5 fair 6-sided dice are rolled. What is the probability of obtaining 2 ones and the sum of the remaining three = 10? Thank you for help.
I don't know how to solve it mathematically, but a short computer code gives the following number:
190 / 6^5 =2.4434 %.
Let us if a mathematician can arrive at the same number using permutations and combinations.
+118608 Thanks guest,
It is always very helpful when someone gives the answer via computer code.
Even better still when the method used has been stated clearly. 
It is not that hard to do it mathematically.
I just looked at how many triplets would add to 10. There is only 6 of them
\( 6,3,1,1,1 \qquad \qquad \frac{5!}{3!}=\frac{120}{6}=20 \;ways \\ 6,2,2,1,1 \qquad \qquad \frac{5!}{2!2!}=\frac{120}{4}=30 \;ways \\ 5,4,1,1,1 \qquad \qquad \frac{5!}{3!}=\frac{120}{6}=20 \;ways \\ 5,3,2,1,1, \qquad \qquad \frac{5!}{2!}=\frac{120}{2}=60 \;ways \\ 4,4,2, 1,1 \qquad \qquad \frac{5!}{2!2!}=\frac{120}{4}=30 \;ways \\ 4,3,3 ,1,1 \qquad \qquad \frac{5!}{2!2!}=\frac{120}{4}=30 \;ways \)
20+30+20+60+30+30 = 190
Altogether there are 6^5 = 7776 combionations
190/7776 = 0.0244341563786008
Just as guest's computer program already computed.
