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Smoker Tree Diagram

  1. Dec 6, 2004 #1
    Tree Diagram

    Hello again!

    I'm stuck and need a little push.

    Referring to the diagram: out of a group of smokers, 20% are heavy smokers, 30% are light smokers, and 50% are non-smokers. A light smoker is twice as likely to die as a non-smoker but half as likely as a heavy smoker.

    H: heavy smoker
    L: light smoker
    N: non-smoker
    D: die

    So, I created the tree diagram (see attached) and understand the relationship. I come up with the following.

    [tex]P(D|L)=2P(D|N)=\frac{1}{2}P(D|H)[/tex]

    [tex]P(H|D)=\frac{P(H)P(D|H)}{P(D)}[/tex]

    and likewise for P(L|D) and P(N|D)

    I need to figure out what P(H|D) is given the info and the tree I created.

    Can someone nudge me in the right direction?

    Thanks,
    dogma
     

    Attached Files:

    Last edited: Dec 7, 2004
  2. jcsd
  3. Dec 7, 2004 #2
    waiter...check please

    I think I got it…can someone check me, please?

    The probability of a person being a heavy smoker given that he died:

    [tex]P(H|D)=\frac{P(H \cap D)}{P(D)}=\frac{P(H)P(D|H)}{P(D)}[/tex]

    The probability that a person in this study dies:

    [tex]P(D)=P(H \cap D)+P(L \cap D)+P(N \cap D)[/tex]

    [tex]= P(H)P(D|H)+P(L)P(D|L)+P(N)P(D|N)[/tex]

    …but since [tex] P(D|L)= 2P(D|N)= \frac{1}{2}P(D|H)[/tex]

    …so

    [tex]P(D)= P(H)P(D|H)+\frac{1}{2}P(L)P(D|H)+\frac{1}{4}P(N)P(D|H)[/tex]

    …then

    [tex] P(H|D)=\frac{P(H)P(D|H)}{ P(H)P(D|H)+\frac{1}{2}P(L)P(D|H)+\frac{1}{4}P(N)P(D|H)}[/tex]

    …the P(D|H)'s cancel out leaving a very simple equation to plug and chug

    P(H|D) = 42.1%
     
  4. Dec 10, 2004 #3
    Your answer checks.
     
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