- #1

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according to the book. P(Y=1)=P[(X1=1)and(X2>=1)]+P[(X2=1)and(X1>=2)]

I don't understand why is the asymmetry is even possible. The way I look at it, it's

P(Y=1)=P[(X1=1)and(X2>1)]+P[(X2=1)and(X1>1)]

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- Thread starter torquerotates
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- #1

- 207

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according to the book. P(Y=1)=P[(X1=1)and(X2>=1)]+P[(X2=1)and(X1>=2)]

I don't understand why is the asymmetry is even possible. The way I look at it, it's

P(Y=1)=P[(X1=1)and(X2>1)]+P[(X2=1)and(X1>1)]

- #2

Stephen Tashi

Science Advisor

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according to the book. P(Y=1)=P[(X1=1)and(X2>=1)]+P[(X2=1)and(X1>=2)]

The way I look at it, it's

P(Y=1)=P[(X1=1)and(X2>1)]+P[(X2=1)and(X1>1)]

In the case of discrete random variables, your way omits the event "X1 = 1 and X2 = 1". In the case of continuous random variables (where that event has zero probability), your method is correct.

The book's way is correct for the case of discrete random variables that only take integer values. It wouldn't be correct for continuous random variables.

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