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,

$P(x; m)$ = $e^{-m}$ $\frac{m^{x}}{x!}$

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.

Mean, $m = 2$

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

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.

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 $0.42319008113$.

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 $0.006737947$.

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 $0.268850169$.

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 $0.54134113295$.