# Probability/Statistics - Independent Random Variables

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## Homework Equations

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P(X∈A,Y∈B)=P(X∈A)×P(Y∈B)

## The Attempt at a Solution

If X and Y are independent then:
P(X∈A,Y∈B)=P(X∈A)×P(Y∈B)

I am confused how I can find P(X∈A) and P(Y∈B) .

I tried doing 0.1 * 0.2 = P(X∈A)
and 0.9 = (P(Y∈B))

and multiply those to get .018, but that is inccorect.

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LCKurtz
Homework Helper
Gold Member

[/B]

## Homework Equations

[/B]
P(X∈A,Y∈B)=P(X∈A)×P(Y∈B)

## The Attempt at a Solution

If X and Y are independent then:
P(X∈A,Y∈B)=P(X∈A)×P(Y∈B)

I am confused how I can find P(X∈A) and P(Y∈B) .

I tried doing 0.1 * 0.2 = P(X∈A)
##\{0,3\} = \{0\}\cup \{3\}##, not their intersection. How do you calculate the probability of a disjoint union?

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##\{0,3\} = \{0\}\cup \{3\}##, not their intersection. How do you calculate the probability of a disjoint union?
Wouldn't it just be the sum of the probabilities?

LCKurtz
Homework Helper
Gold Member
Wouldn't it just be the sum of the probabilities?
Yes. Is that what you calculated?

Yes. Is that what you calculated?
Um not yet. So Would P(X∈{0,3}) = P(X∈{0}) + P(X∈{3}) ?
so that = .1 + .2 = .3
AND P(Y∈{9}) = .9

So P(X∈{0,3},Y∈{9}) = .27?

.27 is still inccorect. I'm still not too sure how to solve this

LCKurtz
Homework Helper
Gold Member
.27 is still inccorect. I'm still not too sure how to solve this
##.27## looks correct to me unless I misunderstood something about the notation. But now your graphic has disappeared, so who knows. Why do you think it is incorrect?

##.27## looks correct to me unless I misunderstood something about the notation. But now your graphic has disappeared, so who knows. Why do you think it is incorrect?
I was correct. The dumb homework program just didn't take .27 as an answer... It wanted 0.27... -_- Thanks!!!

Ray Vickson