Question

The average number of cars arriving at a particular red light each day is 4. Assuming a Poisson distribution, calculate the probability that on a given day, less than 3 cars will arrive at the red light.

Answer 1

By condition, $\lambda = 4$ .

Let X represent the number of such cars.

Then

$P(X = 0) = \frac{{{\lambda ^0}}}{{0!}}{e^{ - \lambda }} = {e^{ - 4}}$

$P(X = 1) = \frac{{{\lambda ^1}}}{{1!}}{e^{ - \lambda }} = 4{e^{ - 4}}$

$P(X = 2) = \frac{{{\lambda ^2}}}{{2!}}{e^{ - \lambda }} = 8{e^{ - 4}}$

Then the wanted probability is

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 13{e^{ - 4}} \approx 0.238$

Answer: $P(X < 3) \approx 0.238$