Probability Unfair Coins I have two coins, yielding heads at $P(H) = a$ and the other with $P(H) = b$.
Is it possible for the following to be equal? If so, what should be values for $a$ and $b$? 
(a) $p_1$=neither is head
(b) $p_2$=exactly one of the coins is a head
(c) $p_3$=both are heads
I am not sure where I am going with this. I am thinking that:
$p_1= (1-a)(1-b)$,
$p_2=a(1-b)+b(1-a)$
$p_3=ab$
Is this correct? In addition, I see the coins as definitely unfair, but should they be double headed/tailed? 
I found: 
$p_1=p_3 \iff a+b=1$
Is it safe to describe the possible values of $a$ and $b$ as complements? 
 A: "I am not sure where I am going with this. I am thinking that the p1= (1-a)(1-b) p2=a∪b and p3=ab."
p1 and p2 are correct.  "p2=a∪b" is nonsense because p2 is a number, not a set.  You have already said that the probability that both are heads (p1) is (1- a)(1- b) and that the probability neither is heads (p2) is ab.  The only thing left is "one is heads and the other isn't" or "exactly one is heads".  So the probability of that is 1- (1- a)(1- b)- ab= 1- 1+ a+ b- ab- ab= a+ b- 2ab.
"In addition, I see the coins as definitely unfair, but should they be double headed/tailed?"
If you think that the only way coins can be unfair is to be "double headed" or "double tailed" you are very unimaginative. (Or haven't spent enough time flipping coins!) It is not at all difficult to weight coins or grind down their edges so that the are more likely to come up one rather than the other.
A: To satisfy (a),(b) and (c) you need first (a)=(c), which as you say, means
$$a+b=1$$
This also means that $1-2ab=ab$ (from (b)=(c)), so $ab=\frac 13$.
Substitute $a=1-b$, we get $b(1-b)=\frac 13$.
However $b(1-b)$ has a local maxima at $b=\frac 12$, and this equates to $\frac 14$, meaning that no real solution is possible.
