# Homework Help: Probability (electric circuit) - need confirmation of my solution

1. Jun 23, 2010

### dperkovic

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

2. Relevant equations
For independent events:

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

3. 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 ?

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2. Jun 23, 2010

### 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.

3. Jun 23, 2010

### dperkovic

I see....so, formula must be:

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

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

4. Jun 23, 2010

### njama

Yes, that's true.

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

Last edited: Jun 23, 2010
5. Jun 23, 2010

### Staff: Mentor

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.

6. Jun 23, 2010

### njama

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

7. Jun 23, 2010

### dperkovic

Thank you both, Njama & Mark44.