Question

The monthly worldwide average number of airplane crashes of commercial airlines is $3.5.$What is the probability that there will be

(a) at least $2$such accidents in the next month;

(b) at most$1$accidents in the next month?

Explain your reasoning!

Short answer

The probability that there will be at least $2$such accidents in the next month is$0.864.$

The probability that will be at most $1$accidents in the next month is$0.1359.$

Step-by-step solution

Given Information (Part-a)

Given in the question that the probability there will be at least $2$such accidents in the next month . The monthly worldwide average number of airplane crashes of commercial airlines can be approximated by Poisson distribution as it is a rare event.

Given $\lambda =3.5$

Solution of the Problem (Part-b)

At least $2$accidents $\Rightarrow 2$or more accidents

$\Rightarrow 1-P[X\le 1]$

$=1-\left[\frac{e^{-35}(3.5{)}^0}{0!}+\frac{(3.5{)}^1{e}^{-35}}{1!}\right]$

$=1-[0.0302+0.1057]$

We get,

$=0.864$

Final Answer (Part-a)

The probability that there will be at least $2$such accidents in the next month is$0.864$.

Given Information (Part-b)

Given in the question that the probability that will be at most $1$accidents in the next month. The monthly worldwide average number of airplane crashes of commercial airlines can be approximated by Poisson distribution as it is a rare event.

Given$\lambda =3.5$

Solution of the Problem (Part-b)

At most 1 accident $=0$or $1$accident

$=P(X=0)+P(X=1)$

$={\mathrm{e}}^{-3.5}+3.5e^{-3.5}$

We get,

$=0.1359$

Final Answer (Part-b)

The probability that will be at most$1$accidents in the next month is$0.1359.$