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I believe that the answer is affirmative, as there is a reputable-appearing subscription-hidden link on the web claiming such, and because, although I have not seen that proof, I have (I believe) a proof of my own (which I’m pretty sure is similar to the hidden one).

Here is the link to the subscription-hidden article.

And here’s my take on the question that I have posed:

Using elementary Probability Theory, and an appeal to symmetry and continuity, it is easy to prove the Binomial Theorem. First, we note that it is easy to establish that $\binom{n}{k}$ = C(n,k), by first (easily) establishing the formula for P(n,k), and then (easily) modifying it to give C(n,k). We can easily show that the binomial coefficients must be C(n,k) for the expansion of ${(x + y)}^n$, for positive integers x and y, by noting the following situation: If an urn contains x white balls and y black balls, then, given that (exactly) $n$ balls are selected, with replacement, the events: 0. picking n white balls 1. picking n – 1 white balls and 1 black ball 2. picking n – 2 white balls and 2 black balls … n – 1. picking 1 white ball and n – 1 black balls n. picking n black balls are mutually-exclusive and exhaustive events, and so the sum of their probabilities is unity. However, it is obvious that the number of ways in which event k can occur is C(n,k). Since all the terms have a denominator of ${(x + y)}^n$, we can multiply both sides of the equation by ${(x + y)}^n$ to get the Binomial Theorem. The case for arbitrary integers x and y follows by symmetry, and the case for arbitrary real numbers x and y follows by continuity. Done.

Notation: $\binom{n}{k}$ is $\frac{n!}{(k!)(n – k)!}$, C(n,k) is the number of combinations of n things taken k at a time, and P(n,k) is the number of permutations of n things taken k at a time.

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I hate to nitpick, but you are using the word "easy" and "obvious" far too many times. I also don't understand exactly what statement you're referring to as the binomial theorem, nor do I understand what distinction you appear to be making between this proof and the usual proof. – Qiaochu Yuan May 8 '11 at 1:42
up vote 10 down vote accepted

The (almost one-page) paper does have a "probabilistic proof" of the binomial theorem, in the sense that it is a proof of the binomial theorem using tools from the theory of probability, and it follows the basic outline of the argument you give above (which is also correct, by the way), in slightly more advanced terminology. So the answer to your question is yes. However, I don't find the paper particularly evocative. When I think of how the binomial theorem is related to probability, I think of this (I know it's not a proof, but still):

enter image description here

(see the wikipedia article on the Galton box)

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Nice picture! :D – El'endia Starman May 8 '11 at 1:20
Much thanks!!!! – Mike Jones May 8 '11 at 4:52

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