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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

- Thread starter Punch
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- Thread starter
- #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

- Jan 26, 2012

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

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

CB