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# As part of a game, four people each must secretly choose an

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Re: As part of a game, four people each must secretly choose an  [#permalink]

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31 Jul 2013, 04:51
Another way:

4/4 * 3/4 * 2/4 * 1/4 = 3/32 = 0.09375 => 9% approx
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17 Oct 2013, 13:46
cstefanita wrote:
Possible combinations of each choosing a diffrent number (favorable events):
4*3*2*1=24

Total possible combinations (total events): 4*4*4*4=256

Favorable events/total events = 24/256 approx 10%

Is this a correct approach?

Yes it is. I also did it this way.
Correct approach but your approx is flawed

Should be 9%, careful with that
Cheers
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Re: As part of a game, four people each must secretly choose an  [#permalink]

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25 Nov 2013, 13:46
tarek99 wrote:
As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the approximate likelihood that all four people will choose different numbers?

A. 9%
B. 12%
C. 16%
D. 20%
E. 25%

4!/4^4 = 24/25 = 9% approx

Cheers
J
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Re: As part of a game, four people each must secretly choose an  [#permalink]

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27 Jan 2014, 04:52
tarek99 wrote:
As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the approximate likelihood that all four people will choose different numbers?

A. 9%
B. 12%
C. 16%
D. 20%
E. 25%

Although this is a very basic question and there are several responses i would like to reply elaborately.

There are 4 persons and each has to choose different numbers. we can have 2 approach.

1. Plain and simple counting

Lets say there are 4 persons : A,B,C,D and they need to choose 4 numbers such that no 2 person choose the same number.

How many choice does A have, if he is the first one to choose. It should be 4. There are 4 numbers and A does not have any constraint. So A can choose in 4 ways
How many choice does B have, if he is the second one to choose. It should be 3. There are in all 4 numbers. But A has already chosen 1, so B has only 3 numbers to choose from.
How many choice does C have, if he is the Third one to choose. It should be 2. There are in all 4 numbers. But A and B has already chosen 2 numbers, so C has only 2 numbers to choose from.
How many choice does D have, if he is the fourth one to choose. It should be 1. There are in all 4 numbers. But A, B and C has already chosen 3 numbers, so D has only 1 number left to choose from.

Now,
Probability of A = Number of favorable outcomes / Total number of outcomes.
Number of favorable outcome = 4 --> As we calculated above.
Total number of outcomes = 4. There are only 4 numbers to choose from.
so P(A) = 4/4
Similarly P(B) = 3/4, P(C) = 2/4, P(D) = 1/4

Now all the persons have to choose the ball for the same event. Its like A chooses AND B chooses AND C chooses AND D chooses. Since we use 'and' we need to multiply.

Hence probability = 4/4 * 3/4*2/4*1/4 = 3/32 ~ 9%

2. Applying combinations

Lets say there are 4 persons : A,B,C,D and they need to choose 4 numbers such that no 2 person choose the same number.

Now this is an unordered arrangement. It does not matter if the 4 chosen numbers are 1,2,3,4 or 2,1,3,4. We are only concerned with the fact that numbers chosen should be different. We are not bothered about a particular order and hence combination applies.

Having said that,

In how many ways can person A choose 1 number from available 4 number? Try it.
4c1

In how many ways can person B choose 1 number from available 4 number such that he does not choose the one which A already chased. Effectively B can choose only 1 from 3 numbers.
3C1

Similarly in how many ways can C and D choose?
C : 2C1 and D 1C1

what will be P(A)?
4c1/4c1. Total number of outcomes = 4C1

what will be P(B), P(C), P(D) ?
B : 3C1/4c1, C : 2C1/4c1, D : 1C1/4c1

Try out the overall probability it should be the same as #1 above
P(A) * P(B) * P (C) * P(D) = 4C1/4C1 * 3C1/4C1 * 2C1/4C1 * 1C1/4C1 = 3/32

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Re: As part of a game, four people each must secretly choose an  [#permalink]

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10 Feb 2015, 05:13
tarek99 wrote:
As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the approximate likelihood that all four people will choose different numbers?

A. 9%
B. 12%
C. 16%
D. 20%
E. 25%

(4/4)*(3/4)*(2/4)*(1/4)=.0937=9%
OA - A
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As part of a game, four people each must secretly choose an  [#permalink]

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30 Mar 2016, 12:04
Primarily no of ways we can select a number for a person = 4 ways since there are 4 nos

so if different nos are to be chosen
if 1st person selects a no then he has all 4 chances

next how many are available fr 2nd one........obviously 3 remaining
so second one has 3 ways

similarly third one has 2 ways

final person is left with only one number

since all of this happen correspondingly and are inter related events they are multiplied
and hence numerator of favorable outcomes is 4*3*2*1 = 24

now we have to consider all the possible cases where all nos can be same or different
that means irrespective of what one chooses does not bother other one and everyone can choose anyone of the 4 nos
hence 4 is multiplied for 4 times for all possible cases

hence

$$prob=\frac{fav}{total}=\frac{{24}}{{4*4*4*4}}$$

since all the choices are in percentages, we convert his one also in percentage by multiplying the prob by 100 i.e., 9% approximately.

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Re: As part of a game, four people each must secretly choose an  [#permalink]

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04 Aug 2016, 11:36
Bunuel wrote:
B. Exactly 2 people choose same number and other 2 choose different numbers - {a,a,b,c}:

$$P(B)=\frac{C^2_4*4*P^2_3}{4^4}=\frac{144}{256}$$.

$$C^2_4$$ - # of ways to choose which 2 persons will have the same number;
$$4$$ - # of ways to choose which number it will be;
$$P^2_3$$ - # of ways to choose 2 different numbers out of 3 left for 2 other persons when order matters;

Bunuel, can you please explain how you got this?
...and how can we do this by simple enumeration/probability way?

Many thanks!
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As part of a game, four people each must secretly choose an  [#permalink]

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18 Nov 2016, 06:00
Bunuel wrote:
tarek99 wrote:
As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the approximate likelihood that all four people will choose different numbers?

a) 9%
b) 12%
c) 16%
d) 20%
e) 25%

SOLUTIONS FOR ALL SCENARIOS

When four people choose an integer between 1 and 4, inclusive 5 cases are possible:

A. All choose different numbers - {a,b,c,d};
B. Exactly 2 people choose same number and other 2 choose different numbers - {a,a,b,c};
C. 2 people choose same number and other 2 also choose same number - {a,a,b,b};
D. 3 people choose same number - {a,a,a,b};
E. All choose same number - {a,a,a,a}.

Some notes before solving:
As only these 5 cases are possible then the sum of their individual probabilities must be 1: $$P(A)+P(B)+P(C)+P(D)+P(E)=1$$

$$Probability=\frac{# \ of \ favorable \ outcomes}{total \ # \ of \ outcomes}$$

As each person has 4 options, integers from 1 to 4, inclusive, thus denominator, total # of outcomes would be 4^4 for all cases.

A. All choose different numbers - {a,b,c,d}:

$$P(A)=\frac{4!}{4^4}=\frac{24}{256}$$.

# of ways to "assign" four different objects (numbers 1, 2, 3, and 4) to 4 persons is 4!.

B. Exactly 2 people choose same number and other 2 choose different numbers - {a,a,b,c}:

$$P(B)=\frac{C^2_4*4*P^2_3}{4^4}=\frac{144}{256}$$.

$$C^2_4$$ - # of ways to choose which 2 persons will have the same number;
$$4$$ - # of ways to choose which number it will be;
$$P^2_3$$ - # of ways to choose 2 different numbers out of 3 left for 2 other persons when order matters;

C. 2 people choose same number and other 2 also choose same number - {a,a,b,b}:

$$P(C)=\frac{{C^2_4*\frac{4!}{2!2!}}}{4^4}=\frac{36}{256}$$.

$$C^2_4$$ - # of ways to choose which 2 numbers out of 4 will be used in {a,a,b,b};
$$\frac{4!}{2!2!}$$ - # of ways to "assign" 4 objects out of which 2 a's and 2 b's are identical to 4 persons;

D. 3 people choose same number - {a,a,a,b}:

$$P(D)=\frac{C^3_4*4*3}{4^4}=\frac{48}{256}$$.

$$C^3_4$$ - # of ways to choose which 3 persons out of 4 will have same number;
$$4$$ - # of ways to choose which number it will be;
$$3$$ - options for 4th person.

E. All choose same number - {a,a,a,a}:

$$P(E)=\frac{4}{4^4}=\frac{4}{256}$$.

$$4$$ - options for the number which will be the same.

Checking: $$P(A)+P(B)+P(C)+P(D)+P(E)=\frac{24}{256}+\frac{144}{256}+\frac{36}{256}+\frac{48}{256}+\frac{4}{256}=1$$.

Hope it's clear.

Bunuel

Why for B, order is matter whereas for C order is not matter?

If we work out B like C, why it will be wrong?
$$4C1.3.2.\frac{4!}{2!} = 288$$
$$4C1$$ - # of ways to choose which 1 number out of 4 will be used in {a,a,b,c};
$$3$$ - # of option for the 3rd person
$$2$$ - # of option for the 4th person
$$\frac{4!}{2!}$$ - # of ways to "assign" 4 objects out of which 2 a's are identical to 4 persons;
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Re: As part of a game, four people each must secretly choose an  [#permalink]

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02 Oct 2017, 03:27
Bunuel wrote:
jainsaurabh wrote:
Hi Bunuel,

For the case C, I am going wrong somewhere. Could you point out where please?

# of ways to chose 2 people from the group = 4C2
# of ways for this group to select one number out of 4 numbers = 4
# of ways to select 2 people out of remaining 2 = 2C2
# of ways for this group to select a number from the remaining 3 = 3

Hence total number of ways = 4C2*4*2C2*3 = 72.

# of ways to divide group of 4 into two groups of 2 when order of the groups does not matter is $$\frac{C^2_4*C^2_2}{2!}$$ and then you can do 4*3 for numbers;

But if you do just $$C^2_4*C^2_2$$ then you get the # of divisions of group of 4 into two groups of 2 when the order of the groups matters (you'll have group XY and also group YX with this fromula) then for numbers you should use $$C^2_4$$.

So your formula needs to be divided by 2! in any case to get rid of the duplications.

Check the following links for more on this issue:
http://gmatclub.com/forum/combination-a ... ng#p767453
http://gmatclub.com/forum/probability-8 ... al%20teams
http://gmatclub.com/forum/ways-to-divid ... al%20teams
http://gmatclub.com/forum/combinations- ... al%20teams

Hope it's clear.

Bunuel
Thanks for amazing explanation.
However, I am still confused about case D.
If I apply the method of dividing group of 4 into 2 groups of 1 and 3 when order of the groups does not matter, does it mean that I have to divide it by 2! too?
So, in D, it should be:
( (4C2 * 1C1)/2! ) * 4 * 3 = 24

And if I solve it by # ways of choosing number (as your solution for C), it will be:
4C2 * (4!/3!1!) = 24

Can you explain more? Thank you!
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Re: As part of a game, four people each must secretly choose an  [#permalink]

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02 Oct 2017, 07:04
tarek99 wrote:
As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the approximate likelihood that all four people will choose different numbers?

A. 9%
B. 12%
C. 16%
D. 20%
E. 25%

Total no. of ways in which the integers can be selected when there is no restriction = 4*4*4*4
Total no. of ways in which the integers can be selected when people have to take different integers = 4*3*2*1

So, percentage = 4*3*2*1 * 100 /(4*4*4*4) = (3/32)*100 = 9.375% ~ 9 %

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Re: As part of a game, four people each must secretly choose an  [#permalink]

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07 Mar 2018, 23:45
alphabeta1234 wrote:
Bunuel wrote:
tarek99 wrote:
As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the approximate likelihood that all four people will choose different numbers?

a) 9%
b) 12%
c) 16%
d) 20%
e) 25%

SOLUTIONS FOR ALL SCENARIOS

B. Exactly 2 people choose same number and other 2 choose different numbers - {a,a,b,c}:

$$P(B)=\frac{C^2_4*4*P^2_3}{4^4}=\frac{144}{256}$$.

$$C^2_4$$ - # of ways to choose which 2 persons will have the same number;
$$4$$ - # of ways to choose which number it will be;
$$P^2_3$$ - # of ways to choose 2 different numbers out of 3 left for 2 other persons when order matters;

C. 2 people choose same number and other 2 also choose same number - {a,a,b,b}:

$$P(C)=\frac{{C^2_4*\frac{4!}{2!2!}}}{4^4}=\frac{36}{256}$$.

$$C^2_4$$ - # of ways to choose which 2 numbers out of 4 will be used in {a,a,b,b};
$$\frac{4!}{2!2!}$$ - # of ways to "assign" 4 objects out of which 2 a's and 2 b's are identical to 4 persons;

Bunuel,

I am teribbly sorry to revive this old post but I am still trying to find out to out compute options (B) and (C) that you have laid out.

For (B) this was my thought process- - {a,a,b,c}:

(4C1)(4C2)(3C1)(2C1)(2C1)(1C1)=288 ( You got 144, which is half my answer, why do I need to divide by 2?)

(# of ways to pick the paired number)(# of ways to place the pair within the 4 slots) (# of ways to pick the first non-pair number)(# of ways to place first non-pair number in the remaining two slots)(# of ways to pick the second non-pair number)(# of ways to place the second non-pair in the last slot)

For (C) this was my thought process - {a,a,b,b}:

(4C1)(4C2)(3C1)(2C2)=72

(# of ways to pick the first pair)(# of ways to place first pair number in the 4 slots)(# of ways to pick the second pair )(# of ways to place second pair in the remaining 2 slots)

What am I doing wrong here? Help Bunuel!
Thank you!

Hi.

Actually this comes from a very basic question.

I had a similar doubt and after some thinking , here is my explanation.

If 2 people have to pick 2 numbers out of 3. How many ways can you do it?

When order matters , you will do it 2P23 when order doesn't matter you will do it in 2C3 ways.

Hope you are with me till this point.

Now for second part is 2C3 = 1C3*2C1?

Ans is No, because you are double counting the cases , hence 2C3= 1C3*2C1/2

Now comes the main part

Once you are taking 1C3*1C2*1C2*1C1

You have already considered the arrangement by picking 1C3*1C2 you don't have to have another 1C2 , simply because you are picking something doesn't mean there has to be a "C" attached to it.

Say you have A and B picking from 1,2 and 3.

Essentially by your approach you are saying

* You picked A , (1C2) and A can select any one out of 3 numbers ( 1C3)
I.e A1,A2,A3 or B1,B2,B3 (6 cases considered)
Consider it like a 2X3 matrix
|1 2 3|- 3 elements
|A B|- 2 elements

3*2 elements already considers all possible cases.

Now if you pick 1C2 the second time , you are picking A2 again in second pick if B1 is picked first. But please see above A2 is already picked as a case with B1. You don't have to do it another time.

It might take some time to digest as it did to me.
Hope practise solves this issue.

Hope to see a kudos for this!!

Thanks
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Re: As part of a game, four people each must secretly choose an  [#permalink]

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06 Oct 2018, 08:08
Each person chooses a distinct numbers/Each person chooses a number of his or her choice => (4X3X2X1)/(4X4X4X4) = 3/32 => 9/96 = ~9%
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Re: As part of a game, four people each must secretly choose an  [#permalink]

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28 Feb 2019, 19:20
Bunuel wrote:
tarek99 wrote:
As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the approximate likelihood that all four people will choose different numbers?

a) 9%
b) 12%
c) 16%
d) 20%
e) 25%

SOLUTIONS FOR ALL SCENARIOS

When four people choose an integer between 1 and 4, inclusive 5 cases are possible:

A. All choose different numbers - {a,b,c,d};
B. Exactly 2 people choose same number and other 2 choose different numbers - {a,a,b,c};
C. 2 people choose same number and other 2 also choose same number - {a,a,b,b};
D. 3 people choose same number - {a,a,a,b};
E. All choose same number - {a,a,a,a}.

Some notes before solving:
As only these 5 cases are possible then the sum of their individual probabilities must be 1: $$P(A)+P(B)+P(C)+P(D)+P(E)=1$$

$$Probability=\frac{# \ of \ favorable \ outcomes}{total \ # \ of \ outcomes}$$

As each person has 4 options, integers from 1 to 4, inclusive, thus denominator, total # of outcomes would be 4^4 for all cases.

A. All choose different numbers - {a,b,c,d}:

$$P(A)=\frac{4!}{4^4}=\frac{24}{256}$$.

# of ways to "assign" four different objects (numbers 1, 2, 3, and 4) to 4 persons is 4!.

B. Exactly 2 people choose same number and other 2 choose different numbers - {a,a,b,c}:

$$P(B)=\frac{C^2_4*4*P^2_3}{4^4}=\frac{144}{256}$$.

$$C^2_4$$ - # of ways to choose which 2 persons will have the same number;
$$4$$ - # of ways to choose which number it will be;
$$P^2_3$$ - # of ways to choose 2 different numbers out of 3 left for 2 other persons when order matters;

C. 2 people choose same number and other 2 also choose same number - {a,a,b,b}:

$$P(C)=\frac{{C^2_4*\frac{4!}{2!2!}}}{4^4}=\frac{36}{256}$$.

$$C^2_4$$ - # of ways to choose which 2 numbers out of 4 will be used in {a,a,b,b};
$$\frac{4!}{2!2!}$$ - # of ways to "assign" 4 objects out of which 2 a's and 2 b's are identical to 4 persons;

D. 3 people choose same number - {a,a,a,b}:

$$P(D)=\frac{C^3_4*4*3}{4^4}=\frac{48}{256}$$.

$$C^3_4$$ - # of ways to choose which 3 persons out of 4 will have same number;
$$4$$ - # of ways to choose which number it will be;
$$3$$ - options for 4th person.

E. All choose same number - {a,a,a,a}:

$$P(E)=\frac{4}{4^4}=\frac{4}{256}$$.

$$4$$ - options for the number which will be the same.

Checking: $$P(A)+P(B)+P(C)+P(D)+P(E)=\frac{24}{256}+\frac{144}{256}+\frac{36}{256}+\frac{48}{256}+\frac{4}{256}=1$$.

Hope it's clear.

Hello Bunuel,

your explanations are always good, But can you please check whether I got it correct or not?

1*3/4*2/4*1/4 gives 3/32

when we multiply 3/32*100 gives 9.37(approx.)...which is close to 9%

is it the right way to solve this type of questions?
Re: As part of a game, four people each must secretly choose an   [#permalink] 28 Feb 2019, 19:20

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