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^-Î» * Î»^x)/(x!) for x = 0,1,2, ..., , where Î» is the parameter which indicates the average number of events in the given time interval.)

Given that 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.

Assuming the number of cars arriving at the two parking lots are independent we have total number of cars arriving X is Poisson with parameter 3+2 = 5

X is Poisson with mean = 5

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

= P(X=3) = 0.1404

b) the probability that less than 3 cars will arrive at the parking lot in a given hour

= P(X<3)

= P(0)+P(1)+P(2)

= 0.1247

warylucknow warylucknow · Answer 2 · 2020-02-12T07:20:47+01:00

Answer:

(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:

[tex]X\sim Poisson(3)\\Y\sim Poisson (2)[/tex]

The probability function of a Poisson distribution is:

[tex]P(U=u)=\frac{\lambda^{u}e^{-\lambda}}{u!}[/tex]

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:

[tex]P(U=3)=\frac{5^{3}e^{-5}}{3!}\\=\frac{125\times0.00674}{6} \\=0.1404[/tex]

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:

[tex]P(U<3)=P(U\leq 2)=P(X=0)+P(X=1)+P(X=2)\\=\frac{5^{0}e^{-5}}{0!}+\frac{5^{1}e^{-5}}{1!}+\frac{5^{2}e^{-5}}{2!}\\=0.00674+0.03369+0.08422\\=0.12465\\\approx0.1247[/tex]

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