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Math Help - Having trouble with this probability question...

  1. #1
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    Having trouble with this probability question...

    A poll found that the probability that a student chosen at random had visited the arcade within the past week was 0.3. The probability that a student chosen at random played games on a home-based game machine was 0.5. The probability that a student did both was 0.2. What was the probability that a student played on an arcade machine or on a home-based machine?

    Thanks in advance!
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  2. #2
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    I believe the probability of either event A or B occurring is

    P(A) + P(B) - P(A + B),

    which in this case is

    0.3 + 0.5 - 0.2 = 0.6.
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  3. #3
    MHF Contributor
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    Quote Originally Posted by PiratePrincess View Post
    A poll found that the probability that a student chosen at random had visited the arcade within the past week was 0.3. The probability that a student chosen at random played games on a home-based game machine was 0.5. The probability that a student did both was 0.2. What was the probability that a student played on an arcade machine or on a home-based machine?

    Thanks in advance!
    Assuming that a student who visits the arcade always plays on an arcade machine, then

    0.3=P(plays on arcade machine)=P(plays only on arcade machine)+P(plays on arcade and home machine)

    0.5=P(plays on home machine)=P(plays at home only)+P(plays at home and at arcade)

    0.3+0.5=P(arcade only)+P(home only)+2P(both)

    0.8=P(plays either one)+P(both)

    P(plays either one)=0.8-0.2
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