What is the probability that I will roll a 2, 6 or more times when I throw a single die 10 times?

Do you know the formula for binomial probability? Given the probability of a single event p, the probability of it happening k times in n attempts is p(k) = C(n,k)p^k(1-p)^(n-k). So, do you know the probability of rolling a 2 for a single roll? Use that value for p, and then find the probability of rolling a two 6 out of 10 times, plus the probability of rolling a to 7 out of 10 times, plus 8 out of 10, plus 9 out of 10, plus 10 out of 10. Post back with what you get for an answer and we'll check it for you.

probability of rolling a 2 in a single role = 1/6
so p = 1/6 k=6; 7; 8; 9; 10 n= 10

P(6)= C(10,6)1/6^6 (1-1/6)^(10-6)

C(n,k) = n!/(k!(n-k)!)
C(10,6) = 10!/(6!(10-4)!)
C(10,6) = 3628800/ 720*720
= 3628800/ 3628800
= 1

P(6)= (1/46656) * (5/6)^4
= 0.00001033635
so 0.001%

this is the answer I am coming up with using the formula. Does this look right? Ive tried to put in all the steps of my calculation in case I've made mistakes:S

Thank you so so much by the way!! this is really helping me out a LOT!

C(n,k) = n!/(k!(n-k)!)
C(10,6) = 10!/(6!(10-4)!)

The second line is incorrect. Given n=10 and k = 6 you should have:

C(10,6)= \frac {10!}{6!4!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = \frac {5040}{24} = {210

So P(6) = 210(1/6)^6(5/6)^4 = 0.00217 = 0.217%

Now repeat the calculation for P(7), P(8), P(9), and P(10) and add them all up to get the total probability of six or more 2's in ten tries.