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Is my Handout wrong? Stumped Conditional Probability

  1. Sep 20, 2009 #1
    Conditional Probability...

    My handout says this...

    P(m) = .4
    P(w) = .5
    P(m|w) = .7

    Find:
    P(MnW) = ?
    P(w|m) = ?
    P(m or w) = ?

    P(MnW) = .4 x .5 = .2

    P(w|m) = .35/.4 = .875

    P(m or w) = ?
     
    Last edited: Sep 20, 2009
  2. jcsd
  3. Sep 20, 2009 #2
    Re: Is my Handout wrong? Stumped... Conditional Probability... Need help urgent!

    P(m|w) is given. You can get P(w|m) from Bayes Theorem:

    P(m|w)=P(w|m)P(m)/P(w) so P(m|w)P(w)/P(m) = P(w|m) =(0.7)(0.5)/(0.4)= 0.875
     
  4. Sep 20, 2009 #3
    Re: Is my Handout wrong? Stumped... Conditional Probability... Need help urgent!

    I edited my post right before you posted :).

    Now I am trying to figure out P(m or w)

    Is it just .4 + .5?
     
    Last edited: Sep 20, 2009
  5. Sep 20, 2009 #4
    Re: Is my Handout wrong? Stumped... Conditional Probability... Need help urgent!

    Do you know how to add probabilities? Are these probabilities independent of each other?
     
    Last edited: Sep 20, 2009
  6. Sep 20, 2009 #5
    Re: Is my Handout wrong? Stumped... Conditional Probability... Need help urgent!


    I do now, I just figured out I was missing 2 pages in my book.... I looked on my online copy of it and found it....


    P(A) + P(B) - P(AnB) = .4 + .5 - .35 = .55 correct?
     
  7. Sep 20, 2009 #6
    Re: Is my Handout wrong? Stumped... Conditional Probability... Need help urgent!

    No..
     
  8. Sep 20, 2009 #7
    Re: Is my Handout wrong? Stumped... Conditional Probability... Need help urgent!


    Lets try this again...

    since I have (b) correct I will ignore it

    (a) p(MnW) = .4 x .5 = .2
    (c) p(MuB) = .4 + .5 - .2 = .7

    is this correct?

    sorry for not getting this, stats is not my strong point.
     
  9. Sep 20, 2009 #8
    Re: Is my Handout wrong? Stumped... Conditional Probability... Need help urgent!

    How about arithmetic? Yes this is correct given m in the absence of prior w and w in the absence of prior m but is this realistic? What about your conditional probabilities?
     
  10. Sep 20, 2009 #9
    Re: Is my Handout wrong? Stumped... Conditional Probability... Need help urgent!


    I guess I am entirely confused. The book is making it worse and reading info online is confusing me.

    I am getting 2 bits of information regarding (a), which affects (c).

    for
    1)
    P(MnW) = p(M) x p(W) = 0.2
    but I also see
    2)
    P(MnW) = p(M) x p(W|M) = 0.35
    P(WnM) = p(w) x p(M|W) = 0.35

    So for this case which is it? I dont understand when it use each one...
     
  11. Sep 20, 2009 #10
    Re: Is my Handout wrong? Stumped... Conditional Probability... Need help urgent!

    OK For P(m or w); since they are conditional on each other we have P(m|w) + P(w|m) - P(m|w)P(w|m) = (0.7)+(0.875)-(.6125)=0.9625

    Now do P(m ^ w).
     
  12. Sep 20, 2009 #11
    Re: Is my Handout wrong? Stumped... Conditional Probability... Need help urgent!

    Thank you

    This is what I am understanding from the book.
    If they are dependent on each other, use this formula
    p(m^w) = p(m) x p(w | m)

    So....
    0.875 x 0.4 = 0.35
    or
    0.5 x 0.7 = 0.35

    Correct?

    -------------------

    If they were independent of each other, I would multiply them... ie; 0.4 x 0.5 = 0.2, correct?
     
  13. Sep 20, 2009 #12
    Re: Is my Handout wrong? Stumped... Conditional Probability... Need help urgent!

    Yes, but because they are conditional on each other, they are shouldn't be treated as independent unless you have some special situation.

    Regarding P(m^w), your equations are P((w|m)^(m)), etc. For P(m^w), I would use P(m|w) x P(w|m) = (0.875)(0.7)=0.6125.

    How would you combine terms of the form P(x)P(x|y), P(x)P(y|x), P(y)P(x|y), P(y)P(y|x)?
     
    Last edited: Sep 20, 2009
  14. Sep 20, 2009 #13
    Re: Is my Handout wrong? Stumped... Conditional Probability... Need help urgent!


    Interesting... I actually do not see this anywhere in my book, so can you explain why I would be using P(m|w) x P(w|m) instead of P((w|m)^(m)) ?

    How would I combine those terms? I don't follow, wouldn't I just multiply them?


    I REALLY appreciate all your help, you have no idea.
     
  15. Sep 20, 2009 #14
    Re: Is my Handout wrong? Stumped... Conditional Probability... Need help urgent!

    Because P(w|m)^m) is a different question than P(m^w). You were asked for P(m^w), right?
     
    Last edited: Sep 20, 2009
  16. Sep 20, 2009 #15
    Right. I was asked for P(m^w), I think I got it confused reading your question.

    so for P(m^w), I would do the P(m|w) x P(w|m) = (0.875)(0.7)=0.6125 for this question.
     
  17. Sep 20, 2009 #16
    Is this a homework question? I would do it this way because we established the non-independence of P(m) and P(w). However, if this is homework, you might treat them as independent since the expression P(m^w) doesn't tell you about the conditional probabilities by itself. Also you have as givens P(m)=0.4 and P(w)=0.5. So perhaps you should play it safe and answer P(m^w)=0.2 and P(m or w)=0.7. It's obviously wrong given the context, but correct from a "frequentist" point of view where prior probabilities are ignored. You can always defend the context free answer.
     
    Last edited: Sep 20, 2009
  18. Sep 20, 2009 #17
    The question from the text is:

    Probability of Mark and Wilma both go riding a bike together.
     
  19. Sep 21, 2009 #18
    Still not clear. P(m),Mark rides with or without Wilma, P(w), Wilma rides w/wt Mark; P(m|w) Mark will ride with Wilma when Wilma's riding?
     
  20. Sep 21, 2009 #19
    Sorry I posted that from my phone, it was not the whole problem.

    The probability of Mark riding his bike is 0.4
    Probability of Wilma riding her bike is 0.5
    The probability of mark riding his bike given that Wilma does is 0.7

    Find:
    P(M^W) both riding their bikes
    P(W|M)
    P(M or W) riding their bike

    I hope this makes it more clear. Thank you :)
     
  21. Sep 21, 2009 #20
    Then P(m^w)=0.2 and P(m or w)=0.7. During the 20% of the time they are both riding, they are riding together 61.25% of the time.

    EDIT: Now that I understand the question, the above is incorrect. See my response #22 below.
     
    Last edited: Sep 21, 2009
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