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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:
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
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 (n) and what the probability of success in one trial (p) might be. Hooray!
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