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!
 A: 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*}
A: *

*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.
$$P(D) = \frac{1}{2} + \frac{1}{6} - \frac{1}{2} \times \frac{1}{6} = \frac{7}{12}$$
A: Here's a fairly straightforward solution:
If you think about it, 3 heads occurring OR at least one head occurring looks like $A \cup B = \{3\} \cup \{1,2,3\} = \{1,2,3\}$. Thus $\bar{A} = \{0\}$ and it is trivially proved from the axioms of probability that $\mathbb{P}(A) = 1- \mathbb{P}(\bar{A}) = 1- \mathbb{P}(\{0 \text{ heads}\})$. 
Because the three tosses are independent, for some string of heads and tails $S$, $\mathbb{P}(S) = \prod \mathbb{P}(S_i)$. 
Therefore $\mathbb{P}(A) = 1-\mathbb{P}(0) = 1- {\left(\frac{1}{2}\right )}^3 = \frac{7}{8}$
