The average number of defectives in 300 motors is μ = 0.01 × 300 = 3

The probability of getting `5` defectives is:

`P(X)=(e^-3 3^5)/(5!)=0.10082`

NOTE: This problem looks similar to a binomial distribution problem, that we met in the last section.

If we do it using binomial, with `n = 300`, `x = 5`, `p = 0.01` and `q = 0.99`, we get:

P(X = 5) = C(300,5)(0.01)5(0.99)295 = 0.10099

We see that the result is very similar. We can use binomial distribution to approximate Poisson distribution (and vice-versa) under certain circumstances.

Histogram of Probabilities

Poisson probability chart