In a Poisson probability problem, the rate of errors is one every two hours. To find the probability of three defects in four hours,
A) μ = 1, x = 4.
B) μ = 2, x = 3.
C) μ = 3, x = 2.
D) μ = 3, x = 6.

Question

Quizplus · Accepted Answer

We use the Poisson probability formula:

P(x; μ) = e^(-μ) * μ^x / x!

where x = number of events and μ = expected number of events.

For four hours, the expected number of errors is:

μ = (rate of errors) * (time) = (1 error/2 hours) * (4 hours) = 2 errors

So, plugging in these values, we get:

P(3; 2) = e^(-2) * 2^3 / 3! = 0.1804

Therefore, the probability of three defects in four hours is 0.1804 or approximately 18.04%.

Out of the given answer choices, only option B has μ = 2 and x = 3.

In a Poisson Probability Problem, the Rate of Errors Is