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

The probability of getting 5 defectives is:

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