Probability of coins and boys example is unsolvable I have a couple of problems that are giving a headache.
P1)
A coin is tossed three times. Let $A = \{\text{Three head occurs}\}$, $B = \{\text{At least one head occurs}\}$
Find $P(A\cup B)$.
Now $P(A) = 1/8$  , $P(B) = 1-1/8 = 7/8$ so
$$P(A \cup B) = P(A)+P(B)-P(A\cap B) = 1/8+7/8 - (1/8)\cdot(7/8)=57/64$$ 
However, the textbook answer is 7/8. Where am i messing up?
P2)
A class has $10$ boys and $4$ girls. If $3$ students are selected at random, what is the probability that all will be boys?
I'm totally clueless with these problems.
 A: You calculated $P(A)$ and $P(B)$ correctly.  But you cannot in general calculate $P(A\cup B)$ as $P(A) + P(B)$, because there might be events where both $A$ and $B$ have occurred, and $P(A) + P(B) $ counts those events twice, once as part of $P(A)$ and once as part of $P(B)$.
Suppose, for example, that the cafeteria serves cake on four days of each week, and serves soup on four days of each week.  You cannot conclude from this that the cafeteria serves cake or soup on eight days of each week; that is absurd, because weeks have only seven days.  To get the right answer you also have to know how many days the cafeteria serves cake and soup.
In your question you have $A = \text{three heads}$ and $B = \text{at least one head}$.  But in this case event $B$ contains event $A$; every instance of event $A$ is also an instance of event $B$, because you cannot have three heads without also having at least one head.  So in this case $P(A\cup B) = P(B) = \frac78$, which matches the answer in your book.
In this case the problem was easy, because event $A$ was a subset of event $B$.  In general it may not be so simple. A technique that always works is to  calculate $$P(A \cup B) = P(A) + P(B) - P(A\cap B)\tag{$\star$}$$ where the $P(A) + P(B)$ counts the overlapping events twice, and then subtracting $P(A\cap B)$ subtracts the overlapping events once; then the overlapping events are counted only once in total.  If you apply that formula in this case, you have:
$$\begin{array}{clc}
P(A) & \text{three heads} & \frac18 \\
P(B) & \text{at least one head} & \frac78\\
P(A\cap B) & \text{three heads }\textbf{and}\text{ at least one head} & \frac18
\end{array}
$$
and using $(\star)$, you get $\frac 18+\frac78 - \frac18 = \frac78$, which again matches the answer in your book. 
Note that you cannot in general calculate $P(A\cap B) $ as $P(A)\cdot P(B)$; that only applies when $A$ and $B$ are independent events.  
A: The intersection of A and B is A. Lets suppose 
$B_1$: 1 head occurs
$B_2$: 2 heads occurs
$B_3$: 3 heads occcurs
Then $P(B)=P(B_1)+P(B_2)+P(B_3)$, where $P(B_3)=P(A)$
Thus $P(A \cap B)=P(A)=\frac{1}{8}$ 
