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

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New post 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

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

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New post 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.
[Reveal] Spoiler:
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.
[Reveal] Spoiler:
3C1

Similarly in how many ways can C and D choose?
[Reveal] Spoiler:
C : 2C1 and D 1C1


what will be P(A)?
[Reveal] Spoiler:
4c1/4c1. Total number of outcomes = 4C1

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

Try out the overall probability it should be the same as #1 above
[Reveal] Spoiler:
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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New post 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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New post 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.

hence A is the answer
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Re: As part of a game, four people each must secretly choose an [#permalink]

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New post 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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New post 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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New post 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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New post 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 %

Answer A
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Re: As part of a game, four people each must secretly choose an   [#permalink] 02 Oct 2017, 07:04

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