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Math Help - Probability Question

  1. #1
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    Question Probability Question

    A recent study indicates that 14% of the students like to play computer games and that 22% of students fail their exam. Also 60% of those who like to play computer games fail their exam. Given that a student is chosen at random from the population, find, to 3 decimal places, the probability that the chosen student

    (i) does not like to play computer games and fails the exam.
    (ii) likes to play computer games and does not fail the exam
    (iii) likes to play computer games, given that the student does not fail the exam.
    (iv) either likes to play computer games or fails the exam or both.

    I'm having problems with questions I have made bold.

    Could somebody help me with this? I've been given the answers, which are (i) 0.136 and (iv) 0.276 but I can't seem to get them right.
    Thank you very much!


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  2. #2
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    Re: Probability Question

    Quote Originally Posted by pumbaa213 View Post
    A recent study indicates that 14% of the students like to play computer games and that 22% of students fail their exam. Also 60% of those who like to play computer games fail their exam. Given that a student is chosen at random from the population, find, to 3 decimal places, the probability that the chosen student

    (i) does not like to play computer games and fails the exam.
    (ii) likes to play computer games and does not fail the exam
    (iii) likes to play computer games, given that the student does not fail the exam.
    (iv) either likes to play computer games or fails the exam or both.

    I'm having problems with questions I have made bold.

    Could somebody help me with this? I've been given the answers, which are (i) 0.136 and (iv) 0.276 but I can't seem to get them right.
    Thank you very much!


    Venn diagrams will help.

    i) P(does not like to play computer games AND fails the exam) =

    P(fail)-P(fail and likes computer games)P(likes computer games) = 0.22 - 0.6*0.14 = 0.136

    iv) P(either likes to play computer games or fails or both) =

    P(likes computer games) + P(fails) - P(likes computer games and fails) = 0.14+0.22-0.6*0.14 = 0.276

    Draw the Venn diagram for each case and highlight the area of interest for each case above and it will really help you see where these formulas are coming from.
    Thanks from pumbaa213
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  3. #3
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    Re: Probability Question

    I've got it! Thank you so much romsek! Indeed it is clearer with venn diagrams..
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