simple and good problem on probability

[chaitanyareddy]

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

[kvcpk]

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.

[sirisha.g]

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.

[Rahul@gurome]

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

[kvcpk]

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.