probability rolling a dice 5 times I can't solve this problem:
What is the probability that, when rolling a dice 5 times, the number of times when you get a 1 or 2 is greater than the number of times when you get a 6.
any help?

@Stones Yes, but the table is not quite right. If there are three or more 1s or 2s, then it does not matter how many 6s there are. So N5 the number of ways of getting five 1s or 2s is 32. Similarly, N4 the number of ways of getting four is $5\times 2^4\times 4=320$ and N4 the number of ways of getting three is $10\times 2^3\times 4^2=1280$. If we get just one 1/2 then we cannot get any 6s, so N1 is $5\times 2\times 3^4=810$.
That leaves the trickier case of two 1s/2s. If there are no 6s, then we have N2a as $10\times 2^2\times 3^3=1080$. If there is one 6, then we have N2b as $10\times 3\times 2^2\times 1\times 3^2=1080$. Adding those up, we get 4602, so the probability is 767/1296 = 0.5918 approx.
[The coefficients 1, 5, 10 are just the binomial coefficients, because we are looking at the number of ways of picking 5, 4, 3 things from 5. The extra 3 for N2b comes because there are 3 ways of choosing which of the three results not a 1 or 2 is the 6.]
