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Math Help - Statistics

  1. #1
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    Red face Statistics

    For the life of me I cannot start this question to me this does not make sense?
    In a certain school, it is know that 80% of the studentís user the internet for school projects 60%m use e-mail on regular basis, and 90% use the internet for school projects or for e-mail on a regular basis. A student from this school is selected at random. Determine the probability that the student used e-mail, given that the student used the internet for school projects

    Would I go like this:
    1/80 x 1/60 = 2.08 x 10^4 this does not make sense??????
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by superman View Post
    For the life of me I cannot start this question to me this does not make sense?
    In a certain school, it is know that 80% of the studentís user the internet for school projects 60%m use e-mail on regular basis, and 90% use the internet for school projects or for e-mail on a regular basis. A student from this school is selected at random. Determine the probability that the student used e-mail, given that the student used the internet for school projects

    Would I go like this:
    1/80 x 1/60 = 2.08 x 10^4 this does not make sense??????
    No no sense at all.

    Out of a nominal 1000 students

    900 use one or the other
    600 use "e"
    800 use "i"

    so 500 use both

    So of the 800 that use "i" 500 also use "e", so the required probability is:

    500/800 = 5/8

    RonL
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  3. #3
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    Hello, superman!

    In a certain school, 80% of the students use the internet,
    60% use e-mail, and 90% use the internet or for e-mail.

    A student from this school is selected at random.
    Determine the probability that the student used e-mail,
    given that the student used the internet.

    Would I go like this: 1/80 x 1/60 = 2.08 x 10^4
    This does not make sense . It certainly doesn't!
    Since when does 80\% \,=\,\frac{1}{80} ???


    Since you are given a Conditional Probability problem,

    I assume you know this formula: . P(A \cup B) \:=\:P(A) + P(B) - P(A \cap B) .[1]

    . . and Bayes' Theorem: . P(A|B) \:=\:\frac{P(A \cap B)}{P(B)} .[2]


    We are given: . P(I) = 0.80,\;P(E) = 0.60,\;P(I \cup E) = 0.90

    Substitute into [1]: . P(I \cup E) \:=\:P(I) + P(E) - P(I \cap E)

    . . . .and we have: . . . . 0.90\:=\:0.80 + 0.60 - P(I \cap E)

    . . . . . . . . Hence: . P(I \cap E) \:=\:0.50


    We are asked for: . P(E|I)

    Bayes' Theorem: . P(E|I) \:=\:\frac{P(E \cap I)}{P(I)} \:=\:\frac{0.50}{0.80} \:=\:\frac{5}{8}

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