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
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
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