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