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? 

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.
