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: A_1-0.6, B_1-0.4, B_2-0.7, B_3-0.9 and A_2-0.5. Find the probabilty of current break, between M and N.

Homework Equations

For independent events:

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

The Attempt at a Solution

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

So:

B = B_1\cup B_2\cup B_3 = B_1\cdot B_2\cdot B_3

And, finaly, probability is:

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.

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:

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

instead of

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

 

[njama]

Yes, that's true.

Now try again, and use P(A_1\cup \left[ B_1\cap B_2 \cap B_3 \right] \cup A_2)

 

[Mark44]

Yes, that's true.

Now try again, and use P(A_1\cap \left[ B_1\cup B_2 \cup B_3 \right] \cap A_2)

Don't you mean P(A_1\cup \left[ B_1\cap B_2 \cap B_3 \right] \cup A_2)
?
I think you just grabbed the wrong tex expression.

 

[njama]

Don't you mean P(A_1\cup \left[ B_1\cap B_2 \cap B_3 \right] \cup A_2)
?
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.