I have a problem I'm working through.
You roll a fair die m times. What is the probability $P(x)$ that exactly j of the rolls are $>4$ (that is, 5 or 6)?
Through seeing some similar problems, I believe that there is a solution involving inclusion-exclusion. Roughly speaking take the possibility of outcomes and subtract the two specific ones.
Following that line of thinking, I think it should look something like this, but feel I shouldn't be doing $6 \choose 4$ each time, or maybe this result should be subtracted form 1 for the two specific clauses I'm looking for: $6 \choose 4$$6^m$$-$$ 6 \choose 4$$5^m$$+$$6 \choose 4$$4^m$$-$$6 \choose 4$$3^m$$+$$6 \choose 4$$2^m$$-$$6 \choose 4$$1^m$$+$$6 \choose 4$$0^m$
What am I doing wrong here and is this in fact an application of inclusion-exclusion?
Let $X$ = number of rolls > 4. $X\sim$ Bin$(m,\frac{1}{3})$
Using the pdf of a binomial random variable, we have: $$ P(X=j)=p_X(j)=\binom{m}{j}\left(\frac{1}{3}\right)^j\left(\frac{2}{3}\right)^{m-j} $$
See: https://en.wikipedia.org/wiki/Binomial_distribution#Probability_mass_function