Probability (electric circuit) - need confirmation of my solution

[dperkovic]

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 ?

 

[njama]

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.

 

[dperkovic]

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] ?

 

[njama]

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]

 

[Mark44]

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.

 

[njama]

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.

 

[dperkovic]

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

Thank you both, Njama & Mark44.