About 10.84% is the answer.
The exact value is (5^10)7(11^5)/(6^18)
First, determine the probability that you'll get at least one 6 on a single throw of two dice. Of the 36 ways that two dice can fall, eleven of them have at least one 6. So the probability is 11/36 for a single throw.
Now what's the probability of exactly 1 throw out of 10 having at least one 6? That would be
+ 11/36 * (25/36)^9
+ 25/36 * 11/36 * (25/36)^8
+ (25/36)^2 * 11/36 * (25/36)^7
...
+ (25/36)^9 * 11/36
or
11/36 * (25/36)^9 * 10
In general the probability of exactly n events out of m having a condition with probability p happen will be
p^n*(1-p)^(m-n) multiplied by the number of ways you can select n out of m items.
And since the formula for selecting n of m items is m!/(n!(m-n)!), the overall formula becomes
m!/(n!(m-n)!)p^n*(1-p)^(m-n)
so
10!/(5!5!)*(11/36)^5*(25/36)^5
= 252*(161051/60466176)*(9765625/60466176)
= (5^10)7(11^5)/(6^18)
~ 0.108402426
So the exact answer is
(5^10)7(11^5)/(6^18)
and the approximate answer is
0.108402426
which is about 10.84%
