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Math Help - Probability Problem

  1. #1
    Junior Member
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    Probability Problem

    Hey,

    Just having a problem with one question:

    A man drives to work the same way every day. There are two sets of traffic lights. The prob. of stopping at the first light is .35, the prob. of stopping at the second light is .65, and the prob. of stopping at at least one light is .70.

    a) Find prob. of stopping at both (I got that... 0.30)
    b) Find the prob of stopping at the first but not the second
    c) Find the prob of stopping at exactly one light.

    Just confused about the last two. I have a small idea, but any help would be super appreciated!!
    Thanks,
    Craig.
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  2. #2
    MHF Contributor
    Grandad's Avatar
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    Hello cjh824
    Quote Originally Posted by cjh824 View Post
    Hey,

    Just having a problem with one question:

    A man drives to work the same way every day. There are two sets of traffic lights. The prob. of stopping at the first light is .35, the prob. of stopping at the second light is .65, and the prob. of stopping at at least one light is .70.

    a) Find prob. of stopping at both (I got that... 0.30)
    b) Find the prob of stopping at the first but not the second
    c) Find the prob of stopping at exactly one light.

    Just confused about the last two. I have a small idea, but any help would be super appreciated!!
    Thanks,
    Craig.
    Clearly these two events are not independent: the probability that he stops at the second light depends upon whether or not he stops at the first.

    I have drawn a probability tree showing the respectively probabilities. For the first light they are

    (1) Stops: 0.35 = \frac{7}{20}

    (2) Doesn't stop: 0.65 = \frac{13}{20}

    For the second light they are (depending on whether (1) or (2) has occurred)

    (1) Stops \frac67; doesn't stop \frac17

    (2) Stops \frac{7}{13}; doesn't stop \frac{6}{13}

    (a) To confirm your first answer, the probability that he stops at both is therefore \frac{7}{20}\times\frac{6}{7}= \frac{6}{20} = 0.3

    The remaining probabilities can be worked out by multiplying these fractions. Thus:

    Stops, Doesn't Stop: \frac{7}{20}\times \frac17 = \frac{1}{20}

    Doesn't stop, stops: \frac{13}{20}\times\frac{7}{13}=\frac{7}{20}

    Doesn't stop, doesn't stop: \frac{13}{20}\times\frac{6}{13}= \frac{6}{20}= 0.3

    So for (b), the answer is \frac{1}{20}= 0.05

    And for (c), it's \frac{1}{20}+\frac{7}{20}=\frac{8}{20}=0.4

    Grandad
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