- #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)]

- Thread starter torquerotates
- Start date

- #1

- 207

- 0

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)]

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 way I look at it, it's

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

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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