Suppose that A and B are mutually exclusive events for which $P(A) = 0.3$ and $P(B) = 0.5$ Suppose that A and B are mutually exclusive events for which $P(A) = 0.3$ and $P(B) = 0.5$. What is the probability that 
(a)  either A or B occurs? 
(b)  A occurs but B does not? 
(c)  both A and B occur? 

Trying to understand how to solve this problem, I solved  (a) and got $0.15$ (by $0.3 \cdot0.5$) for the probability for either of them to occur. However, the back of my text stated the answer for the problem was 
$\dfrac{3\cdot4\cdot4\cdot3}{{14 \choose 4}} = 0.1439$.
Could someone help explain to me what I'm doing wrong? This is my first stats class.
 A: To have mutually exclusive events means if one of those events occurs, the others cannot occur. Therefore, for the intersection of mutually exclusive events $A_\mathrm{i}$ with $\mathrm{i} \in \{1, \dots, \mathrm{n} \ | \ \mathrm{n} \in \mathbb{N}   \setminus  \{1\}  \}$ holds $\   \bigcap_1^n A_\mathrm{i} = \emptyset$. This implies $P[\bigcap_1^n A_\mathrm{i}] = P[\emptyset] = 0$. In general, the probability of the union of two events is $P[B\bigcup C] = P[B] + P[C] - P[B\bigcap C]$ . Hence, for mutually exclusive events holds $P[\bigcup_1^n A_\mathrm{i}] = \sum\limits_1^n P[A_i]$. Knowing this, you can apply it to your tasks:
a) $P[A\bigcup B] = P[A] + P[B] = 0.3 + 0.5 =0.8$
b) Occurence of A implies no occurence of B$ \implies P[A] = 0.3$
c) $P[A\bigcap B] = P[\emptyset] = 0$.
As already was suggested in the comments, the solutions in your textbook are not right and inappropriate for this kind of task. 
Moreover, what you did in the calculation of a) was assuming $A$ and $B$ are independent and interpreting "either" as "and". Note, in probability theory, the term "or" indicates the union of events and the term "and" indicates the intersection. Therefore, it holds:
$P[A$ or $B] \ge P[A$ and $B]$ $ \iff $$P[A \cup B] \ge P[A \cap B]$.
