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

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Manager
Joined: 03 Sep 2005
Posts: 135
As part of a game, four people each must secretly choose an [#permalink]

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16 Sep 2005, 10:43
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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%

I tried to solve this by doing (1/4)(1/3)(1/2)(1), but that didn't work. Can someone explain why that is incorrect? My thought process was that there is a 1/4 chance the first person will pick a certain number, then since there are only 3 numbers left, there is a 1/3 chance, etc... Now I'm all confused. Help?
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Life is not measured by the number of breaths you take, but by the number of moments that take your breath away.

Manager
Joined: 03 Sep 2005
Posts: 135

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16 Sep 2005, 10:52
need700+, can you please explain your logic?

need700+ wrote:
i get A

4/4*3/4*2/4*1/4=24/256=9.385%

_________________

Life is not measured by the number of breaths you take, but by the number of moments that take your breath away.

Senior Manager
Joined: 27 Aug 2005
Posts: 331
Location: Montreal, Canada

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16 Sep 2005, 11:12
This may be a bit less mathematical but I also got A.

Probability = desired outcomes / total outcomes.

Total outcomes = 4^4 = 256.

Desired outcomes = 6*4 = 24

To get to the 24 desired outcomes, I simply went about it methodically. If the first person drew a 1, how many ways are there for the other three to avoid repeats. There are 6 ways: 1234, 1243, 1324, 1342, 1423, and 1432. The same thing repeats when the first person chooses 2, 3 or 4. So there are 6x4 or 24 desired outcomes.

24/256 = 3/32 = ~9%
VP
Joined: 22 Aug 2005
Posts: 1112
Location: CA
Re: PS: Permutation Game [#permalink]

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16 Sep 2005, 11:27
trulyblessed 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%

I tried to solve this by doing (1/4)(1/3)(1/2)(1), but that didn't work. Can someone explain why that is incorrect? My thought process was that there is a 1/4 chance the first person will pick a certain number, then since there are only 3 numbers left, there is a 1/3 chance, etc... Now I'm all confused. Help?

A.

corrections:
1/4 chance the first person will pick a certain number

you dont have to find probability of first person "picking a number". The question asks about probability of picking DIFFERENT (than the rest 3) numbers. Now, lets solve your ways:

- chance for first person to pick a DIFFERENT number is: 1. As he is first person, his number is always different.

- chance for second person to chose DIFFERENT number = 3/4
- thirs: 2/4
- fourth : 1/4

p = 1 * 3/4 * 1/2 * 1/4 = 3/32 = .0937 or ~9 %
Intern
Joined: 19 Aug 2005
Posts: 41

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16 Sep 2005, 11:57
here is how i did it

we are letting the first person to choose any number among 4 so his chance is 4/4
the second person can choose any 3 numbers among 4.. so his chances is 3/4
the third person can choose any 2 numbers among 4.. so his chance is 2/4
and the last person had only 1 number to choose frm 4 so his chance is 1/4

multiplying all these i get ~9%
Manager
Joined: 03 Sep 2005
Posts: 135

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16 Sep 2005, 12:33
Wow, you guys are soooo awesome. I really appreciate your help!!!
_________________

Life is not measured by the number of breaths you take, but by the number of moments that take your breath away.

16 Sep 2005, 12:33
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# As part of a game, four people each must secretly choose an

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