The key formula here will be the binomial distribution. It allows you to calculate the probability that you will obtain a specified result a specified number of times, given the number of total trials, and provided that there are only two possible outcomes. The formula is:
__N!__(p)[itex]\stackrel{k}{}[/itex](1-p)[itex]\stackrel{(N-k)}{}[/itex] = P
k!(N-k)!
where N=number of trials
p=probability that the specified outcome will occur in one trial
P=probability that the specified outcome will occur k times
So for instance, if you are dealing with a fair coin, and you want to know the probability that a heads will be flipped 2 times in 6 tries,
N=4 (your number of tries)
k=2 (you want the probability that it will turn up heads twice)
p=0.50 (a fair coin has a 50% chance of landing on heads in one flip)
P= probability that you will get 2 heads after 6 flips (in this particular situation P=0.234)
This should be enough information to help you find out the probability that 4 heads will turn up in 5 flips. However, the question asks "What is the probability that you will get at most 4 heads?". This is just the probability of getting 4, plus the probability of getting three, etc.
The standard deviation is just the square root of the average squared distance from the average value (so if angled brackets denote the average of a certain quantity and [itex]\sigma[/itex] is the standard deviation, [itex]\sigma[/itex]=[itex]\sqrt{<(x-<x>)^{2}>}[/itex] )
What this means for you is that you should first find the average number of heads using the probability that 1 head will be flipped in 5 tries, the probability that 2 heads will be flipped in 5 tries, and so on (up to 5 heads in 5 tries). This average corresponds to <x> in the formula above. Then calculate the average squared difference between each of these probabilities and the mean (average [itex](x-<x>)^{2}[/itex]) and take the square root of the result.
*There are actually formulas that would speed up some of these calculations, (such as the standard deviation calculation), but since you haven't listed any known equations, here's a minimal formula solution.