A manufacturing concern employing a large number of workers finds that, over a period of time, the average absentee rate is 2 workers per shift. The probability that exactly 2 workers will be absent in a chosen shift at random is

Hint: According to the question we have to determine the probability that exactly 2 workers will be absent in a chosen shift at random when a manufacturing concern employing a large number of workers finds that, over a period of time, the average absentee rate is 2 workers per shift. So, first of all we have to use the Poisson distribution rule as explained below:
Poisson distribution rule: The Poisson distribution is the discrete probability distribution of the number of given events occurring in a given time period, given the average number of times the event occurs over that time period.

Formula used: $ \Rightarrow P(x;\mu ) = \dfrac{{{e^{ - 1}}{\mu ^x}}}{{x!}}...............(A)$
Where, x is the actual number of successes that occur in a specified region, $\mu $is the mean number of successes that occur in a specified region, and $P(x;\mu )$is the Poisson probability that exactly x success occurs in a Poisson experiment.

Complete step-by-step solution:
Step 1: First of all we have to determine the mean number of successes is $\mu $and as in the question $\mu $= 2 and x = 2 (number of 2 workers)
Step 2: Now, we have to substitute all the values in the formula (A) as mentioned in the solution hint. Hence,
$ \Rightarrow P(2;2) = \dfrac{{{e^{ - 2}}{2^2}}}{{2!}}$
Step 3: Now, to obtain the values of expression we have to solve the expression (1) as obtained in the solution step 2. Hence,
$
   \Rightarrow P(2;2) = \dfrac{{{e^{ - 2}}2 \times 2}}{{2 \times 1}} \\
   \Rightarrow P(2;2) = 2{e^{ - 2}} \\
 $

Final solution: Hence, with the help of the formula (A) as mentioned in the solution hint we have obtained the probability that exactly 2 workers will be absent in a chosen shift at random is $P(2;2) = 2{e^{ - 2}}$

Note: The Poisson distribution is a discrete probability for the counts of events that occur randomly in a given interval of time and we have to let X as the number of events in a given interval and then if the mean number of events per interval is$\mu $.
In Poisson distribution e is a mathematical constant where, $e = 2.718282$ and $\mu $ is the parameter of the distribution.