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Someone pls solve this question...

by akshatgupta87 » Wed Apr 06, 2011 10:20 am
Can someone explain the logic behind this question.

Q) 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%

TIA
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by sourabh33 » Wed Apr 06, 2011 12:27 pm
is Ans 16%

The probability of all choosing a different no = 4 x (1/4) x (1/3) x (1/2) x (1/1) = 1/6 = 16.67%
a> Multiplied by 4 since 4 permutations possible.

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by akshatgupta87 » Wed Apr 06, 2011 12:38 pm
Answer is A> 9%
But i want to know the logic behind it.
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by Geva@EconomistGMAT » Wed Apr 06, 2011 12:40 pm
akshatgupta87 wrote:Can someone explain the logic behind this question.

Q) 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%

TIA
~Akshat
Break down proabability questions into a series of events, and calculate the probability of sucess of each event given that the previous ones were a success. The trick is to define "success" properly.

For the first guy, we really don't care which number he chooses, so success is a guaranteed 1. You can also think of it as "4 wanted outcomes" out of "4 possible outcomes, or 4/4.
Given that the first guy chose a number, the success of the second event is "not the number chosen by the first. the next guy has only 3 "wanted" outcomes - the other three numbers not chosen by the first - from a possible four.
For the third guy, only 2 wanted outcomes out of 4.
For the fourth guy, only 1/4.
Thus, the probability we're looking for is 1* 3/4 * 2/4 * 1/4 = 6/64 - slightly less than 10%, so the answer is A.
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by GMATGuruNY » Wed Apr 06, 2011 12:46 pm
akshatgupta87 wrote:Can someone explain the logic behind this question.

Q) 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%

TIA
~Akshat
The first person can choose any of the 4 numbers.

P(2nd number is different) = 3/4. (Out of the 4 numbers, we can't select the first number chosen, leaving us 4-1=3 good options.)
P(3rd number is different) = 2/4. (Out of the 4 numbers, we can't select the first 2 numbers chosen, leaving us 4-2=2 good options.)
P(4th number is different) = 1/4. (Out of the 4 numbers, we can't select the first 3 numbers chosen, leaving us 4-3=1 good option.)

Since we want all of the events above to happen together, we multiply the fractions:
3/4 * 2/4 * 1/4 = 3/32 = 9/96 ≈ 9%.

The correct answer is A.
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by pankajks2010 » Thu Apr 07, 2011 4:43 am
akshatgupta87 wrote:Can someone explain the logic behind this question.

Q) 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%

TIA
~Akshat
Well, the detailed text book approach can be like this:

Total number of ways in which 4 numbers can be chosen by 4 people with repetition is: 4*4*4*4=256
Now, 4 people can choose 4 different numbers in 4*3*2*1=24 ways

Now, the required probability is 24/256 which can be expressed in % as (24/256)*100~9.4%
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