The monthly worldwide average number of airplane crashes of commercial airlines is $2.2$. What is the probability that there will be (a) more than $2$such accidents in the next month? (b) more than $4$such accidents in the next $2$months? (c) more than $5$such accidents in the next $3$months? Explain your reasoning!

In a month, the average number of airplane crashes of commercial airlines is $2.2$. Let, N be the random variable that marks the number of plane crashes in a certain month. Such that N has approx. Pois $\left(2.2\right)$distribution.

Since the average number of crashes is $2.2$. Then we need to use Poisson approximation

Thus,

$$\begin{gathered} P\left(N>2\right)=1-P\left(N=0\right)-P\left(N=1\right)-P\left(N=2\right) \\ =1-e^{-2.2}-2.2e^{-2.2}-\frac{2.2^2}{2!}{e}^{-2.2} \\ \approx 0.3773 \end{gathered}$$

In a month, the average number of airplane crashes of commercial airlines is $2.2$. Let, N be the random variable that marks the number of plane crashes in two months. Since the average number of plane crashes in one month is $2.2$. the average number of plane crashes in two months is $4.4$.

N₁ has approx. Pois $\left(4.4\right)$distribution.

Since the average number of crashes in two months is $4.4$. then we need to use the Poisson approximation.

Thus,

$$\begin{gathered} P\left(N_1>4\right)=1-P\left(N_1=0\right)-P\left(N_1=2\right)-P\left(N_1=3\right)-P\left(N_1=4\right) \\ =1-\sum _{k=0}^4\frac{4.4^k}{k!}{e}^{-4.4} \\ \approx 0.44882 \end{gathered}$$

In a month, the average number of airplane crashes of commercial airlines is $2.2$. Let, N be the random variable that marks the number of plane crashes in three months. Since the average number of plane crashes in one month is $2.2$. then the average number of plane crashes in three months is $6.6$.

N₁ has approx. Pois $\left(6.6\right)$distribution. Since the average number of crashes in three months is $6.6$.

Then we need to use the Poisson approximation.

Thus,

$$\begin{gathered} P\left(N_2>5\right)=1-P\left(N_2=0\right)-P\left(N_2=1\right)-P\left(N_2=2\right)-P\left(N_2=3\right)-P\left(N_2=4\right)-P\left(N_2=5\right) \\ =1-\sum _{k=0}^5\frac{6.6^k}{k!}{e}^{-6.6} \\ \approx 0.64533 \end{gathered}$$