BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Difficult Math Problem #87 - Probability

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 354
Joined: Tue Jun 27, 2006 9:20 pm
Thanked: 11 times
Followed by:5 members

Difficult Math Problem #87 - Probability

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.
Top
Source: — Problem Solving |
Senior | Next Rank: 100 Posts
Posts: 76
Joined: Tue Dec 26, 2006 11:26 am
Followed by:3 members

by maxim730 » Wed Jan 17, 2007 3:14 pm
4!?
Top
Newbie | Next Rank: 10 Posts
Posts: 5
Joined: Mon Oct 23, 2006 7:19 pm

by sumitkrishna » Thu Jan 18, 2007 9:08 pm
1/24 ???
Top
Newbie | Next Rank: 10 Posts
Posts: 3
Joined: Tue Dec 26, 2006 10:41 pm

by sarika.k » Thu Jan 18, 2007 9:36 pm
3/32 ?
Top
User avatar
Legendary Member
Posts: 2469
Joined: Thu Apr 20, 2006 12:09 pm
Location: BtG Underground
Thanked: 85 times
Followed by:14 members

by aim-wsc » Fri Jan 19, 2007 1:56 am
c'mon guys come up with explanation.

why do you think so?
Top
User avatar
Senior | Next Rank: 100 Posts
Posts: 41
Joined: Fri Dec 15, 2006 9:23 am
Location: Greece
Followed by:1 members

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
Top
Master | Next Rank: 500 Posts
Posts: 354
Joined: Tue Jun 27, 2006 9:20 pm
Thanked: 11 times
Followed by:5 members

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
Top
Legendary Member
Posts: 627
Joined: Thu Jun 23, 2011 9:12 am
Thanked: 4 times
Followed by:1 members

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
Mayank
Top
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

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.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 496
Joined: Tue Jun 07, 2011 5:34 am
Thanked: 38 times
Followed by:1 members

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?
Top
Post new topic Post Reply
• Page 1 of 1