# Probability/Statistics - Independent Random Variables

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1. Apr 15, 2017

### Marcin H

1. The problem statement, all variables and given/known data

2. Relevant equations

P(X∈A,Y∈B)=P(X∈A)×P(Y∈B)

3. 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.

Last edited by a moderator: May 8, 2017
2. Apr 15, 2017

### LCKurtz

$\{0,3\} = \{0\}\cup \{3\}$, not their intersection. How do you calculate the probability of a disjoint union?

Last edited by a moderator: May 8, 2017
3. Apr 15, 2017

### Marcin H

Wouldn't it just be the sum of the probabilities?

4. Apr 15, 2017

### LCKurtz

Yes. Is that what you calculated?

5. Apr 15, 2017

### Marcin H

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?

6. Apr 15, 2017

### Marcin H

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

7. Apr 15, 2017

### LCKurtz

$.27$ 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?

8. Apr 15, 2017

### Marcin H

I was correct. The dumb homework program just didn't take .27 as an answer... It wanted 0.27... -_- Thanks!!!

9. Apr 16, 2017

### Ray Vickson

So, it wanted '0.27' followed by three decimal points (or periods), then a minus sign, an underscore sign and another minus sign?