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

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As part of a game, four people each must secretly choose an [#permalink] New post 11 Jun 2006, 12:20
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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%

pls explain...
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Re: Permutation: Choose numbers [#permalink] New post 11 Jun 2006, 12:53
acfuture 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%

pls explain...


= 4!/4^2=9% approx...
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 [#permalink] New post 11 Jun 2006, 12:56
Yep A too.

Reqd. prob = 4/4*3/4*2/4*1/4 = 3/32 =~ 9%
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 [#permalink] New post 11 Jun 2006, 14:23
giddi77 wrote:
Yep A too.

Reqd. prob = 4/4*3/4*2/4*1/4 = 3/32 =~ 9%


Can you explain this a bit firther? I am really weak in Probablity.
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 [#permalink] New post 11 Jun 2006, 16:57
OA is 9%

but can one of you explain it in more detail... I still don't get it...

thanks
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 [#permalink] New post 11 Jun 2006, 17:10
pesquadero wrote:
giddi77 wrote:
Yep A too.

Reqd. prob = 4/4*3/4*2/4*1/4 = 3/32 =~ 9%


Can you explain this a bit firther? I am really weak in Probablity.


First person can pick any of the 4 numbers. Hence prob = 4/4
Second person has to pick any of the remaining 3 numbers. Hence his/her prob = 3/4
Similarly, third person, prob = 2/4
Fourth person ... prob = 1/4

Total prob = 4/4*3/4*2/4*1/4
HTH
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- Bernard Edmonds

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 [#permalink] New post 12 Jun 2006, 01:09
giddi77 wrote:
pesquadero wrote:
giddi77 wrote:
Yep A too.

Reqd. prob = 4/4*3/4*2/4*1/4 = 3/32 =~ 9%


Can you explain this a bit firther? I am really weak in Probablity.


First person can pick any of the 4 numbers. Hence prob = 4/4
Second person has to pick any of the remaining 3 numbers. Hence his/her prob = 3/4
Similarly, third person, prob = 2/4
Fourth person ... prob = 1/4

Total prob = 4/4*3/4*2/4*1/4
HTH


Alternative way of looking at it -

Each person can choose from four numbers. So total number of ways = 4*4*4*4 = 256

Number of ways in which each person selects one unique number is similar to a permutation of 4 unique objects (order is important as each person is unique) = 4P4 = 24

Probability = 24/256 = 3/32
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 [#permalink] New post 12 Jun 2006, 01:14
Total number of combinations: 4^4 = 256

# of ways all 4 choose different number = 4! = 24


P = 24/256 = 9%
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 [#permalink] New post 12 Jun 2006, 01:18
Probabilty of first person choosing a number = 4/4
Probability of 2nd person choosing a number not chosen by 1st person = 3/4
Similarily 3rd person = 2/4
4th person = 1/4

Total P = 3/32

Liklihood = 3/32*100 ~9%
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 [#permalink] New post 12 Jun 2006, 10:10
A. 9%

The # of ways each person can pick a number between 1 & 4 (1,2,3,4 ) = 4

Four people can pick a number in 4x4x4x4

For each person to pick a unique number # of ways = 4x3x2x1

Prob = (4x3x2x1)/(4x4x4x4) = (3/32)

Approx. prob = 9%
  [#permalink] 12 Jun 2006, 10:10
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