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As a part of a game, 4 people each choose one number from 1

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New post 16 Apr 2006, 18:07
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As a part of a game, 4 people each choose one number from 1 to 4. What is the likelihood that all people will choose different numbers?
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New post 16 Apr 2006, 18:22
3/32?

If right, i will try to explain...
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New post 16 Apr 2006, 20:26
I guess some info should be added ))Do they return the numbers after choosing ?
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New post 16 Apr 2006, 20:28
willget800 wrote:
3/32?

If right, i will try to explain...


I think this is correct. The probability would be 4!/4^4.

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New post 16 Apr 2006, 21:09
willget800 wrote:
3/32?

If right, i will try to explain...


Please do :) I think answer should be 1/24 ... but I am not sure...
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New post 17 Apr 2006, 07:06
People answer is
6/4^3 = 0.09
Can someone explain?

As a part of a game, 4 people each choose one number from 1 to 4. What is the likelihood that all people will choose different numbers?

This is what I personally feel

Good event = all 4 people getting same number. can be 4 possibilities

Total events = 4*4*4

P= 4/4*4*4 = 1/16
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New post 17 Apr 2006, 07:21
chillpill wrote:
People answer is
6/4^3 = 0.09
Can someone explain?

As a part of a game, 4 people each choose one number from 1 to 4. What is the likelihood that all people will choose different numbers?

This is what I personally feel

Good event = all 4 people getting same number. can be 4 possibilities

Total events = 4*4*4

P= 4/4*4*4 = 1/16


4! / 4^4 = 4*3! / 4*4^3 = 3! / 4^3 = 6 / 4^3 = 0.09
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New post 17 Apr 2006, 08:14
As a part of a game, 4 people each choose one number from 1 to 4. What is the likelihood that all people will choose different numbers?

1st chance choose out of 4
2nd out of 3
3rd out of 2
4th the remanining one

different selections = 4! = 24

total selections = 4*4*4*4 = 256

P(different selections) = 24/256 = 3/32
  [#permalink] 17 Apr 2006, 08:14
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As a part of a game, 4 people each choose one number from 1

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