A Binomial bound for the CDF of the Hypergeometric distribution?

• Let $$H \sim Hyp(N,K,n)$$, where $$Hyp$$ denotes the hypergeometric distribution, $$N$$ the number of objects, $$K$$ the number of "good" objects, and $$n$$ the number of draws.
• I am interested in a particular bound for $$\mathbb{P}(H \leq x)$$.
• Let $$B_x \sim Bi\left(n, \frac{K-x}{N-x}\right)$$, where $$Bi$$ denotes the Binomial distribution.

Intuitively, if no more than $$x$$ of the $$n$$ draws associated with the Hypergeometric distribution are successful (i.e., result in drawing a "good" object), the probability of a success in each of these draws never falls below $$\frac{K-x}{N-x}$$. Therefore, the following inequality might seem like a reasonable conjecture:

(1)$$\qquad$$ $$\mathbb{P}(H\leq x) \leq \mathbb{P}(B_x \leq x)$$, $$\qquad$$ for all $$x \leq K$$.

I've looked online a little bit and couldn't find any reference to (1). Maybe this inequality is easy to prove or disprove, but I haven't been able to.

Beside the intuitive "argument" above, her is some (arguably very limited) suggestive evidence that (1) might be true.

Example 1

Suppose that $$N =4$$, $$K=2$$, and $$n = 2$$. Then,

• $$\mathbb{P}(H \leq 0) = (1/2)*(1/3)= 1/6$$
• $$\mathbb{P}(B_0 \leq 0) = (1/2)*(1/2)= 1/4$$

Also,

• $$\mathbb{P}(H \leq 1) = (1/2)*(1/3) + (1/2)*(2/3) + (1/2)*(2/3)= 5/6$$
• $$\mathbb{P}(B_1 \leq 1) = (2/3)*(2/3) + (1/3)*(2/3) + (2/3)*(1/3)= 8/9$$

Example 2

In Mathematica, plotting the difference between the two CDFs for a couple of values of the paramaters and $$x$$

DiscretePlot[
Table[CDF[HypergeometricDistribution[n, 50, 100],
k], {n, {10, 20, 50}}] -
Table[CDF[BinomialDistribution[n, (50 - k)/(100 - k)],
k], {n, {10, 20, 50}}] // Evaluate, {k, 0, 32}, PlotRange -> All]


yields

Some things I found difficult when trying to prove the inequality:

1. As far as I know, there is no really convenient formula for the CDF's of Binomial and (even less so) of Hypergeometric distributions.
2. If the inequality holds, it certainly does not hold "pointwise", in the sense that we don't have $$\mathbb{P}(H = y) \leq \mathbb{P}(B_x = y)$$ for all $$y \in \{0,\dots, x\}$$. So if the inequality holds, it really has to do with the whole sum of the PMFs from $$0$$ to $$x$$, which I find hard to play with.

My questions:

1. Can someone provide a counter-examples or a proof of (1)?
2. If (1) is true, is there a good reference for it that I could cite?