Let $p$ be given, $0 < p < 1$. Let $n$ and $k$ be positive integers such that $np \leqq k \leqq n$, and let $B_n(k) = \sum^n_{r=k} \binom{n}{r} p^rq^{n-r}$, where $q = 1 - p$. It is shown that $B_n(k) = \big\lbrack\binom{n}{k} p^kq^{n - k}\big\rbrack qF(n + 1, 1; k + 1; p),$ where $F$ is the hypergeometric function. This representation seems useful for numerical and theoretical investigations of small tail probabilities. The representation yields, in particular, the result that, with $A_n(k) = \big\lbrack\binom{n}{k}p^kq^{n - k + 1}\big\rbrack \lbrack(k + 1)/(k + 1 - (n + 1)p)\rbrack$, we have $1 \leqq A_n(k)/B_n(k) \leqq 1 + x^{-2}$, where $x = (k - np)/(npq)^{\frac{1}{2}}$. Next, let $N_n(k)$ denote the normal approximation to $B_n(k)$, and let $C_n(k) = (x + \sqrt{q/np}) \sqrt{2\pi} \exp \lbrack x^2/2 \rbrack$. It is shown that $(A_nN_nC_n)/B_n \rightarrow 1$ as $n \rightarrow \infty$, provided only that $k$ varies with $n$ so that $x \geqq 0$ for each $n$. It follows hence that $A_n/B_n \rightarrow 1$ if and only if $x \rightarrow \infty$ (i.e. $B_n \rightarrow 0$). It also follows that $N_nN_n \rightarrow 1$ if and only if $A_nC_n \rightarrow 1$. This last condition reduces to $x = o(n^{1/6})$ for certain values of $p$, but is weaker for other values; in particular, there are values of $p$ for which $N_n/B_n$ can tend to one without even the requirement that $k/n$ tend to $p$.