Probability (electric circuit) - need confirmation of my solution

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Homework Help Overview

The problem involves calculating the probability of a current break in an electric circuit consisting of five elements, where the failures of the elements are independent. The probabilities of failure for each element are provided, and the original poster attempts to determine the conditions under which the current will break.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the conditions for the current to break, questioning the original poster's interpretation of the relationships between the elements A and B. There is a focus on whether the current breaks when any A element fails or when all B elements fail.

Discussion Status

There is an ongoing clarification regarding the correct formulation of the probability expression. Participants are actively correcting each other’s interpretations and suggesting revisions to the mathematical expressions used.

Contextual Notes

Participants are navigating the implications of independent events and the correct application of union and intersection in the context of probability. There is a noted confusion regarding the correct logical structure of the expressions involved.

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 probability of current break, between [tex]M[/tex] and [tex]N[/tex].
attachment.php?attachmentid=26624&stc=1&d=1277294014.gif


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 probability 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 ?
 

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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.
 
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] ?
 
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:
njama said:
Yes, that's true.

Now try again, and use [tex]P(A_1\cap \left[ B_1\cup B_2 \cup B_3 \right] \cap A_2)[/tex]
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.
 
Mark44 said:
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.

Yes, that's right. Thanks for the correction. I replaced \cup with \cap and vice versa.
 
njama said:
Yes, that's right. Thanks for the correction. I replaced \cup with \cap and vice versa.

Thank you both, Njama & Mark44.
 

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