Flawed understanding of independent events Suppose we pick a bit string of length 4 at random, all bit strings equally likely. Consider the following
events:
$E_1$: the string begins with 1.
$E_2$: the string ends with 1.
$E_3$: the string has exactly two 1’s
Are $E_1$ and $E_2$ independent? Justify your answer.
My reasoning was that:
$Pr(E_2) * Pr(E_1) = .5 * .5 = .25$ 
$Pr(E_1∩E_2) =$ $4\choose2$$.5^2*.5^2 = .375$
$.375$ does not equal $.25$, so I said they were not independent. However, the states: 
Yes, $Pr(E_1 ∩ E_2) = .25 =.5 * .5 = Pr(E_1) * Pr(E_2)$
I was wondering why my reasoning was flawed?
 A: I'm not sure that your understanding of independence is flawed, it's just your calculation of $P(E_1\cap E_2)$.  To choose a string in $E_1\cap E_2$ you have to choose the second digit ($2$ possibilities) and choose the third digit ($2$ possibilities).  So there are $2^2$ strings, not $\binom42$, and we get
$$P(E_1\cap E_2)=2^2(0.5)^2(0.5)^2=0.25\ .$$
A: 
My reasoning was that:
$\Pr(E_2) \cdot \Pr(E_1) = 0.5 \cdot 0.5 = 0.25$ 
$\Pr(E_1\cap E_2) = {4\choose 2}\cdot 0.5^2\cdot 0.5^2 = .375$

The first is correct, but the second is in error.
What you have calculated the the probability of having exactly two 1 anywhere in the bit string (and two 0 elsewhere).   That isn't what you wanted.   This is actually: $\Pr(E_3)$
$$E_3~\neq~ E_1\cap E_2$$
What you needed to calculate was the probability of having 1 in the first and last positions and anything in the middle two positions.
$$\begin{array}{rcl}\Pr(E_1) =& 0.5\cdot 1\cdot 1\cdot 1~~~ &= 0.5 \\ \Pr(E_2) =& ~~~1\cdot 1\cdot 1\cdot 0.5 &= 0.5 \\ \Pr(E_1\cap E_2) =& 0.5\cdot 1\cdot 1\cdot 0.5 &= 0.25\end{array}$$
Now, as stated, you have found $\Pr(E_3)$ and similarly we find:
$$\begin{array}{rcl}\Pr(E_3) =& \binom{4}{2}\cdot 0.5^4 &= 0.375 \\ \Pr(E_1\cap E_3) =& \binom{3}{1}\cdot 0.5^4 &= 0.1875 \\ \Pr(E_2\cap E_3) =& \binom{3}{1}\cdot 0.5^4 &= 0.1875 \\ \Pr(E_1\cap E_2\cap E_3) =& ~~~~~~0.5^4 &= 0.0625\end{array}$$
So, what does this all tell you about the independencies of the three events?
A: It's not that hard to look at all possible strings. There are $2^4= 16$ "bit strings of length 4".  The are 
0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111,
1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111.
"E1: the string begins with 1"
  Okay there are 8 of those: 
1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111.
"E2: the string ends with 1."
There are 8 of those:
0001, 0011, 0101, 0111, 1001, 1011, 1101, 1111.
4 of the strings in E1 also end with a "1" so P(E2)= 8/16= 1/2 and P(E2| E1)= 4/8= 1/2.  Those are equal so E1 and E2 are independent.
E3: the string has exactly two 1’s"
there are $\begin{pmatrix}4 \\ 2 \end{pmatrix}= \frac{4!}{2!2!}= 6$ of them
0011, 0101, 0110, 1001, 1010, 1100.
