Poisson Distribution Examples - Word Problems
The Poisson distribution is used when the mean of occurrence of a certain event is given for a certain time period and the probability is needed to be calculated for a certain value in the same time period. Let the mean be m, then the probability for the event to occur $x$ times in the given time period will be,
$P(x; m)$ = $e^{-m}$ $\frac{m^{x}}{x!}$
Example 1: 

The mean value for an event X to occur is 2 in a day. Find the probability of event X to occur thrice in a day.

Solution: 

Mean, $m = 2$

Probability of the event to occur thrice, $P(3; 2)$ = $e^{-2}$ $\frac{2^{3}}{3!}$ = $0.1804465$ 

Poisson Distribution Word Problems

Problem 1: 

A shop sells five pieces of shirt everyday, then what is the probability of selling three shirts today?

Solution:


Mean value for 1 day, $m = 5$

Probability of selling $3$ shirts, $P(3; 5)$ = $e^{-5}$ $\frac{5^{3}}{3!}$ = $\frac{0.006737947 \times 125}{5!}$ = $0.00701869$

Hence, the probability of selling three shirts is $0.00701869$ when at the average $5$ shirts are being sold each day.
Problem 2: 

If three persons, on an average, come to ABC company for job interview, then find the probability that less than three people have come for interview on a given day.

Solution:


The mean for Poisson random variable, $m = 3$

$P(x < 3; 3) = P(0; 3) + P(1; 3) + P(2; 3)$

$P(0; 3)$ = $e^{-3}$ $\frac{3^{0}}{0!}$ = $0.04978706837$

$P(1; 3)$ = $e^{-3}$ $\frac{3^{1}}{1!}$ = $0.1493612051$

$P(2; 3)$ = $e^{-3}$ $\frac{3^{2}}{2!}$ = $0.22404180766$

Hence,

$P(x < 3; 3)$ = $P(0; 3) + P(1; 3) + P(2; 3)$
                   
                    = $0.04978706837 + 0.1493612051 + 0.22404180766$
                    
                    = $0.42319008113$

The probability of less than three persons coming for interview on a certain day is $0.42319008113$.
Problem 3: 

Number of calls coming to the customer care center of a mobile company per minute is a Poisson random variable with mean 5. Find the probability that no call comes in a certain minute.

Solution:


The mean value, $m = 5$

We need to find the probability of getting zero calls when 5 calls are known to come every minute.

$P(0; 5)$ = $e^{-5}$ $\frac{5^{0}}{0!}$ = $0.006737947$

Hence, the probability of getting zero calls in a minute is $0.006737947$.
Problem 4:

There are five students in a class and the number of students who will participate in annual day every year is a Poisson random variable with mean 3. What will be the probability of more than 3 students participating in annual day this year?

Solution:


Mean for Poisson random variable, m = 3

$P(x > 3; 3) = P(4; 3) + P(5; 3)$

$P(4; 3)$ = $e^{-3}$ $\frac{3^{4}}{4!}$ = $0.16803135574$

$P(5; 3)$ = $e^{-3}$ $\frac{3^{5}}{5!}$ = $0.10081881344$

Hence, $P(x > 3; 3)$ = $P(4; 3) + P(5; 3) = 0.268850169$

The probability of getting more than three students participating is $0.268850169$.
Problem 5:

The deals cracked by an agent per day is a random Poisson variable with mean 2. Given that each day is independent of other day, find the probability of getting 2 deals cracked on first day and 1 deal to be cracked the next day.

Solution:


The probability of getting 2 deals in a day is $P(2; 2)$ and the probability of getting 1 deal is $P(1; 2)$.

The probability of getting 2 deals on first day and one deal on second day = $P(2; 2)$ $\times$ $P(1; 2)$

$P$ = $e^{-2}$ $\frac{2^{2}}{2!}$ $\times$ $e^{-2}$ $\frac{2^{1}}{1!}$

    = $0.27067056647 + 0.27067056647 = 0.54134113295$

The probability the first day two deals are cracked and the second day one deal is cracked is $0.54134113295$.