A car dealership sells cars at an average rate of 3 cars per day. What is the probability that exactly 5 cars will be sold on a given day, assuming a Poisson distribution?
The Poisson distribution is used to model the number of events occurring within a fixed interval of time or space, given a known average rate of occurrence (λ). In this case, λ represents the average number of cars sold per day, which is 3.

The probability mass function (PMF) of the Poisson distribution is given by:

P(X=k) = (e^(-λ) * λ^k) / k!

Where X is the random variable representing the number of events (cars sold), k is the specific number of events we want to find the probability for (5 cars), e is the base of the natural logarithm (approximately 2.71828), and λ is the average rate of occurrence (3 cars per day).

Plugging in the values, we get:

P(X=5) = (e^(-3) * 3^5) / 5!
           ≈ (0.0498 * 243) / 120
           ≈ 12.0426 / 120
           ≈ 0.1004

So, the probability that exactly 5 cars will be sold on a given day is approximately 0.1004 or 10.04%.