A parking lot has two entrances. Cars arrive at entrance I according to a Poisson distribution at an average of 3 per hour and at entrance II according to a Poisson distribution at an average of 2 per hour. What is the probability that a total of 3 cars will arrive at the parking lot in a given hour? What is the probability that less than 3 cars will arrive at the parking lot in a given hour? (Assume that the numbers of cars arriving at the two entrances are independent. (Note: The probability mass function of Poisson distribution is p(x) = (e^-λ * λ^x)/(x!) for x = 0,1,2, ..., , where λ is the parameter which indicates the average number of events in the given time interval.)

(a) the probability that a total of 3 cars will arrive at the parking lot in a given hour is 0.1404.

(b) The probability that less than 3 cars will arrive at the parking lot in a given hour is 0.1247.

Step-by-step explanation:

Let X = cars arriving through entrance I and Y = cars arriving through entrance II.

Given:

The probability function of a Poisson distribution is:

It is also provided that the events X and Y are independent.

Let U = X + Y

Then E (U) = E (X) + E(Y) = 3 + 2 = 5.

The random variable U also follows a Poisson distribution with parameter λ = 5

(a)

The probability that a total of 3 cars will arrive at the parking lot in a given hour is:

Thus, the probability that a total of 3 cars will arrive at the parking lot in a given hour is 0.1404.

(b)

The probability that less than 3 cars will arrive at the parking lot in a given hour is:

Thus, the probability that less than 3 cars will arrive at the parking lot in a given hour is 0.1247.