Question

Consider a Poisson distribution with a mean of two occurrences per time period. What is the probability of observing exactly three occurrences in a given time period?

1) 0.180
2) 0.271
3) 0.406
4) 0.541

Answer 1

The probability of observing exactly three occurrences in a Poisson distribution with a mean of two is approximately 0.180, which is calculated using the provided Poisson distribution formula.

Step-by-step explanation:

The question is asking for the probability of observing exactly three occurrences in a given time period from a Poisson distribution with a mean (μ) of two occurrences. To find this probability, we use the formula:

P(X = k) = (μ^k * e^{-μ}) / k!, where μ is the mean number of occurrences, e is the base of the natural logarithm, and k is the number of occurrences we want to find the probability for.

Plugging in our values:

P(X = 3) = (2^3 * e^{-2}) / 3! = (8 * e^{-2}) / (3*2*1) = (8 / e^2) / 6 ≈ 0.180 (using a calculator).

Therefore, the probability of observing exactly three occurrences is approximately 0.180, which is option 1).