Answer: a. 0.1252
b. 0.04736
Step-by-step explanation:
Let X denotes a the number of soldiers killed by horse kicks in a year such that it has a Poisson distribution with mean [tex]\lambda=0.61[/tex].
Poisson distribution formula:
[tex]P(x; \lambda) = \dfrac{(e^{-\lambda}) (\lambda^ x)}{ x!}[/tex]
a. [tex]P(X>1)=1-P(X\leq1)[/tex]
[tex]=1- (P(X=0)+P(X=1))\\\\=1-( \dfrac{(e^{-0.61}) ((0.61)^0)}{ 0!}+\dfrac{(e^{-0.61}) ((0.61)^1)}{ 1!})\\\\= 1-( \dfrac{(0.54335) }{ 1}+\dfrac{(0.54335) (0.61)}{ 1})\\\\\approx0.1252[/tex]
Hence, the probability of more than 1 death in a corps in a year= 0.1252
b. For five years, Mean = [tex]\mu =\lambda \times 5= 0.61\times5 =3.05[/tex]
[tex]P(X=0)=e^{-3.05}\dfrac{3.05^0}{0!}[/tex]
[tex]\\\\=(0.04736)\dfrac{1}{1}\approx0.04736[/tex]
Hence, the probability of no deaths in a corps over five years = 0.04736
