# The monthly worldwide average number of airplane crashes of commercial

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 more than 5 such accidents in the next 3 months? Explain your reasoning!

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Cherry McCormick
Define N as the random variable that marks the number of plane crashes in a certain month. We are going to use Poisson approximation since we are given that the average number of crashes is 2.2. Now, we have that
N pois
a) $$\displaystyle{P}{\left({N}{>}{2}\right)}={1}-{P}{\left({N}={0}\right)}-{P}{\left({N}={1}\right)}-{P}{\left({N}={2}\right)}$$
$$\displaystyle={1}-{e}^{{-{2.2}}}-{2.2}{e}^{{-{2.2}}}-{\frac{{{2.2}^{{2}}}}{{{2}!}}}{e}^{{-{2.2}}}\approx{0.3773}$$
b) Since the average number of crashes in one month is 2.2, the average number of crashes in two months is 4.4. Hence, if we say that $$\displaystyle{N}_{{1}}$$ is the number of crashes in two months, we have that $$\displaystyle{N}_{{1}}$$ pois. Thus
$$\displaystyle{P}{\left({N}_{{1}}{>}{4}\right)}={1}-{P}{\left({N}_{{1}}={0}\right)}-{P}{\left({N}_{{1}}={1}\right)}-{P}{\left({N}_{{1}}={2}\right)}-{P}{\left({N}_{{1}}={3}\right)}-{P}{\left({N}_{{1}}={4}\right)}$$
$$\displaystyle={1}-{\sum_{{{k}={0}}}^{{4}}}{\frac{{{4.4}^{{k}}}}{{{k}!}}}{e}^{{-{4.4}}}\approx{0.44882}$$
c) If we say that $$\displaystyle{N}_{{2}}$$ marks the number of crashes in next three months, using the same argument as in a) have that $$\displaystyle{N}_{{2}}$$. Hence
$$\displaystyle{P}{\left({N}_{{2}}{>}{5}\right)}={1}-{\sum_{{{k}={0}}}^{{5}}}{\frac{{{6.6}^{{k}}}}{{{k}!}}}{e}^{{-{6.6}}}\approx{0.64533}$$