Question

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?

Answer 1

Answer:

(a) more than 2 such accidents in the next month $\approx 0.3773$

(b) more than 4 such accidents in the next 2 months $\approx 0.44882$

(c) more than 5 such accidents in the next 3 more than 5 such accidents in the next 3 months $\approx 0.64533$

Step-by-step explanation:

Let  N be the  Random variable that marks the number of crashes in certain month.

Now let us use Poisson distribution since we are given with average number of crashes that is N \sim Pois(2.2)

(A)  more than 2 such accidents in the next month

Probability(more than 2 such accidents in the next month)=P(N>2)

P(N>2)=1-P(N=0)-P(N=1)-P(N=2)

=>$1-e^-{2.2}-2.2e^{-2.2}-\frac{2.2^2}{2!}e^{2.2}$

=> $\approx 0.3773$

B) more than 4 such accidents in the next 2 months

since the average number of crashes in 1 month is 2.2, the average number of crashes in two months is 4.4. hence, if we say that $N_1$ is the number of crashes  in 2 months, we have that $N\sim$Pois(4.4)

Thus,

Probability(more than 4 such accidents in the next 2 months)=P($N_1>4$)

=$1-P(N_1=0)-P(N_1=1)-P(N_1=2)=P(N_1=3)-P(N_1=4)$

$1-\sum_{k=1}^{4} \frac{4.4^{k}}{k !} e^{-4.4}$

=> $\approx 0.44882$

C)  more than 5 such accidents in the next 3 more than 5 such accidents in the next 3 months

If we say that $N_2$ marks the number of crashes in the next 3 months , using the same argument as in (a) we have that  a $N\sim$Pois(6.6)

Hence

P($N_2>5$)=$1-\sum_{k=0}^{5} \frac{6.6^{k}}{k !} e^{-6.6}$

=>$\approx 0.64533$