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-   -   Dang Probability! (https://www.askmehelpdesk.com/showthread.php?t=142659)

  • Oct 19, 2007, 05:13 PM
    Ail Bane
    Dang Probability!
    I know this seems like an easy problem but I can't get it figured out.

    The problem is:

    If P(high) = .3, P(low) = .7,
    P(favorable | high) = .9,
    P(unfavorable | low) = .6,

    then P(favorable) =

    Normally I can figure stuff like this out, but usually I have more info than that. Well, there's always a first time.

    I came up with .3, 27, and .1. I don't know what I'm doing wrong.
  • Oct 19, 2007, 06:09 PM
    terryg752
    U = Unfav, F = Fav, H = High, L = Low

    P(F/H) = .9

    Hence P(U/H) = .1

    [ Formula: P(A & B) = P(A) P(B/A) ]

    P(U & H) = P(H) P(U/H) = .03

    P(U & L) = P(L) P(U/L) = .42

    P(U) = .03 + .42 = .45

    HENCE: P(F) = 1 - .45 = .55
  • Oct 19, 2007, 07:00 PM
    Ail Bane
    Thanks for the help but does that mean that P(F) is .55?
  • Oct 19, 2007, 10:55 PM
    terryg752
    Correct!
  • Oct 20, 2007, 12:09 AM
    Ail Bane
    Thanks, you're so helpful. Sorry I didn't see the .55 the first time.
    I'm not sure if I could have done that on my own.

    Its getting close to quiz time at the university so I needed it.

    PS: I rated your answer.;)
  • Oct 20, 2007, 03:54 AM
    terryg752
    Thanks, this was quite an interesting problem.

    You are welcome.

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