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simple and good problem on probability

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by chaitanyareddy » Fri Aug 20, 2010 11:31 pm
Hi this is a simple and very good question on probability. Just need to think differently to answer this question.

Posting this for your reference.

Q) As a part of a game, four people each much secretly chose an integer between
1 and 4 inclusive. What is the approximate likelihood that all four people will
chose different numbers?

Soln: The probability that the first person will pick unique number is 1 (obviously)
then the probability for the second is 3/4 since one number is already picked by the
first, then similarly the probabilities for the 3rd and 4th are 1/2 and 1/4 respectively.
Their product 3/4*1/2*1/4 = 3/32
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by kvcpk » Sat Aug 21, 2010 5:39 am
Hi Chaitanya,

Are you sure about the solution?
I think you are wrong.

Question asks fo r the probability that all four people will chose different numbers.
Pick can be like this too
A-1
B-1
C-1
D-2

This is also a valid pick. your count doesnt include this.
"Once you start working on something,
don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
Chanakya quotes (Indian politician, strategist and writer, 350 BC-275BC)
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by sirisha.g » Sat Aug 21, 2010 5:53 am
kvcpk wrote:Hi Chaitanya,

Are you sure about the solution?
I think you are wrong.

Question asks fo r the probability that all four people will chose different numbers.
Pick can be like this too
A-1
B-1
C-1
D-2

This is also a valid pick. your count doesnt include this.
each person should pick a unique number. A,B,C picking the same number isn't possible. So, chaitanyareddy is right.
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by Rahul@gurome » Sat Aug 21, 2010 5:54 am
Another approach:

The number of ways in which all four people can select different numbers is 4!.
The number of ways in which all four can select any number is 4*4*4*4 = 4^4.
So the required probability is 4!/(4^4) = 3/32
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by kvcpk » Sat Aug 21, 2010 6:03 am
chaitanyareddy wrote:Hi this is a simple and very good question on probability. Just need to think differently to answer this question.

Posting this for your reference.

Q) As a part of a game, four people each much secretly chose an integer between
1 and 4 inclusive. What is the approximate likelihood that all four people will
chose different numbers?

Soln: The probability that the first person will pick unique number is 1 (obviously)
then the probability for the second is 3/4 since one number is already picked by the
first, then similarly the probabilities for the 3rd and 4th are 1/2 and 1/4 respectively.
Their product 3/4*1/2*1/4 = 3/32
I misinterpreted the question. 3/32 should be right for this.
"Once you start working on something,
don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
Chanakya quotes (Indian politician, strategist and writer, 350 BC-275BC)
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