There are two entrances to a movie theater. Customers arrive atEntrance I accord

Question

There are two entrances to a movie theater. Customers arrive atEntrance I according to
a Poisson distribution averaging 3 per hour. Customers arrive atEntrance II according
to a Poisson distribution averaging 4 per hour. What is theprobability that 3 customers
visit the movie theater in a given hour? (Assume arrivals areindependent)

Explanation / Answer

There are twoentrances to a movie theater. Customers arrive at Entrance Iaccording to a Poisson distribution averaging 3 per hour. Customers arrive atEntrance II according to a Poisson distribution averaging 4 per hour. What is theprobability that 3 customers visit the movie theater in a given hour? (Assume arrivals areindependent) We first determine the distributionof the total of the Random Variables for customers arriving at Entrance #1 "(X_1)" and Entrance #2"(X_2)": {Random Variable "(X_1)"} ~ Poisson{ (_1)= 3 }   (    {Random Variable Y = (X_1) + (X_2)} ~ Poisson{ = (3) + (4) } ---->    {Random Variable Y = (X_1) + (X_2)} ~ Poisson{ = 7 } ---->    Prob{Exactly 3Customers Visit In Given Hour} =                 = Prob{Y = (X_1) +(X_2) = 3}                 = Poisson{Y = 3, =7}                 = (^k)*Exp[-]/(k!)                 = ((7)^(3))*Exp[-(7)]/(3!)                 = 0.05213 .

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