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!

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*Suppose a fair coin, meaning that there is $\frac{1}{2}$ chance of obtaining heads, and also $\frac{1}{2}$ chance of obtaining tails.


$A \cup B$ means the union of case A and case B, and $P(A \cup B)$ is the probability of EITHER of these cases happening. Since obtaining at least one head, case B, includes obtaining three heads (case A), we just need to consider $P(B)$, as case A is part of case B. Hence, $P(A \cup B) = P(B)$.
The probability of obtaining at least one head also includes the probability of obtaining two and three heads. You can apply the binomial probability distribution to obtain the answer, but we can simplify this. The probability of obtaining at least one head, $P(B)$, is also equivalent to one minus the probability of obtaining no heads. The probability of obtaining no heads, say $P(C)$, is equivalent to the probability of obtaining three tails, and it can be calculated as follows:
$$P(C) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}$$
Then, we calculate $P(B)$:
$$P(B) = 1 - P(C) = 1 - \frac{1}{8} = \frac{7}{8}$$
And finally,
$$P (A \cup B) = P(B) = \frac{7}{8}$$


*Getting a head or a 4 does not include getting both a head AND a four.


Say the probability of getting a head or a 4 is $P(D)$. Besides adding together the probability of obtaining a head and the probability of obtaining a 4 (which is what you did), you need to subtract the probability of obtaining a head and a four.
$$P(D) = \frac{1}{2} + \frac{1}{6} - \frac{1}{2} \times \frac{1}{6} = \frac{7}{12}$$
