 # The monthly worldwide average number of airplane crashes of commercial Idilwsiw2 2021-11-12 Answered
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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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
$$\displaystyle{P}{\left({N}_{{2}}{>}{5}\right)}={1}-{\sum_{{{k}={0}}}^{{5}}}{\frac{{{6.6}^{{k}}}}{{{k}!}}}{e}^{{-{6.6}}}\approx{0.64533}$$