Originally Posted by

**Jhevon** first, do not put questions in quotes. notice that they do not show up when i quote you? this makes it annoying.

anyway, the trick here is to realize that you have a binomial distribution here.

Lets say you are repeating an experiment where each outcome is independent of the others. moreover, each outcome can be considered as a "success" or "failure". Then the probability of $\displaystyle k$ successes in $\displaystyle n$ trials, is given by:

$\displaystyle P(k) = {n \choose k} p^kq^{n - k}$

where $\displaystyle p$ is the probability of success and $\displaystyle q = 1 - p$ is the probability of failure, and $\displaystyle {n \choose k} = _nC_k$.

Now, for your first problem: $\displaystyle n = 4, ~k=3$ and for the second, $\displaystyle n = 6,~k = 4$. plug it in the formula i gave you to derive the desired expressions.

for the last problem, set the expression in the first problem equal to 2 times the expression in the second and simplify to get to what you want