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Math Help - Quick probability question

  1. #1
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    Quick probability question

    Hello everyone,

    I'm finishing up some of my stat HW right now and I have hit a wall. There are two questions that are on the tip of my tongue and I can't answer them! It's maddening!

    Anyway, here is the the first question...in question:

    Suppose the P(E) = .6 , P(F) = .3, and that E and F are independent events. What is the probability that the student knows the answer to the first question and does NOT know the answer to the second question?

    (E=prob of first question and F= prob of second)

    I really feel like the solutions are obvious, but nothing is coming to me.

    Question two is:

    P(E) = .6 and P(F) = .5 and P(E and F) = .4

    Given the above, what is the probability that the student knows the answer to EXACTLY one of the questions?




    Thanks!
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  2. #2
    MHF Contributor
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    Dec 2008
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    South Coast of England
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    Hello Nohg

    Welcome to Math Help Forum!
    Quote Originally Posted by Nohg View Post
    Hello everyone,

    I'm finishing up some of my stat HW right now and I have hit a wall. There are two questions that are on the tip of my tongue and I can't answer them! It's maddening!

    Anyway, here is the the first question...in question:

    Suppose the P(E) = .6 , P(F) = .3, and that E and F are independent events. What is the probability that the student knows the answer to the first question and does NOT know the answer to the second question?

    (E=prob of first question and F= prob of second)

    I really feel like the solutions are obvious, but nothing is coming to me.

    Question two is:

    P(E) = .6 and P(F) = .5 and P(E and F) = .4

    Given the above, what is the probability that the student knows the answer to EXACTLY one of the questions?




    Thanks!
    1) If p(F)=0.3, then p(F') (= the probability that the student does not know the answer to the second question) = 1-0.3=0.7.

    And if the events are independent, p(E \cap F')=p(E)p(F') = 0.6 \times 0.7 = 0.42.

    2) p(E \cup F) = p(E) + p(F) -p(E\cap F) = 0.6+0.5-0.4=0.7

    But this is an inclusive OR; in other words, it is the probability that the student knows the answer to question 1 or question 2 or both. So if we want the probability that he knows the answer to exactly one question, we subtract from this the probability that he knows the answer to both. Thus:

     p(E or F but not both )=p((E \cup F) \cap (E\cap F)') = 0.7-0.4=0.3

    Grandad
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