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.

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 $P(A \cap B)=P(A)=\frac{1}{8}$ 
