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Suppose there are 80 vesicles available in a given synapse, and that the
probability of any given one of them fusing with the membrane and releasing
its load of ACh is p (like all probabilities, 0 < p < 1).
Therefore, if the probability that something will happen is p, then the
probability that it won't happen is (1-p). Suppose we choose 4 of 80
vesicles to react, and the remaining 80 - 4 = 76 not to react: what's the
probability that things will work out just this way? Each of the 80 vesicles has a probability associated with it (p for 4 of them, and (1-p) for the other 76). Since the way any one of them responds to an impulse is independent of the responses of any of the others, probability theory tells us the overall probability of having the chosen 4 disgorge their ACh, and the other 76 not, is: P(specific 4 yes, other 76 no) = p4(1-p)76, which is the product of the 80 associated probabilities. But probability theory also tells us that from a collection of N objects, there are N!/[(k!(N-k)!] ways of choosing exactly k out of N (k < N, and N! is the factorial of N, N! = N(N-1)(N-2)...1). So there are many ways of choosing exactly 4 vesicles from a collection of 80, and each such way has the probability associated to it given above. Therefore, the overall probability of choosing exactly 4 (but no specific 4) out of 80 to react is: P(any 4 yes, the other 76 no) = [80!/(4! 76!)]{p4(1-p)76}. But there are several problems with this formula, not the least of which is the difficulty one has performing such a calculation on a calculator, and even some computers. My particular calculator is even unable to handle the factorial of any N bigger than 70, much less N in the thousands. More significantly, the number of vesicles in a real-world situation is unknowable, as is the probability p. And this is where we opt to replace this so-called binomial probability with an approximation, the Poisson probability, which will prove much easier to apply. |