The question generate event with $6.75\space p^2q$, $20\space p^3q^2$, $3.9\space pq$? prompted me to think more generally about which expressions of this form can be a probability in a finite coin experiment. Since the answer turns out to be rather nice, I thought I'd post it here as an answer to the general question.
We have a coin that shows heads with probability $p$ and tails with probability $q:=1-p$, and the question is whether for given $\mu\in\mathbb R^+$ and $i,j\in\mathbb N$ (not both zero) there is an event in an experiment with a fixed finite number of tosses of the coin whose probability is $\mu p^iq^j$ for all $p\in[0,1]$.
First, $\mu$ needs to be an integer: The probability of any event in an experiment with a fixed finite number of tosses is a polynomial in $p$ with integer coefficients; and $\mu p^iq^j=\mu p^i(1-p)^j$ is a polynomial in $p$ with coefficient $\mu$ for $p^i$, so $\mu$ must be integer for the two to be equal.
The rest of the “only if” direction is also readily proved. Since $\mu p^iq^j$ is a probability for all $p$, it must be $\le1$ for all $p$. If there were $p$ such that $\mu p^iq^j=1$, the event would have to include all elementary events, which implies $\mu p^iq^j=1$ for all $p$, which only occurs in the trivial case $i=j=0$ that was excluded in the question.
Now, to prove the “if” direction, consider an experiment with $i+j+l$ tosses. If we can select $\mu\binom lk$ elementary events with $i+k$ heads and $j+l-k$ tails for all $k=0,\ldots,l$, then the probability of the event composed of all these elementary events is
$$ \sum_{k=0}^l\mu\binom lkp^{i+k}q^{j+l-k}=\mu p^iq^j(p+q)^l=\mu p^iq^j\;. $$
There are $\binom{i+j+l}{i+k}$ such elementary events, so we can choose $\mu\binom lk$ of them if
$$ \mu\binom lk\le\binom{i+j+l}{i+k} $$
for all $k$. If we cancel factors in the three pairs of corresponding factorials, this becomes
$$ \mu(k+i)\cdots(k+1)(l-k+j)\cdots(l-k+1)\le(l+i+j)\cdots(l+1)\;, $$
and then dividing through by $l^{i+j}$ and denoting $\frac kl\in[0,1]$ by $\alpha$ leads to
$$ \mu\left(\alpha+\frac il\right)\cdots\left(\alpha+\frac1l\right)\left(1-\alpha+\frac jl\right)\cdots\left(1-\alpha+\frac 1l\right)\le\left(1+\frac{i+j}l\right)\cdots\left(1+\frac1l\right) $$
and thus to
$$ \mu\alpha^i(1-\alpha)^j+O\left(\frac{(i+j)^2}l\right)\le1\;. $$
Thus, we can satisfy this inequality for sufficiently large $l$ if $\mu\alpha^i(1-\alpha)^j\lt1$ for all $\alpha\in[0,1]$.