Poisson Distribution: Prob of <=3 Wrong Connections in Building

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SUMMARY

The discussion focuses on calculating the probability of having at most 3 wrong connections in a building with two independent automatic telephone exchanges, A and B. The number of wrong connections for exchange A follows a Poisson distribution with a parameter of 0.5, while exchange B has a parameter of 1. The correct approach to find the probability P(X+Y≤3 | X≥2) involves using the formula P(X=3)P(Y=0) + P(X=2)P(Y≤1) divided by P(X≥2). The initial attempts to calculate this probability were incorrect due to misapplication of the Poisson distribution properties.

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Punch
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A building has 2 independent automatc telephone exchanges A and B. The number X of wrong connections for A in anyone day is a poisson variable with parameter 0.5 and the number Y of wrong connections for B in any one day is a poisson variable with parameter 1.

Calculate in any particular day, the probability that there will be at most 3 wrong connections in the building given X≥2

I tried using P(X=2)P(Y=0)+P(X=2)P(Y=1)+P(X=3)P(Y=0) but the answer was wrong
 
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Punch said:
A building has 2 independent automatc telephone exchanges A and B. The number X of wrong connections for A in anyone day is a poisson variable with parameter 0.5 and the number Y of wrong connections for B in any one day is a poisson variable with parameter 1.

Calculate in any particular day, the probability that there will be at most 3 wrong connections in the building given X≥2

I tried using P(X=2)P(Y=0)+P(X=2)P(Y=1)+P(X=3)P(Y=0) but the answer was wrong

\[P(X+Y\le 3|X\ge 2)=\frac{P(X=3)P(Y=0)+P(X=2)P(Y\le 1)}{P(X\ge 2)}\]CB
 

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