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As you can probably imagine, a couple hundred years ago, calculating binomial probabilities could get difficult, specifically, *n*! and 1 - *p* to the power of *n* - *X*.

Until a guy named Poisson, who was studying the problem of Napoleon's cavaliers getting killed by their own horses, found a simpler approximation for the binomial formula, which improved as *n* got larger:

\dfrac{e^{-\mu}\mu^X}{X!}

*prob*(*X|n*) =

, where \(\mu = n\times p \).

Take a binomial problem with a large number of trials. Let's say the probability of success in one trial is *p* = 0.03, and *n* = 200.

The mean (average), \(\mu\), would be *np* = 200 x 0.03 = 6.

What's the probability of actually getting *X* = 5 successes in 200 trials?

The probability is

\dfrac{e^{-\mu}\mu^X}{X!} = \dfrac{e^{-6}6^5}{5!} =

\dfrac{0.00247875217 \times 7776}{120} = 0.16062

Compare this to the correct binomial probability, 0.16225, and you see it's a pretty good approximation.

It gets better as *n* gets larger.

Though the Poisson is only an approximation to the always-correct binomial, there are many practical applications in business where we prefer Poisson.

That's because the Poisson can do something the binomial can't: **The Poisson can work with averages without knowing n and p.** It's far easier in the real world to get average occurrences of something over time or within a space than it is to know how many trials (

EXAMPLE: On average, 2 birds a day fly into my office window, thinking it's a way in. Stupid birds.

I'm expecting important visitors tomorrow, and I don't want stupid birds to frighten them. What's the probability that 0 birds will fly into my window tomorrow?

Notice that it's impossible to say what *n* and *p* are in this situation. You'd have to know how many birds are at play on any given day that might, but don't, make a run at my office window. So you could never honestly use the binomial formula in this case. You need Poisson.

Such is the case with many real-world scenarios. You can't say what *n* and *p* might be, but you can sure get an average.

What I'm counting (successes) are birds running into the window. The average is \(\mu\) = 2 per day. The number I'm interested in is *X* = 0. What's the probability I'll actually get 0 tomorrow?

\(\dfrac{e^{-\mu}\mu^X}{X!}\) = \(\dfrac{e^{-2}2^0}{0!}\) = 0.13533