simple and good problem on probability

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chaitanyareddy: Senior | Next Rank: 100 Posts; Posts: 30; Joined: Fri Feb 27, 2009 12:45 am; Thanked: 1 times; GMAT Score:620

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

Source: Beat The GMAT — Problem Solving | Fri Aug 20, 2010 11:31 pm

kvcpk: Legendary Member; Posts: 1893; Joined: Sun May 30, 2010 11:48 pmFollowed by: 7 members; Thanked: 215 times

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.

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sirisha.g: Senior | Next Rank: 100 Posts; Posts: 85; Joined: Thu Jul 15, 2010 7:50 pm; Thanked: 1 times; GMAT Score:690

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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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

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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kvcpk: Legendary Member; Posts: 1893; Joined: Sun May 30, 2010 11:48 pmFollowed by: 7 members; Thanked: 215 times

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.

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