How to find the probability of three friends out of five friends coming to pick you up? I am working on a probability course with following problem:

You are stranded by the road, so you decide to call some of your friends. You know they are not very trustworthy– in fact, each one will only come with probability 0.1– so you decide to call five of them and ask them to come get you without telling them that you called the others.
a. If each of your friends decides to go get you independently from the others, what is the probability that three of your friends come?
b. You realize that the independence assumption does not make sense (your friends tend to be in similar moods). Lower bound the probability that you are left stranded.

For point a., I think the solution should be: $\mathbb{P}(\text{3 friends coming}) = 0.1 + 0.1 +0.1$. Because, as everything is independent, its just adding up individual probabilities of friends showing up.
What do you think?
For point b., I am not sure about the lower bound. Any inputs would be great.
 A: 
(a)

You add probabilities in case of mutually exclusive events and multiply them in case of independent event. If $3$ friends are coming to your rescue, the $3$ events (each corresponding to a friend coming) are not mutually exclusive at all, in fact they are happening together so for (a) the probability of three specified friends coming is $0.1^3$. And as lulu notes below the probability of any three friends coming is $\binom{5}{3}0.1^30.9^2$.   

(b)

Suppose you had two friends and the events they come are denoted by $A_1,A_2$. The probability of you getting stranded is
$$
P(A_1^c\cap A_2^c)=1-P(A_1\cup A_2)
$$
So to get the lower bound you need to maximize $P(A_1\cup A_2)$.
$$
P(A_1\cup A_2)=P(A_1)+P(A_2)-P(A_1\cap A_2)
$$
To maximize $P(A_1\cup A_2)$ we have to set $P(A_1\cap A_2)=0$ which gives the lower bound as $0.8$. So we just saw that
$$
P(A_1\cup A_2)\le P(A_1)+P(A_2)\\
\implies P(A_1\cup A_2\cup A_3)\le P(A_1\cup A_2)+P(A_3)\le P(A_1)+P(A_2)+P(A_3)\\
\implies\cdots
\implies P\left(\bigcup_{j=1}^5A_j\right)\le\sum_{j=1}^5P(A_j)
$$
So in your original problem a lower bound is
$$
1-\sum_{j=1}^5P(A_j)=1-0.5=0.5
$$
A: The probability of exactly $3$ of your friends showing up independently requires that the other two do not show up, and the number of ways of choosing $3$ from the $5$: $$0.1^3 \cdot 0.9^2 {5 \choose 3} = 0.00081 \cdot 10 = 0.0081$$
If your friends do not make decisions independently then, worst case, they make identical decisions so you either get all $5$ turning up (probability $0.1$) or none of them (probability $0.9$). This is clearly worst case - $0.9$ chance of being stranded - because there is no improvement over calling a single friend.  
Best case is that your friends coordinate (somewhat), and only at most one of your friends will turn up - then the probability of getting rescued is then $5\times 0.1=0.5$ and the probability of being stranded is also $0.5$, which is its lower bound.
