Question #189840

On average, 3 students per month cry in happiness. What is the probability that in any given month,


a) exactly 5 students will cry in happiness

b) fewer than 3 students will cry in happiness

c) at least 2 students will cry in happiness


Answers (Decimal Format; 4 decimal places)


1
Expert's answer
2021-05-07T14:26:00-0400

Given, Average students cry in a month, λ=3\lambda=3


Let X denote the number of students cry in happiness


(i)P(X=5)=eλ.λ55!=e3355!=0.1008(i) P(X=5)=\dfrac{e^{-\lambda}.\lambda^5}{5!}=\dfrac{e^{-3}3^5}{5!}=0.1008


(ii)P(X<3)=P(X=0)+P(X=1)+P(X=2)(ii) P(X<3)=P(X=0)+P(X=1)+P(X=2)


=eλ.λ11!+eλ.λ2!+eλ.λ22!=e3300!+e3311!+e3.322!=0.048978+0.14936+0.22404=0.4224=\dfrac{e^{-\lambda}.\lambda^1}{1!}+\dfrac{e^{-\lambda}.\lambda^2}{!}+\dfrac{e^{-\lambda}.\lambda^2}{2!}\\[9pt]=\dfrac{e^{-3}3^0}{0!}+\dfrac{e^{-3}3^1}{1!}+\dfrac{e^{-3}.3^2}{2!}\\[9pt]=0.048978+0.14936+0.22404=0.4224


(iii)P(X2)=1P(X<2)=1[P(X=0)+P(X=1)](iii) P(X\ge 2)=1-P(X<2)=1-[P(X=0)+P(X=1)]


=1[eλλ00!+eλλ11!]=1[0.048978+0.14936]=10.198338=0.8016=1-[\dfrac{e^{-\lambda}\lambda^0}{0!}+\dfrac{e^{-\lambda}\lambda^1}{1!}] \\[9pt] =1-[0.048978+0.14936]\\[9pt]=1-0.198338=0.8016


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