It is currently 26 Jun 2017, 19:23

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Probability - Dice -- Need Clarification

Author Message
CEO
Joined: 15 Aug 2003
Posts: 3454
Probability - Dice -- Need Clarification [#permalink]

### Show Tags

24 Aug 2003, 15:47
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Hi ,

This is problem that has been solved by a member of this forum...i think its not quite correct...can you please check?

(1) One has three fair dice and rolls them together. What is the probability of having two same numbers and one different? (for example, 2-2-6, 1-3-3, 6-5-6, and so on)

His solution :

1) Total no. of possibilities = 6*6*6 = 216
No. of events where all the three are equal = 6
No. of events when none are eqaul = 6*5*4 = 120
Therefore the no. of events when two are equal and third different
= 216-6-120 = 90

Therefore Probability that 2 are equal and third is different = 90 / 216 = 5/12

I think that the solution 90 contains numbers that have been counted more than once. Since no positions are unique..166 is as good as 661 which is as good as 616..since it fulfills the conditions using the same set of numbers..

IMO, 90 should be divided by 3 to take care of these numbers.
so the favorable results are 90/3 = 30

The required probability is 30/216 = 5/36

Thanks
Praetorian
GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Re: Probability - Dice -- Need Clarification [#permalink]

### Show Tags

24 Aug 2003, 17:01
praetorian123 wrote:
Hi ,

This is problem that has been solved by a member of this forum...i think its not quite correct...can you please check?

(1) One has three fair dice and rolls them together. What is the probability of having two same numbers and one different? (for example, 2-2-6, 1-3-3, 6-5-6, and so on)

His solution :

1) Total no. of possibilities = 6*6*6 = 216
No. of events where all the three are equal = 6
No. of events when none are eqaul = 6*5*4 = 120
Therefore the no. of events when two are equal and third different
= 216-6-120 = 90

Therefore Probability that 2 are equal and third is different = 90 / 216 = 5/12

I think that the solution 90 contains numbers that have been counted more than once. Since no positions are unique..166 is as good as 661 which is as good as 616..since it fulfills the conditions using the same set of numbers..

IMO, 90 should be divided by 3 to take care of these numbers.
so the favorable results are 90/3 = 30

The required probability is 30/216 = 5/36

Thanks
Praetorian

the original solution is correct. You need to count all of the "double counts" because the denominator of the fraction is a permutation, thus it includes all of the double counts also.

Another way of solving:

We can roll any number for 1st die.

1/6 chance to match it with 2nd die, 5/6 chance that the 3rd will not match both 1st and 2nd.

5/6 chance not to match it, 2/6 chance to match either 1st or 2nd.

(1/6 * 5/6) + (5/6 * 2/6) = 15/36 = 5/12
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Re: Probability - Dice -- Need Clarification   [#permalink] 24 Aug 2003, 17:01
Display posts from previous: Sort by