A coin is tossed 3 times. What is the probability of getting 3 heads or at least 1 head? 
  
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*A coin is tossed $3$ times. Let $A=[3 \text{ heads occur}]$ and $B = [\text{at least 1 head occurs}]$. What is $P(A \cup B)$?
  

This is a SAT MATH 2 questions from Barron's book. The answer is $\frac{7}{8}$ but I do not understand why. 


  
*If a coin is flipped and one die is thrown, what is the probability of getting a head or a $4$? 
  

For this one, I tried doing $\frac{1}{2} + \frac{1}{6} =\frac{2}{3}$, $\frac{1}{2}$ being the probability of getting a head and $\frac{1}{6}$ being the probability of getting a $4$. Since the problem asks for either or, I added the two. But the answer is $\frac{7}{12}$.
Can anyone please explain how to arrive at the answers? Thank you!

This is a common mistake. You likely memorized 
$$P(A\cup B) = P(A)+P(B)$$
but this is only true if $A$ and $B$ are disjoint: $A\cap B = \varnothing$. 
Instead, we have by inclusion-exclusion
$$P(A\cup B) = P(A)+P(B)-P(A\cap B).$$
Also recall that $$P(AB) = P(A)P(B)$$
if $A$ and $B$ are independent.


*

*Notice that $A\cap B = A$ and so
$$P(A\cup B) = P(A)+P(B)-P(A) = P(B).$$
Use the complement,
$$P(B) = 1-P(\bar B) = 1-P(\text{No heads}) = 1-\left(\frac{1}{2}\right)^3 = \frac{7}{8}.$$

*Let $A$ be the event that you flip a head and let $B$ be the event that you land a four. Assume they are independent. Then
\begin{align*}
P(A\cup B) &= P(A)+P(B)-P(AB) \\
&= P(A)+P(B)-P(A)P(B) \\
&= \frac{1}{2}+ \frac16-\frac{1}{2}\cdot\frac{1}{6} \\
&= \frac{7}{12}.
\end{align*}
