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Math Help - Bayes Probability - Rain and Umbrellas

  1. #1
    Junior Member
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    Australia
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    Bayes Probability - Rain and Umbrellas

    I do not have the answers for this question, and was hoping if anyone could provide feedback on my thinking.

    In London, half of the days have some rain. The weather forecaster
    is correct 2/3 of the time, i.e., the probability that it rains, given that she has
    predicted rain, and the probability that it does not rain, given that she has
    predicted that it won’t rain, are both equal to 2/3. When rain is forecast,
    Mr. Pickwick takes his umbrella. When rain is not forecast, he takes it with
    probability 1/3. Find

    (a) the probability that Pickwick has no umbrella, given that it rains.
    If rain is predicted, regardless of the result - Pickwick will take the umbrella. So prob =0
    If rain is not predicted and it rains and he does not have an umbrella = \frac{1}{3}\times \frac{2}{3}=\frac{2}{9}

    (b) the probability that it doesn’t rain, given that he brings his umbrella.
    When I used the tree diagram with a three step [rain(1/2)/not rain(1/2), prediction and result, umbrella/no umbrella] I am confused what the numerator is.

    Thank you in advance for any assistance provided
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  2. #2
    Newbie
    Joined
    Jul 2010
    Posts
    1

    Right, it's a harder problem than most realize

    This is a very favorite problem of mine.
    Right, it’s not easy to see all the conditionals.
    You actually have to do some algebra to get that the probability of her forecasting rain is ˝ ,
    and that the probability of her having forecasted rain given that it rains = the probability of her not having forecasted rain given that it does not rain = 2/3
    Once this is done, you can easily fill in the tree.

    See my website http://duport.com
    for a hint on the full solution.
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