(a) $\{$ Head, Tail $\}$

(b) $\{1,2,3,4,5,6\}$

(c) $\{(1 \cap H e a d),(1 \cap$ Tail $), \ldots,(6 \cap$ Head $),(6 \cap$ Tail $)\}$

Clearly $\mathrm{P}(A)=\frac{1}{2}=\mathrm{P}(B)$. We can assume that the two events are independent, so

$$

\mathrm{P}(A \cap B)=\mathrm{P}(A) \mathrm{P}(B)=\frac{1}{4}

$$

Alternatively, we can examine the sample space above and deduce that three of the twelve equally likely events comprise $A \cap B$. Also, $\mathrm{P}(A \cup B)=\mathrm{P}(A)+\mathrm{P}(B)-\mathrm{P}(A \cap B)=\frac{3}{4}$, where this probability can also be determined by

noticing from the sample space that nine of twelve equally likely events comprise $A \cup B$.