this = 4C3 = 4.
Out of these three that we just selected, let's choose the one that is going to repeat. No of ways of doing this = 3C1 = 3
So now we have 3 numbers, and we know which one repeats and which one doesn't.
Now we will go arranging. Imagine the four people standing in a row. That gives us four spaces. We have to arrange something like a,a,b,c in 4 places. No. of ways of doing this = 4!/(2!) (we divide by 2! since a is repeating).

Hence overall probability = 4*3*4!/(4^4*2!) = 9/16 = 56.25%. Still E.

Let me know if this helps :)
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by Brent@GMATPrepNow » Tue Jun 05, 2012 7:44 am
gmatter2012 wrote:As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the likelihood Exactly 2 people choose same number and other 2 choose different numbers ?

a) 9%
b) 12%
c) 20%
d) 40%
e) 56%
The denominator: Total number of outcomes = 4x4x4x4 = 256

The numerator:
Stage 1: Choose 2 people to select the same integer. This can be accomplished in 4C2 ways (6 ways)
Stage 2: Choose an integer (1, 2, 3, or 4) for these 2 people to select. This can be accomplished in 4 ways
Stage 3: Choose a different integer for one of remaining people. Since one of the integers is already accounted for, this stage can be accomplished in 3 ways
Stage 4: Choose a different integer for the last person. Since two of the integers are already accounted for, this stage can be accomplished in 2 ways
So, by the Fundamental Counting Principle (FCP), the numerator = (6)(4)(3)(2) = 144

So, the probability = 144/256
=9/16
Since this fraction is greater than 1/2, the answer must be E.

Cheers,
Brent

To learn more about the Fundamental Counting Principle, check out this video: https://www.gmatprepnow.com/module/gmat-counting?id=775
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by myselfhari » Tue Jun 05, 2012 10:08 am
Gross blunder by me in my solution.. very sorry.

Damn me...
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by bubbliiiiiiii » Tue Jun 05, 2012 10:20 am
Thanks Brent. I missed stage 1 :)
Brent@GMATPrepNow wrote:
gmatter2012 wrote:As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the likelihood Exactly 2 people choose same number and other 2 choose different numbers ?

a) 9%
b) 12%
c) 20%
d) 40%
e) 56%
The denominator: Total number of outcomes = 4x4x4x4 = 256

The numerator:
Stage 1: Choose 2 people to select the same integer. This can be accomplished in 4C2 ways (6 ways)
Stage 2: Choose an integer (1, 2, 3, or 4) for these 2 people to select. This can be accomplished in 4 ways
Stage 3: Choose a different integer for one of remaining people. Since one of the integers is already accounted for, this stage can be accomplished in 3 ways
Stage 4: Choose a different integer for the last person. Since two of the integers are already accounted for, this stage can be accomplished in 2 ways
So, by the Fundamental Counting Principle (FCP), the numerator = (6)(4)(3)(2) = 144

So, the probability = 144/256
=9/16
Since this fraction is greater than 1/2, the answer must be E.

Cheers,
Brent

To learn more about the Fundamental Counting Principle, check out this video: https://www.gmatprepnow.com/module/gmat-counting?id=775
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Pranay
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by Stuart@KaplanGMAT » Tue Jun 05, 2012 10:21 am
Here's another way you can solve:

Prob = # desired outcomes/total number of possibilities

there are 4*4*4*4 = 256 total possibilities

Now on to desired outcomes!

The first person can pick any number, so she has 4 possibilities.
We want the second to be the same as the first, so she has 1 possibility.
We want the third to be different, so she has 3 possibilities.
We want the fourth to be different, so she has 2 possibilities.

4*1*3*2 = 24

Now after the first person chooses, the other 3 picks can be in any order. So, there are 3! = 6 different ways to arrange those 3 picks.

24 * 6 = 144

So the probability of what we want is 144/256 = more than half = (E)!

(Remember: the answer choices can be your best friends on test day! Don't do unnecessary calculations!)
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by SoftwareDrone » Mon Jun 25, 2012 12:16 pm
Brent@GMATPrepNow wrote:
gmatter2012 wrote:As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the likelihood Exactly 2 people choose same number and other 2 choose different numbers ?

a) 9%
b) 12%
c) 20%
d) 40%
e) 56%
The denominator: Total number of outcomes = 4x4x4x4 = 256

The numerator:
Stage 1: Choose 2 people to select the same integer. This can be accomplished in 4C2 ways (6 ways)
Stage 2: Choose an integer (1, 2, 3, or 4) for these 2 people to select. This can be accomplished in 4 ways
Stage 3: Choose a different integer for one of remaining people. Since one of the integers is already accounted for, this stage can be accomplished in 3 ways
Stage 4: Choose a different integer for the last person. Since two of the integers are already accounted for, this stage can be accomplished in 2 ways
So, by the Fundamental Counting Principle (FCP), the numerator = (6)(4)(3)(2) = 144

So, the probability = 144/256
=9/16
Since this fraction is greater than 1/2, the answer must be E.

Cheers,
Brent

To learn more about the Fundamental Counting Principle, check out this video: https://www.gmatprepnow.com/module/gmat-counting?id=775
I am thinking I may need to watch the video, because I am totally lost. I do not understand any of the stages mentioned. :-(
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by sana.noor » Mon Oct 28, 2013 10:38 am
what is the right answer? Anurag says its C and Brent says its E. i am confused. however, i also end up with E but two experts with different answers.
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