Answer: (a) 0.8641
(b) 0.1359
Step-by-step explanation:
Given : The monthly worldwide average number of airplane crashes of commercial airlines [tex]\lambda= 3.5[/tex]
We use the Poisson distribution for the given situation.
The Poisson distribution formula for probability is given by :-
[tex]P(X=x)=\dfrac{e^{-\lambda}\lambda^x}{x!}[/tex]
a) The probability that there will be at least 2 such accidents in the next month is given by :-
[tex]P(X\geq2)=1-(P(X=1)+P(X=0))\\\\=1-(\dfrac{e^{-3.5}(3.5)^0}{0!}+\dfrac{e^{-3.5}(3.5)^1}{1!})\\\\=1-(0.1358882254)=0.8641117746\approx0.8641[/tex]
b) The probability that there will be at most 1 accident in the next month is given by :-
[tex]P(X\leq1)=(P(X=1)+P(X=0))\\\\=\dfrac{e^{-3.5}(3.5)^0}{0!}+\dfrac{e^{-3.5}(3.5)^1}{1!}\\\\=0.1358882254\approx0.1359[/tex]
