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

AI Thread Summary
The discussion revolves around calculating the probability of at most 3 wrong connections in a building with two independent telephone exchanges, A and B, modeled as Poisson variables. Exchange A has a parameter of 0.5, while exchange B has a parameter of 1. The user initially attempted to calculate this probability using a specific formula but found the answer to be incorrect. The correct approach involves using the conditional probability formula, specifically P(X+Y≤3|X≥2), which incorporates the probabilities of different combinations of wrong connections. The conversation emphasizes the importance of correctly applying the Poisson distribution in this context.
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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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