# Examples of using the Poisson distribution

### Example 3.5

Assuming the number of defects in a part isλ= 5, What is the expected number of defects in a part? What is the probability of two defects?

Up to two defects?

Expected number of defects = 5

Probability of two defects = P{x = 2, X = 5) = e-552/2! = 0.0842 Probability of up to two defects = P(0, 1,2) = e-5(l + 5 + 25/2) = 0.12

### Example 3.6

Assuming that the number of defects in a production line during a single hour is \ = 4. What is the probability that six defects will occur in that hour?

P(x = 6, λ = 4) = e-446/6! = 0.1042

### Example 3.7

Assuming the probability of obtaining a defective product is 0.01， what is the probability of obtaining at least three defective products out of a lot of 100, using binomial and Poisson distributions?

For binomial distribution:

The result of the Poisson distribution is in good agreement with the value of the binomial distribution for small p and large n, but much easier to compute.