Question

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 10 passengers per minute. a. Compute the probability of no arrivals in a one-minute period. Round your answer to six decimal places. b. Compute the probability that three or fewer passengers arrive in a one-minute period. Round your answer to four decimal places. c. Compute the probability of no arrivals in a 15-second period. Round your answer to four decimal places. d. Compute the probability of at least one arrival in a 15-second period. Round your answer to four decimal places.

Answer 1

Concepts and reason

The concept of Poisson distribution is used to solve the problem.

The limiting case of binomial is Poisson distribution.

The Poisson distribution is used to model the number of times events occur in a given interval of time. Poisson distribution has the following conditions:

The first is $n \to \infty$ , that is, the number of trials is indefinite.

The second is $p \to 0$ , which represents that the probability of success is small for each trial.

The third is $np = \lambda$ , which is finite, where $\lambda$ is a positive real number.

The Poisson distribution has the property that its mean and variance are equal to each other, and the value is $\lambda$ .

Fundamentals

A discrete random variable X is said to follow Poisson distribution if its probability mass function is:

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}},x = 0,1,2 \cdots n$

Where $\lambda$ is the parameter of the Poisson distribution.

(a)

Airline passengers are arriving at an airport independently. The mean arrival rate is 10 passengers per minute. Consider the random variable X to represent the number of passengers arriving per minute. The random variable X follows a Poisson distribution. That is,

$X \sim {\rm{Poisson}}\left( {\lambda = 10} \right)$

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

Substitute the value of $\lambda = 10$ in the formula as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{{\left( {10} \right)}^x}}}{{x!}}$

The probability that there are no arrivals in one minute is calculated by substituting $x = 0$ in the formula as,

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}}\\\\ = {e^{ - 10}}\\\\ = 0.000045\\\end{array}$

(b)

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

The probability of the arrival of three or fewer passengers in one minute is calculated by substituting $\lambda = 10$ and $x = 0,1,2,3$ in the formula as,

$\begin{array}{c}\\P\left( {X \le 3} \right) = \sum\limits_{x = 0}^3 {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}} \\\\ = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^1}}}{{1!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^2}}}{{2!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^3}}}{{3!}}\\\\ = 0.000045 + 0.00045 + 0.00227 + 0.00756\\\\ = 0.0103\\\end{array}$

(c)

Consider the random variable Y to denote the passengers arriving in 15 seconds. This means that the random variable Y can be defined as $\frac{X}{4}$ . So,

$\begin{array}{c}\\E\left( Y \right) = E\left( {\frac{X}{4}} \right)\\\\ = \frac{1}{4} \times 10\\\\ = 2.5\\\end{array}$

That is, $Y \sim {\rm{Poisson}}\left( {\lambda = 2.5} \right)$ . So, the probability mass function of Y is,

$P\left( {Y = y} \right) = \frac{{{e^{ - \lambda }}{\lambda ^y}}}{{y!}};x = 0,1,2, \ldots$

The probability that there are no arrivals in the 15-second period can be calculated by substituting the value of $\left( {\lambda = 2.5} \right)$ and y as 0 as:

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = {e^{ - 2.5}}\\\\ = 0.0821\\\end{array}$

(d)

The probability that there is at least one arrival in a 15-second period is calculated as,

$\begin{array}{c}\\P\left( {X \ge 1} \right) = 1 - P\left( {X < 1} \right)\\\\ = 1 - P\left( {X = 0} \right)\\\\ = 1 - \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = 1 - {e^{ - 2.5}}\\\end{array}$

$\begin{array}{c}\\ = 1 - 0.082\\\\ = 0.9179\\\end{array}$

Ans: Part a

The probability of no arrivals in one minute is 0.000045.

Part b

The probability of the arrival of three or fewer passengers in one minute is 0.0103.

Part c

The probability that there are no arrivals in the 15-second period is 0.0821.

Part d

The probability of at least one arrival in a 15-second period is 0.9179.