|
Author |
Message |
|
TAGS:
|
|
|
Senior Manager
Joined: 19 Oct 2004
Posts: 329
Location: Missouri, USA
Followers: 1
Kudos [?]:
2
[0], given: 0
|
Five fair dice with faces marked 1,2,3,4,5 and 6 are rolled. [#permalink]
 08 Nov 2004, 14:03
Five fair dice with faces marked 1,2,3,4,5 and 6 are rolled. Calculate the probability of throwing exactly the same number on at least four dice.
I liked this question quite a lot. Wanted to share it with everyone..
_________________
Let's get it right!!!!
|
|
|
|
|
|
|
Director
Joined: 16 Jun 2004
Posts: 929
Followers: 1
Kudos [?]:
4
[0], given: 0
|
Denomiator = 6^5 = 7776 possibilities.
Now, for the numerator = total outcomes(which is 6^5) - number of outcomes where all four dice show different numbers (which is 6*5*4*3) -number of outcomes where all show same number (which is 6) = 7776-360 -6 = 7410
Prob is 7410/7776 = 3705/3888. I dont think I am sure of this one. OA pls.
|
|
|
|
|
|
Senior Manager
Joined: 19 Oct 2004
Posts: 329
Location: Missouri, USA
Followers: 1
Kudos [?]:
2
[0], given: 0
|
Well thats wrong..... Shall I give out the OA so fast?
_________________
Let's get it right!!!!
|
|
|
|
|
|
Director
Joined: 16 Jun 2004
Posts: 929
Followers: 1
Kudos [?]:
4
[0], given: 0
|
I know that the numbers are getting unfriednly...some mistake...how about 5 answer choices..
|
|
|
|
|
|
Director
Joined: 16 Jun 2004
Posts: 929
Followers: 1
Kudos [?]:
4
[0], given: 0
|
Also, missed on 'ATleast'...anyway give the answer choices...this is a tough one..
|
|
|
|
|
|
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4441
Followers: 10
Kudos [?]:
82
[0], given: 0
|
Wait, not yet. I'm currently working on a report. I'll give an answer within next 15 minutes 
_________________
Best Regards,
Paul
|
|
|
|
|
|
Senior Manager
Joined: 19 Oct 2004
Posts: 329
Location: Missouri, USA
Followers: 1
Kudos [?]:
2
[0], given: 0
|
 No answer choices for this one.. sorry venksune.. 
Its just a Probability sum a like. So I picked it up. IT has a few concepts in it. Lets wait for Paul's Answer.
_________________
Let's get it right!!!!
|
|
|
|
|
|
Director
Joined: 31 Aug 2004
Posts: 619
Followers: 1
Kudos [?]:
8
[0], given: 0
|
First draft with Bernoulli, 6.(5C4.(1/6)^4.(5/6)^1+1/6^5)=26/1296
does not smell so good
Last edited by twixt on 08 Nov 2004, 16:56, edited 3 times in total.
|
|
|
|
|
|
Director
Joined: 16 Jun 2004
Posts: 929
Followers: 1
Kudos [?]:
4
[0], given: 0
|
|
|
|
|
|
|
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4441
Followers: 10
Kudos [?]:
82
[0], given: 0
|
Too late, damn report! 
_________________
Best Regards,
Paul
|
|
|
|
|
|
Intern
Joined: 08 Nov 2004
Posts: 46
Location: Montreal
Followers: 0
Kudos [?]:
0
[0], given: 0
|
The probability of getting four dice to have the same number:
6 / 6^4
So, In the numerator we have: (6/6^4).6= 1/6^2
In the denominator we have 6^5
The result is: (1/6^2)/6^5 = 1/6^7
|
|
|
|
|
|
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4441
Followers: 10
Kudos [?]:
82
[0], given: 0
|
Answer should be 25/7776 after the afore mentioned calculations
_________________
Best Regards,
Paul
|
|
|
|
|
|
Manager
Joined: 07 Nov 2004
Posts: 93
Location: London
Followers: 1
Kudos [?]:
1
[0], given: 0
|
The probability of obtaining at least four "sixes" when rolling five dice is as per Paul's solution that is, C(5,4)*5/(6^5). However, there are six marks on each dice {1,2,3,4,5,6} so that ans is
6*C(5,4)*5/(6^5) = 5*5/(6^4) = 25/1296
My humble opinion 
|
|
|
|
|
|
Manager
Joined: 07 Nov 2004
Posts: 93
Location: London
Followers: 1
Kudos [?]:
1
[0], given: 0
|
Forgot the fact that all five dice could show the same number - In that case ans is
6*[ 5C4*5/(6^5) + 5C5*1/(6^5)] = 13/648 ~ 1/50
|
|
|
|
|
|
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4441
Followers: 10
Kudos [?]:
82
[0], given: 0
|
Oxon, you are right, I forgot to consider that there are 6 possible values (1,2,3,4,5,6) that could be same and also when all 5 are same!
Hence, it should be:
[(1/6)^4 * 5/6 * 5C4 + (1/6)^5] * 6 = 26/6^4 = 13/648
You are absolutely right 
_________________
Best Regards,
Paul
|
|
|
|
|
|
Manager
Joined: 07 Nov 2004
Posts: 93
Location: London
Followers: 1
Kudos [?]:
1
[0], given: 0
|
That was a tough one!
ruhi could you please give all 5 answer choices?
|
|
|
|
|
|
Manager
Joined: 07 Nov 2004
Posts: 93
Location: London
Followers: 1
Kudos [?]:
1
[0], given: 0
|
sorry!
OA would do it!
|
|
|
|
|
|
Director
Joined: 16 Jun 2004
Posts: 929
Followers: 1
Kudos [?]:
4
[0], given: 0
|
Extremely good posts (except mine). Well solved.
|
|
|
|
|
|
Senior Manager
Joined: 19 Oct 2004
Posts: 329
Location: Missouri, USA
Followers: 1
Kudos [?]:
2
[0], given: 0
|
Oxorn is RIGHT. Thats the perfect answer. 
_________________
Let's get it right!!!!
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
close
In the count of three I need either the answer or Paul's response...can't sleep otherwise peacefully.. 
