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Homework Help: Stats - Independence (circuits)

  1. Feb 4, 2006 #1
    Hello, my book has this question, and no examples (very) similar to it, so I am wondering if I did it correct :smile:

    The following circuit operates if and only if there is a path of functional devices from left to right. The probability that deach device functions is as shown. Assume that the probability that a device is functional does not depend on whether or not other devices are functional. What is the probability that the circuit operates?



    So what I did was:

    Let T denote the event that the entire top is true, meaning that everything functions on the top.
    Let B denote the event that the entire bottom is true.

    [itex]P(T \cup B) = 1 - P(T' \cap B')[/itex]
    [itex]P(T' \cap B') = P(T')P(B')[/itex]
    [itex]P(T') = 1 - P(T1 \cap T2 \cap T3) = 1 - P(T1)P(T2)P(T3)[/itex] T1 is the first thing on the top being functional (meaning 0.9), T2 being 0.8, etc
    [itex]P(B') = 1 - P(B1 \cap B2 \cap B3) = 1 - P(B1)P(B2)P(B3)[/itex]
    [itex]P(T') = 1 - (0.9)(0.8)(0.7) = 0.496[/itex]
    [itex]P(B') = 1 - (.95)(.95)(.95) = 0.142625[/itex]
    [itex]P(T' \cap B') = (0.496)(0.142625) = 0.070742[/itex]
    [itex]P(T \cup B) = 1 - 0.070742 = 0.929258[/itex]
    Which is my answer.

    Does that look correct?
    Last edited: Feb 4, 2006
  2. jcsd
  3. Mar 30, 2006 #2

    Hi! Your answer looks all right, but there is an alternative.

    [tex] P(A\cup B) = P(A) + P(B) - P(A\cap B) [/tex]

    So [tex]P(A) = (0.9)(0.8)(0.7)....
    P(B) = (0.95)(0.95)(0.95)....
    P(A\cap B) = (0.9)(0.8)(0.7)(0.95)(0.95)(0.95).... [/tex]

    You will also obtain the same answer of 0.929258.
    Last edited: Mar 30, 2006
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