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Difficult Math Problem #87 - Probability

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by 800guy » Wed Jan 17, 2007 11:33 am
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?

this question comes from the difficult math doc on this website, OA coming after some people answer.
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by maxim730 » Wed Jan 17, 2007 3:14 pm
4!?
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by sumitkrishna » Thu Jan 18, 2007 9:08 pm
1/24 ???
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by sarika.k » Thu Jan 18, 2007 9:36 pm
3/32 ?
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by aim-wsc » Fri Jan 19, 2007 1:56 am
c'mon guys come up with explanation.

why do you think so?
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by thankont » Fri Jan 19, 2007 3:45 am
Choosing the same number is (1/4)^4 since say our chosen # is 2 then prob. of first person choosing 2 is 1/4 etc.
Now here we want different numbers so let first person choose 1 with prob. (1/4), second chooses number 2 with p = 1/4 so I think that probability is the same with the first case and p=(1/4)^4
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OA

by 800guy » Fri Jan 19, 2007 9:31 am
i second aim-wsc's comments..the purpose of sharing these questions is to get more discussion around explanations, sharing of knowledge.

here's the OA:

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 mankey » Wed Sep 21, 2011 9:27 am
Favourable cases: 4! (All assigned different numbers)
Total cases: 4^4 (Since each number can be assigned to any person)

Probability=4!/4^4=3/32

Please let me know if this approach is correct?

Thanks
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by GMATGuruNY » Wed Sep 21, 2011 10:23 am
800guy wrote: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?

this question comes from the difficult math doc on this website, OA coming after some people answer.
The first person can choose any of the four numbers.

P(2nd person chooses a number not yet chosen) = 3/4. (4 total numbers, 3 of them not yet chosen.)
P(3rd person chooses a number not yet chosen) = 2/4. (4 total numbers, 2 of them not yet chosen.)
P(4th person chooses a number not yet chosen) = 1/4. (4 total numbers, 1 of them not yet chosen.)

Since we want all of the events above to happen together, we multiply the fractions:
3/4 * 2/4 * 1/4 = 3/32.
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by sl750 » Wed Sep 21, 2011 11:47 am
I am just curious. If the first number is no longer available for picking, the next person has to chose from the remaining 3 numbers, so how is that the total numbers still remain as 4 and not 3?
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