Patients with infections are randomly selected and the number of distinct species belonging to the Spirochaete phylum (responsible for illnesses such as Lyme disease and syphilis) are recorded. Let X count the number of distinct species found. It is speculated that X follows a Poisson distribution with a mean of 0.25.
What is the probability of observing exactly two distinct species?

Question

Quizplus · Accepted Answer

If X follows a Poisson distribution with mean of 0.25, then the probability of observing exactly x distinct species is given by the Poisson probability mass function P(X=x) = (e^(-0.25) * 0.25^x) / x!. Therefore, the probability of observing exactly two distinct species is:

P(X=2) = (e^(-0.25) * 0.25^2) / 2! = 0.0243

Hence, the correct answer is B.

Patients with Infections Are Randomly Selected and the Number of Distinct