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Fix $0 \le p \le 1$. The cumulative distribution function of the Binomial Distribution $B(n, p)$, which counts the number of successes out of $n$ independent trials, each of which has probability $p$ for success, is $$F_n(k) = \sum_{i\le k} \binom{n}{i} p^i (1-p)^{n-i}.$$

For any fixed $k\ge 0$, is $F_n(k) $ decreasing in $n$, for $n\ge k$?

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Yes: Model the binomial distribution as the amount of successes in $n$ independent coin tosses $X_1, \dots, X_n$ with success probability $p$. Let $B_n$ be the amount of successes in the first $n$ tosses. By definition, $B_n$ is now $B(n,p)$ distributed and $B_{n-1}$ is $B(n-1,p)$-distributed. \begin{align} F_n(k) &= \mathbf P(B_n≤k) \\ &= \mathbf P(B_{n-1}+X_n ≤ k) \\ &= \mathbf P(B_{n-1}≤ k-X_n) \\ &≤ \mathbf P(B_{n-1} ≤ k) \\ &= F_{n-1}(k) \end{align} So $F_n(k)$ is indeed decreasing in $n$.

In words: The probability of $k$ or less successes in $n$ throws must be lower than $k$ or less successes in $(n-1)$ throws, because the former implies the latter.

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  • $\begingroup$ $$\Pr(B_n\leq k)=\frac{n!}{k!(n-k-1)!}\int_p^1t^k(1-t)^{n-k-1}dt$$ is good to know. $\endgroup$ Commented Dec 17, 2023 at 9:32

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