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
successes in
trials, is given by:
where
is the probability of success and
is the probability of failure, and
.
Now, for your first problem:
and for the second,
. 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