Probability/Statistics - Independent Random Variables

[Marcin H]

Homework Statement

[/B]
https://www.physicsforums.com/attachments/screen-shot-2017-04-15-at-12-28-52-pm-png.194886/?temp_hash=4939cc24bd25e6adfbe75458bec6d011

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)
and 0.9 = (P(Y∈B))

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

 

[LCKurtz]

Homework Statement

[/B]
https://www.physicsforums.com/attachments/screen-shot-2017-04-15-at-12-28-52-pm-png.194886/?temp_hash=4939cc24bd25e6adfbe75458bec6d011

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?

 

[Marcin H]

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

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

 

[LCKurtz]

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

Yes. Is that what you calculated?

 

[Marcin H]

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?

 

[Marcin H]

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

 

[LCKurtz]

.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?

 

[Marcin H]

$.27$ 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]

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

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