A ratio of two probabilities I am concerned about the monotonicity of the following ratio
$
f(\eta)=\frac{\sum_{x=K}^{N}\left(\begin{array}{c}
N\\
x
\end{array}\right)\left(q_{G}\eta\right)^{x}\left(1-q_{G}\eta\right)^{N-x}}{\sum_{x=K}^{N}\left(\begin{array}{c}
N\\
x
\end{array}\right)\left(q_{B}\eta\right)^{x}\left(1-q_{B}\eta\right)^{N-x}}
$
For $q_{G}>q_{B}$ and $0<K<N$, some numerical experiments imply that it is decreasing in $\eta$. I wonder how to prove this property? Thanks a lot.
 A: Let $t:=\eta$, $n:=N\in\{1,2,\dots\}$, and $k:=K\in\{1,\dots,n\}$. We need to show that the ratio
$$r(t):=\frac{G(t)}{B(t)}$$
decreases in $t$,
where
\begin{equation}
    B(t):=s(q_B\, t),\quad G(t):=s(q_G\, t),
\end{equation}
\begin{equation}
    s(p):=P(X_{n,p}\ge k),
\end{equation}
and $X_{n,p}$ is a random variable with the binomial distribution with parameters $n,p$; here we must assume that $q_B>0$ and $t\in(0,1/g_G)$, so that $q_G\,t$ and $q_B\,t$ be in the interval $(0,1)$.
Following the comment by Brendan McKay, note that
\begin{equation}
    s'(p)=\frac{n!}{(k-1)! (n-k)!}\, p^{k-1} (1-p)^{n-k}. 
\end{equation}
So, the "derivative ratio"
\begin{equation}
    \rho(t):=\frac{G'(t)}{B'(t)}=C\Big(\frac{1-q_G\,t}{1-q_B\,t}\Big)^{n-k} 
\end{equation}
is decreasing in $t\in(0,1/g_G)$; here, $C$ is a positive real number which does not depend on $t$. Also, $B(0+)=0=G(0+)$.
So, by the special-case l'Hospital-type rule for monotonicity (see e.g. Proposition 4.1),
$r(t)$ is decreasing in $t\in(0,1/g_G)$, as desired.
