Trouble finding probabilities. I am working through some problems about probability and seem to be having trouble working through this one in particular. I'd love some help learning how to go about solving problems such as this.
A website estimates that 19% of people have a phobia regarding public speaking. If three students are assigned to a project group, what's the probability...
a.) That all 3 students have a fear of public speaking.
b.) That none have a fear of public speaking
c.) That at least one of the students has a fear of public speaking.
 A: Use the fact that these students are chosen independently, and so $$\Bbb P(A\text{ and }B\text{ and }C)=\Bbb P(A)\Bbb P(B)\Bbb P(C)$$ For your problem, you could say $A$ is the event that the first student has a fear of public speaking, etc. You can also use this property for part $B$.
For questions where they say something like "at least", think of the opposite - the opposite of at least one student having this fear is for none of them to have this fear. This is easier to calculate, and in fact you have already worked it out in another question. Then the probability you'd want is $1-\Bbb P(\text{the opposite})$.
A: The problem involves the count of 'successes' in a known amount of independent trials with equal success rate.   There are three people, each with a probability of $0.19$ for possessing phobia.
The key point in this problem is to recognise that this count is a Random Variable which follows a Binomial Distribution. 
Familiarity with such will then solve the problem.   In particular, you should know the probability mass function (pmf) for a binomial distribution.
$$X\sim\mathcal{Bin}(n, p) \quad\iff\quad \mathsf P(X=k) ~=~ \binom n k p^k (1-p)^{n-k}~\mathbf 1_{k\in \{0,..,n\}}$$
If you have yet to encounted a binomial distribution, note that this is simply the probability of having a sequence of $k$ successes and $n-k$ failures multiplied by the count of distinct arrangements of that sequence.   (Here $n=3$, $p=0.19$, and a success is "having the phobia").
$$\mathsf P(X=k) ~=~\begin{cases}0.89^3 &:& k=0 \\ 3\cdot 0.19\cdot 0.89^2 &:& k=1\\3\cdot 0.19^2\cdot 0.89 &:& k=2 \\ 0.19^3 &:& k=3\\ 0 & :& \text{otherwise}\end{cases}$$
From there it is simply a matter of determining what probabilities are requested and applying the pmf to evaluate.

PS: The indicator function, $\mathbf 1_{k\in\{0,..,n\}}$, has the value of $1$ when $k$ is an integer between $0$ and $n$, and a value of $0$ otherwise. 
PPS: $\binom nk$ is the binomial coefficient, also written as ${}^n\mathrm C_k$.   It is the count of ways to choose $k$ from a set of $n$ elements (here, people).
A: For this question,:
A) If one kid has a fear then probability of picking him will be $0.19$ while picking one more reduces the probability to $(0.19)*(0.19)$ . For three people: it becomes $(0.19)^3$
 This is from selection via independent events.
B)None have fear: $(1-0.19)^3$
This is via noting that 81% population won't be fearful as such. and following the above principal again.
C) At least one have fear: $1-(1-0.19)^3$
This one is by asking in account that for at least one to be fearful, we may take it as Anti-event when none is fearful. Thus 1-P(part B) 
