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Probability/Statistics - Independent Random Variables

  1. Apr 15, 2017 #1
    1. The problem statement, all variables and given/known data

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


    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. jcsd
  3. Apr 15, 2017 #2

    LCKurtz

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    ##\{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
  4. Apr 15, 2017 #3
    Wouldn't it just be the sum of the probabilities?
     
  5. Apr 15, 2017 #4

    LCKurtz

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    Yes. Is that what you calculated?
     
  6. Apr 15, 2017 #5
    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?
     
  7. Apr 15, 2017 #6
    .27 is still inccorect. I'm still not too sure how to solve this
     
  8. Apr 15, 2017 #7

    LCKurtz

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    ##.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?
     
  9. Apr 15, 2017 #8
    I was correct. The dumb homework program just didn't take .27 as an answer... It wanted 0.27... -_- Thanks!!!
     
  10. Apr 16, 2017 #9

    Ray Vickson

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