Question #215873

Q1

A mine hauls, on average, three dump truck of waste per hour to a waste dump. For a given hour, find the probability that it will haul the following number of truck loads:

a)     At most 3 trucks.

b)     At least 3 trucks.

c)      Five or more 


Expert's answer

Let X=X= the number of trucks per hour: X∼Po(λ).X\sim Po(\lambda).

Given λ=3.\lambda=3.


a)

P(X≤3)=P(X=0)+P(X=1)+P(X=2)P(X\leq 3)=P(X=0)+P(X=1)+P(X=2)

+P(X=3)=e−λ⋅λ00!+e−λ⋅λ11!+P(X=3)=\dfrac{e^{-\lambda}\cdot\lambda^0}{0!}+\dfrac{e^{-\lambda}\cdot\lambda^1}{1!}

+e−λ⋅λ22!+e−λ⋅λ33!+\dfrac{e^{-\lambda}\cdot\lambda^2}{2!}+\dfrac{e^{-\lambda}\cdot\lambda^3}{3!}

=e−3(6+18+27+27)6=13e−3≈0.647232=\dfrac{e^{-3}(6+18+27+27)}{6}=13e^{-3}\approx0.647232


b)

P(X≥3)=1−P(X=0)−P(X=1)−P(X=2)P(X\geq 3)=1-P(X=0)-P(X=1)-P(X=2)

=1−e−λ⋅λ00!−e−λ⋅λ11!−e−λ⋅λ22!=1-\dfrac{e^{-\lambda}\cdot\lambda^0}{0!}-\dfrac{e^{-\lambda}\cdot\lambda^1}{1!}-\dfrac{e^{-\lambda}\cdot\lambda^2}{2!}

=1−e−3(2+6+9)2=1−8.5e−3≈0.576810=1-\dfrac{e^{-3}(2+6+9)}{2}=1-8.5e^{-3}\approx0.576810




c)

P(X≥5)=1−P(X=0)−P(X=1)−P(X=2)P(X\geq 5)=1-P(X=0)-P(X=1)-P(X=2)

−P(X=3)−P(X=4)=1−e−λ⋅λ00!-P(X=3)-P(X=4)=1-\dfrac{e^{-\lambda}\cdot\lambda^0}{0!}

−e−λ⋅λ11!−e−λ⋅λ22!−e−λ⋅λ33!−e−λ⋅λ44!-\dfrac{e^{-\lambda}\cdot\lambda^1}{1!}-\dfrac{e^{-\lambda}\cdot\lambda^2}{2!}-\dfrac{e^{-\lambda}\cdot\lambda^3}{3!}-\dfrac{e^{-\lambda}\cdot\lambda^4}{4!}

=1−e−3(24+72+108+108+81)24=1-\dfrac{e^{-3}(24+72+108+108+81)}{24}


=1−1318e−3≈0.184737=1-\dfrac{131}{8}e^{-3}\approx0.184737


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