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dperkovic
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Homework Statement
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].
Homework 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]
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].
So:
[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 ?