calculating different probabilities of a coin toss I am trying to practice calculating the probabilities of various events regarding a coin toss:
We throw a regular coin $50$ times and define the following events:
a - four heads in a row in the $4$ first tosses
b - four heads in a row in the last $4$ tosses
c - obtaining $20$ total heads out of the $50$ tosses
I am trying to calculate: $p(a \cup b)$ and $p(a^C \cap b \cap c)$.
So the basic probabilities are as following: 
$$p(a)=p(b)=\frac{4!}{4!}\left(\frac{1}{2}\right)^4$$ 
$$p(c)=\frac{50!}{20!30!}\left(\frac{1}{2}\right)^{20}\left(\frac{1}{2}\right)^{30}$$
For $p(a \cup b)$: Since they are disjoint, we can omit the $p(a \cap b)$, and we get that $p(a \cup b) = p(a) + p(b)$.
For $p(a^C \cap b \cap c)$: $p((1-p(a) \times p(c) \times p(b))$.
Is this correct? Would really appreciate your input or corrections.
 A: You have calculated $\Pr(a)$, $\Pr(b)$, and $\Pr(c)$ correctly.  

What is $\Pr(a \cup b)$?

We know that 
$$\Pr(a \cup b) = \Pr(a) + \Pr(b) - \Pr(a \cap b)$$
Your claim that events $a$ and $b$ cannot both occur is false.  Notice that a sequence of $50$ tosses can begin with four heads in a row and end with $4$ heads in a row.  Since these events are independent,
$$\Pr(a \cap b) = \Pr(a)\Pr(b)$$
With that in mind, you can now calculate $\Pr(a \cup b)$.

What is $\Pr(a^C \cap b \cap c)$?

You cannot simply multiply the probabilities that events $a^c$, $b$, and $c$ occur since they are not independent.  Notice that if events $b$ and $c$ both occur, then four of the twenty heads occur in the last four tosses.  Thus, $20 - 4 = 16$ heads occur in the first $46$ tosses.  If event $a$ did occur, event $b$ and $c$ would both occur if and only if exactly $20 - 4 - 4 = 12$ of the middle $50 - 4 - 4 = 42$ tosses are heads.  Thus, the desired probability is found by subtracting the probability that exactly $12$ of the middle $42$ tosses are heads from the probability that exactly $16$ of the first $46$ tosses are heads.    
A: You're right with your probabilities.
$P(a\cap b)=(\frac{1}{2})^{8}=\frac{1}{256}$
This means by $$P(a\cup b)=P(a)+P(b)-P(a\cap b);$$
$$P(a\cup b)=\frac{31}{256}\approx0.121$$

$$P(a' \cap b \cap c)$$
This means your results must meet the following criteria:A: The first four must contain 1 or more tails.
B: 16 of Tosses 5-46 inclusive (42 in all) must be heads
C: All four of 47-50 must be heads.
$$P(A)=\frac{15}{16}$$
$$P(B)=\binom{42}{16}\bigg(\frac{1}{2}\bigg)^{42}$$
$$P(C)=\frac{1}{16}$$
Overall probability is all three events together $$\rightarrow \frac{15}{2^{50}}\binom{42}{16}\approx0.00222$$
