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Math Help - simple probability.

  1. #1
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    simple probability.

    in camelot it never rains on friday, saturday, sunday or monday. the probability that it rains on a given tuesday is (1/5). for each of the two remaining days, wednesday and thursday, the conditional probability that it rains, given that it rained on the previous day is a, and the conditional probability that it rains, given that it did not rain on the previous day is b.

    the answer for the first part of the questions is
    p(raining on wednesday)=1/5(a+4b)
    p(raining on thursday)=1/5(a-b)(a+4b)+b

    b) if X is the event that, in a randomly chosen week, it rains on thursday, Y is the event that it rains on tuesday and y' is the event it does not rain on tuesday, show that
    p(x|y)-p(x|y ')=(a-b)^2

    what i did for question b was:
    p(x|y)-p(x|y)
     =p(x \cap y)/p(y) - p(x \cap y')/p(y')
    but what i did was wrong... so help please
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  2. #2
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    Quote Originally Posted by z1llch View Post
    in camelot it never rains on friday, saturday, sunday or monday. the probability that it rains on a given tuesday is (1/5). for each of the two remaining days, wednesday and thursday, the conditional probability that it rains, given that it rained on the previous day is a, and the conditional probability that it rains, given that it did not rain on the previous day is b.

    the answer for the first part of the questions is
    p(raining on wednesday)=1/5(a+4b)
    p(raining on thursday)=1/5(a-b)(a+4b)+b

    b) if X is the event that, in a randomly chosen week, it rains on thursday, Y is the event that it rains on tuesday and y' is the event it does not rain on tuesday, show that
    p(x|y)-p(x|y ')=(a-b)^2

    what i did for question b was:
    p(x|y)-p(x|y)
     =p(x \cap y)/p(y) - p(x \cap y')/p(y')
    but what i did was wrong... so help please
    What you did (all one line of it, admittedly) is fine. Why do you say it's wrong??

    You know Pr(Y) = 1/5 and Pr(Y') = 4/5. So now all you have to do is calculate \Pr(X \cap Y) and \Pr(X \cap Y').

    To do this I suggest drawing a tree diagram (with branches going from Tuesday to Wednesday and from Wednesday to Thursday). Then it's easy to see that:

    \Pr(X \cap Y) = \frac{1}{5} \, (a) \, (a) + \frac{1}{5} \, (1-a) \, (b) = \frac{1}{5} \, (a^2 + b - ab).

    \Pr(X \cap Y') = \frac{4}{5} \, (b) \, (a) + \frac{1}{5} \, (1-b) \, (b) = \frac{4}{5} \, (ab + b - b^2).

    Now substitute everything into  \frac{\Pr(X \cap Y)}{\Pr(Y)} - \frac{\Pr(X \cap Y')}{\Pr(Y')} and it's blue sky all the way.
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  3. #3
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    thanks! i did that too, but i do it altogether and my equation got a bit messed up and stucked. thanks again for the help, got the answer already
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