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 someone with exactly zero distinct species?

Question

Quizplus · Accepted Answer

The probability of observing exactly k events in a Poisson distribution is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$, where $\lambda$ is the mean number of events. For $k=0$ and $\lambda=0.25$, the probability is $P(X=0) = \frac{e^{-0.25} \cdot 0.25^0}{0!} = e^{-0.25} = 0.7788$.

Patients with Infections Are Randomly Selected and the Number of Distinct