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As part of a game, four people each must secretly choose an [#permalink]
30 Nov 2007, 07:57

10

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

63% (02:20) correct
37% (01:30) wrong based on 487 sessions

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?

out of curiosity guys, sometimes there are questions that we will have to look for the opposite and then subtract that answer from 1. but how come it doesn't work with this problem? for example:

what is the probability that they will choose the same number instead of different numbers:

4/4 * 1/4 * 1/4 * 1/4 = 1/64

so 1 - 1/64 = 63/64 which is not even close to 9%. how come this doesn't work? cause i always confuse between when to use the opposite and when not. anybody?

out of curiosity guys, sometimes there are questions that we will have to look for the opposite and then subtract that answer from 1. but how come it doesn't work with this problem? for example:

what is the probability that they will choose the same number instead of different numbers:

4/4 * 1/4 * 1/4 * 1/4 = 1/64

so 1 - 1/64 = 63/64 which is not even close to 9%. how come this doesn't work? cause i always confuse between when to use the opposite and when not. anybody?

Because

(1 - everyone chooses same number) "Not equal to" (everyone chooses different number)

Infact , (1 - everyone chooses same number) "equal to" (everyone chooses not the same number)

But it still means 2 or 3 people can choose the same number, just not all 4.

The first guy can choose anything chance of success 1/1

Second guy can choose any three of the four numbers chance of success 3/4

Third guy can choose any two of the four numbers chance of success 2/4

Last guy can choose only one of the four numbers chance of success 1/4

Prob: 1 x 3/4 x 2/4 x 1/4 = 6/64 = 9%

Answer A

Excellent and quite fast.

Quote:

(1 - everyone chooses same number) "Not equal to" (everyone chooses different number)

Infact , (1 - everyone chooses same number) "equal to" (everyone chooses not the same number)

But it still means 2 or 3 people can choose the same number, just not all 4.

This is also possible, if we can write the combinations for the cases in which same number is selected by 4 people and then by 3 people and then by 2 people, and subtract the sum total from 1.

People are A,B,C,and D.

Number are 1,2,3,and 4

If 4 people select the same number:

number selected is 1:

A can choose it in 1/4 ways

B can choose it in 1/4 ways

C can choose it in 1/4 ways

D can choose it in 1/4 ways

Therefore: (1/4)*(1/4)*(1/4)*(1/4) = (1/4)^4

Other numbers which can be selected are 2,3,and 4

Therefore total number of ways = 4* [ (1/4)^4] = (1/4)^3 = 1/64

If 3 people select the same number:

number selected is 1:

A can choose it in 1/4 ways

B can choose it in 1/4 ways

C can choose it in 1/4 ways

D can choose any other numbers in 3/4 ways

Therefore: (1/4)*(1/4)*(1/4)*(3/4) = (3/4^4)

Other numbers which can be selected are 2,3,and 4

Therefore total number of ways = 4*(3/4^4) = 3/(4)^3 =3/64

If 2 people select the same number:

number selected is 1:

A can choose it in 1/4 ways

B can choose it in 3/4 ways

C can choose it in 2/4 ways

D can choose any other numbers in 3/4 ways

Therefore: (1/4)*(1/4)*(3/4)*(2/4) = (6/4^4)

Other numbers which can be selected are 2,3,and 4

Therefore total number of ways = 4*(6/4^4) = 6/(4)^3 =6/64

very quickly: total area of probability equals 4*4*4*4=4^4 at the numerator: 4! (if the first choose one number, there are 3 possible alternatives remaining...) so, 4!/4^4=0.093....

Re: PS: Probability II [#permalink]
27 Sep 2009, 00:59

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%

Soln: Chances that all four choose different numbers is Assuming first person chooses any of 4 numbers, second can choose the 1 from the left over 3, third can choose one number from the left over two and the last person chooses the last number. Hence 4 * 3 * 2 * 1 = 24 ways

Re: PS: Probability II [#permalink]
21 Apr 2010, 23:18

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%

Ans:a) 1st person has option no's- (1,2,3,4) - there fore probability of getting a no = 4c1/4c1 = 1

2nd person has option no's any three , he has to choose a no from three no's - there fore probability of getting a no = 3c1/4c1 = 3/4

3rd person has option no's any two , he has to choose a no from two no's -there fore probability of getting a no = 2c1/4c1 = 1/2

4th person has only one option - there fore probability of getting a no = 1c1/4c1 = 1/4 =1*3/4*1/2*1/4 = 3/32 = 9%

Re: PS: Probability II [#permalink]
08 Sep 2010, 03:48

11

This post received KUDOS

Expert's post

11

This post was BOOKMARKED

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}:

\(C^2_4\) - # 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}:

\(C^2_4\) - # 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.

Re: PS: Probability II [#permalink]
10 Sep 2010, 11:33

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

Re: PS: Probability II [#permalink]
10 Sep 2010, 12:13

1

This post received KUDOS

Expert's post

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.

Re: PS: Probability II [#permalink]
10 Sep 2010, 12:14

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.

Re: PS: Probability II [#permalink]
22 Feb 2012, 14:23

1

This post received KUDOS

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}:

\(C^2_4\) - # 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}:

\(C^2_4\) - # 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.

Way cool. Wish I could bookmark just this response. Many times, I bookmark a topic, but forget that it was actually a constituent post way way down that really prompted me to do so...

Re: PS: Probability II [#permalink]
28 Apr 2012, 17:34

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}:

\(C^2_4\) - # 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}:

\(C^2_4\) - # 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!

gmatclubot

Re: PS: Probability II
[#permalink]
28 Apr 2012, 17:34

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