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Math Help - Probability

  1. #1
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    Probability

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  2. #2
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    Quote Originally Posted by mauro21pl View Post
    One more
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    a) x1 can be 0 or 1, and for any value of x1 x2 can be 0 or 1, and for any
    values of x1, x2 x3 can be 0 or 1,...

    there are five x's each of which can be 0 or 1 that is a total of 2^5 outcomes in the sample space.

    b) working through them systematicaly and discarding the duplicates
    they appear to be:

    11000
    11100
    11010
    11001
    11110
    11011
    11101
    1111

    00110
    10110
    01110
    00111
    01111
    10111

    10101

    c) As x4 and x5 are determined there are three bits which may be 0 or 1, so
    the number of events in A is 2^3

    d) AW is the event that 4 and 5 failed and (1 and 2 are both working or
    3 and 4 are both working or 1, 3 and 5 are both working).

    That leaves two possibilitiea 1 and 2 working 3 may be 0 or 1, and 4 an 5
    failed: 11100, 11000.

    RonL
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