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

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Manager
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As part of a game, four people each must secretly choose an [#permalink]

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11 Jun 2006, 13:20
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...
VP
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11 Jun 2006, 13:53
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...
VP
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11 Jun 2006, 13:56
Yep A too.

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

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

- Bernard Edmonds

Manager
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11 Jun 2006, 15:23
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.
Manager
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11 Jun 2006, 17:57
OA is 9%

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

thanks
VP
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11 Jun 2006, 18:10
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
_________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

- Bernard Edmonds

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12 Jun 2006, 02:09
giddi77 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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12 Jun 2006, 02:14
Total number of combinations: 4^4 = 256

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

P = 24/256 = 9%
SVP
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12 Jun 2006, 02:18
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%
VP
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12 Jun 2006, 11:10
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%
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