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Probability (electric circuit) - need confirmation of my solution

  1. Jun 23, 2010 #1
    1. The problem statement, all variables and given/known data
    Electric circuit (as shown in the picture), is made from five elements. Failures of elements are independent, with probabilites: [tex]A_1-0.6[/tex], [tex]B_1-0.4[/tex], [tex]B_2-0.7[/tex], [tex]B_3-0.9[/tex] and [tex]A_2-0.5[/tex]. Find the probabilty of current break, between [tex]M[/tex] and [tex]N[/tex].

    2. Relevant equations
    For independent events:

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

    3. The attempt at a solution
    Current will be break, when any of the [tex]A[/tex] element fail, or if all of the [tex]B[/tex] elements fail. So probabilty of the current break is: [tex]A_1 \cap\left[B_1\cup B_2\cup B_3\right]\cap A_2[/tex].


    [tex]B = B_1\cup B_2\cup B_3 = B_1\cdot B_2\cdot B_3[/tex]

    And, finaly, probability is:

    [tex]A_1 \cap\left[B_1\cup B_2\cup B_3\right]\cap A_2 = A_1 \cap B\cap A_2 = A_1 + A_2 + B - A_1\cdot B - A_1\cdot A_2 - A_2\cdot B + A_1\cdot A_2 \cdot B[/tex].

    Is that correct ?

    Attached Files:

  2. jcsd
  3. Jun 23, 2010 #2
    That's not correct.

    First, give me the correct information. When the current will break?

    I ask you this because you said that ALL of B need to fail, and then you write B1 U B2 U B3.
  4. Jun 23, 2010 #3
    I see....so, formula must be:

    [tex] A_1\cup \left[ B_1\cap B_2 \cap B_3 \right] \cup A_2[/tex]

    instead of

    [tex] A_1\cap \left[ B_1\cup B_2 \cup B_3 \right] \cap A_2[/tex] ?
  5. Jun 23, 2010 #4
    Yes, that's true.

    Now try again, and use [tex]P(A_1\cup \left[ B_1\cap B_2 \cap B_3 \right] \cup A_2)[/tex]
    Last edited: Jun 23, 2010
  6. Jun 23, 2010 #5


    Staff: Mentor

    Don't you mean [tex]P(A_1\cup \left[ B_1\cap B_2 \cap B_3 \right] \cup A_2)[/tex]
    I think you just grabbed the wrong tex expression.
  7. Jun 23, 2010 #6
    Yes, that's right. Thanks for the correction. I replaced \cup with \cap and vice versa.
  8. Jun 23, 2010 #7
    Thank you both, Njama & Mark44.
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