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24 Aug 2003, 15:47
Kudos
Bookmarks
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
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
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
24 Aug 2003, 17:01
Kudos
Bookmarks
praetorian123
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
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
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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