Probability/Statistics - Independent Random Variables

In summary, the conversation discusses the calculation of probabilities for independent events X and Y, using the equation P(X∈A,Y∈B)=P(X∈A)×P(Y∈B). The poster is confused about finding the probabilities of X∈A and Y∈B and suggests calculating them separately and then multiplying them, but this is incorrect. The correct method is to calculate the probability of the disjoint union of X and Y, which is equal to the sum of their individual probabilities. The final answer of P(X∈{0,3},Y∈{9})=0.27 is correct, but the homework program did not accept it without a leading zero and additional punctuation.
  • #1
Marcin H
306
6

Homework Statement


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

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.
 
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  • #2
Marcin H said:

Homework Statement


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

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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  • #3
LCKurtz said:
##\{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?
 
  • #4
Marcin H said:
Wouldn't it just be the sum of the probabilities?
Yes. Is that what you calculated?
 
  • #5
LCKurtz said:
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?
 
  • #6
.27 is still inccorect. I'm still not too sure how to solve this
 
  • #7
Marcin H said:
.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?
 
  • #8
LCKurtz said:
##.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!
 
  • #9
Marcin H said:
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?
 

1. What is the difference between independent and dependent random variables?

Independent random variables are variables that do not influence each other's outcomes, meaning the probability of one variable occurring does not affect the probability of the other variable occurring. Dependent random variables, on the other hand, are variables whose outcomes are influenced by each other, meaning the probability of one variable occurring will affect the probability of the other variable occurring.

2. How do you determine if two random variables are independent?

To determine if two random variables are independent, you can use the mathematical definition of independence, which states that the joint probability of the two variables equals the product of their individual probabilities. In simpler terms, if the probability of both variables occurring together is the same as the probability of each individual variable occurring separately, then the variables are independent.

3. What is the importance of independent random variables in probability and statistics?

Independent random variables are important in probability and statistics because they allow for simpler and more accurate calculations of probabilities and statistical measures. It also allows for the use of certain statistical models and methods that rely on the assumption of independence.

4. Can two random variables be independent if they have a correlation?

Yes, two random variables can be independent even if they have a correlation. Correlation measures the strength and direction of the relationship between two variables, but it does not determine if they are dependent or independent. It is possible for two variables to have a correlation but not influence each other's outcomes, making them independent.

5. How does the concept of independence apply to real-life situations?

The concept of independence can be applied to real-life situations in many ways. For example, in medical research, independent variables can be used to test the effectiveness of a treatment, as the outcome of the treatment should not be influenced by external factors. In finance, independent variables can be used to analyze the performance of different stocks or investments, as the performance of one investment should not be affected by the performance of another. Overall, the concept of independence allows for more accurate and reliable analysis in various fields.

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