Airline passenger arrive randomly and independently at the

Given information: Mean arrival rate of passengers at a major international airport is 10 passengers per minute.

Let's assume that the passenger arrival follows a Poisson distribution with parameter $\lambda =10$, which gives the probability of the number of passengers arriving in a minute period.

1. The probability of exactly 7 passengers arriving in a minute can be calculated using the Poisson distribution as follows:

$P(X=7)=\frac{{\lambda }^x\cdot e^{-\lambda }}{x!}$
         $=\frac{{10}^7\cdot e^{-10}}{7!}$
         $=0.0908$ (rounded to four decimal places)

Therefore, the probability of exactly 7 passengers arriving in a minute is 0.0908.

2. The probability of at most 5 passengers arriving in a minute can be calculated by summing up the probabilities of having 0, 1, 2, 3, 4, or 5 passengers arriving in a minute. This can be written mathematically as:

$P(X\le 5)=\Sigma \frac{{\lambda }^x\cdot e^{-\lambda }}{x!}$, where x = 0 to 5

         $=\left[\frac{{10}^0\cdot e^{-10}}{0!}\right]+\left[\frac{{10}^1\cdot e^{-10}}{1!}\right]$
           $+\left[\frac{{10}^2\cdot e^{-10}}{2!}\right]+\left[\frac{{10}^3\cdot e^{-10}}{3!}\right]$
           $+\left[\frac{{10}^4\cdot e^{-10}}{4!}\right]+\left[\frac{{10}^5\cdot e^{-10}}{5!}\right]$

         $=0.0671$(rounded to four decimal places)

Therefore, the probability of at most 5 passengers arriving in a minute is 0.0671.

3. The probability of having between 8 and 12 passengers arriving in a minute, both inclusive, can be calculated by summing up the probabilities of having 8, 9, 10, 11, or 12 passengers arriving in a minute. This can be written mathematically as:

$P(8\le X\le 12)=\sum \frac{{\lambda }^x\cdot e^{-\lambda }}{x!}$, where x = 8 to 12

                $=\left[\frac{{10}^8\cdot e^{-10}}{8!}\right]+\left[\frac{{10}^9\cdot e^{-10}}{9!}\right]$
                  $+\left[\frac{{10}^{10}\cdot e^{-10}}{10!}\right]+\left[\frac{{10}^{11}\cdot e^{-10}}{11!}\right]$
                  $+\left[\frac{{10}^{12}\cdot e^{-10}}{12!}\right]$

                $=0.3213$ (rounded to four decimal places)

Therefore, the probability of having between 8 and 12 passengers arriving in a minute, both inclusive, is 0.3213.