determining the independence of coin flips i'm a bit confused by this example.
what confuses me the most is how to calculate Pr(E1), Pr(E2), Pr(E3), can someone explain

A fair coin is flipped 3 times. If (F1, F2, F3) denotes a typical flip sequence, let E1 denote the event that at least two of the Fi's are Heads, let E2 denote the event that exactly two of the Fi's are Heads, and let E3 denote the event that all the Fi are the same. Which of the pairs of these three events are independent?
Solution:
The only pair that is independent is E1 and E3.
Since Pr(E1) = (1/2)3 + (3 choose 2)(1/2)3 = 1/2
and
Pr(E3) = 2* (1/2)3 = 1/4, so
Pr(E1 ∩ E3) = Pr(all heads) = 1/8 = Pr(E1)Pr(E3), while Pr(E1 ∩ E2)=Pr(E2) and Pr (E2 ∩ E3)=0
 A: Record the result of the coin flips as a string of length $3$ made of of the letters H and/or T. Such a word records the results, in order, of the coin tosses. For example, HHT says we got a head then a head then a tail.
There are $(2)(2)(2)$ such sequences. They are all equally likely. 
Now consider the event $E_1$, which says that at least two of the results are heads. We can list explicitly the outcomes that make $E_1$ happen. They are HHH, HHT, HTH, and THH. So in $4$ of the $8$ equally likely possibilities, the event $E_1$ happens. Thus $\Pr(E_1)=\frac{4}{8}$.
Let us do this a slightly different way, without explicit listing. There is $1$ way we could have $3$ heads. For $2$ heads, the locations of the H can be chosen in $\binom{3}{2}$ ways. Thus the total number of patterns in which $E_1$ happens is $1+\binom{3}{2}$.
Now let us do the same thing in more probabilistic language. Each string of length $3$ has probability $\left(\frac{1}{2}\right)^3$. There is $1$ string with all $H$, and there are $\binom{3}{2}$ strings with two H and a T, so the probability of $E_1$ is $\left(1+\binom{3}{2}\right)\left(\frac{1}{2}\right)^3$.
Now to check whether two events $A$ and $B$ based on the above coin tossing are independent, all we have to do is to check whether $\Pr(A\cap B)=\Pr(A)\Pr(B)$. If we have equality, then $A$ and $B$ are independent. If we do not have equality, then $A$ and $B$ are not independent.
Since we can determine any of these probabilities by counting, checking for independence is mechanical, and quick. 
