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Thread: Random Needs

  1. #1
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    Random Needs

    Professor Jackson is in charge of a program to prepare people for a high school equivalency exam. Records show that 80% of the students need work in Math, 70% need work in English, and 55% need work in both areas. Compute the probability that a student selected at random needs:

    (a) Math or English
    (b) Math only
    (c) English only
    (d) Both
    (e) None
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  2. #2
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    Re: Random Needs

    Quote Originally Posted by joshuaa View Post
    Professor Jackson is in charge of a program to prepare people for a high school equivalency exam. Records show that 80% of the students need work in Math, 70% need work in English, and 55% need work in both areas. Compute the probability that a student selected at random needs:

    (a) Math or English
    (b) Math only
    (c) English only
    (d) Both
    (e) None
    a) $\mathcal{P}(M\cup E)=\mathcal{P}(M)+\mathcal{P}(E)-\mathcal{P}(M\cap E)$

    b) $\mathcal{P}(M\cap \neg E)=\mathcal{P}(M)-\mathcal{P}(M\cap E)$


    $\mathcal{P}(\neg E)$ is probability of not English.
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    Re: Random Needs

    Thanks Plato. What about (d) and (e)?
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    Re: Random Needs

    Quote Originally Posted by joshuaa View Post
    Thanks Plato. What about (d) and (e)?
    This is your homework. You should do it for yourself.

    Now if you can do parts a & b, with my hints, then you can do the others.
    Because, those are answered in parts a & b.

    Give it a try and show us,
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    Re: Random Needs

    it is obvious that (d) is 0.55 and I got it from the Question itself (55% need work in both areas), then (e) must be 0.45, but I don't understand why the book's answer is 0.44. This is why I was doubting in (d) & (e)
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    Re: Random Needs

    Quote Originally Posted by joshuaa View Post
    it is obvious that (d) is 0.55 and I got it from the Question itself (55% need work in both areas), then (e) must be 0.45, but I don't understand why the book's answer is 0.44. This is why I was doubting in (d) & (e)
    e) $1-\mathcal{P}(M\cup E)$
    Can you explain why?
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    Re: Random Needs

    If you mean to explain why the answer of (e) looks like that, it is difficult to tell you now. I have to mimic the basic rules first!

    Thanks a lot for the help and I am sure the 0.44 in the book is mistyped!
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  8. #8
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    Re: Random Needs

    Suppose there were a total of 100 students. 55% of them, 55 students, need work in both English and Math. 80% of them, 80, need work in Math. Since 55 also need work in English, 80- 55= 25 need work only in math. 70%, 70 students, need work in English but 55 also need work in Math, so that leaves 70- 55= 15 need work only in English. That's a total of 55+ 25+ 15= 95 students who need work in Math or English or both. 100- 95= 5 students do not need either.

    The percentage who need work in
    (a) Math or English is 95/100= 0.95 or 95%.

    (b) Math only is 25/100= 0.25 or 25%

    (c) English only is 15/100= 0.15 or 15%

    (d) Both is 55/100= 0.55 or 55%

    (e) None is 5/100= 0.05 or 5%
    Last edited by HallsofIvy; Oct 3rd 2017 at 11:03 AM.
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  9. #9
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    Re: Random Needs

    Thanks a lot HallsofIvy. Very clear.
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