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    In a class of students 1/3 are left handed and 1/8 wear a dental brace.
    A student is selected at random.

    What is the probability of that student.
    a) is left handed AND wears a brace
    b) is left handed but does not wear a brace
    c) does not wear a brace AND is not left handed

    Note: 1/3 and 1/8 DOES NOT = 1 :/

    Any ideas? for a) I did : 1/3 * 1/8

    The probability of a student not wearing a brace is 7/8 (1-1/8). Therefore the probability of a student being left handed and not wearing a brace is 1/3*7/8. Part c involves the same method.
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Updated: October 13, 2017
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