# Thread: Probability Problem

1. ## Probability Problem

A circuit has 3 components: A,B,C. Both B and C serve the same purpose and it is only necessary that one is operational. But A is distinct and must always work. Given that:

Probability (A fails) = 0.8
Probability (B fails) = 0.4
Probability (C fails) = 0.8

What is the overall probability of failure assuming these are all independent events?

2. Originally Posted by mr_motivator
A circuit has 3 components: A,B,C. Both B and C serve the same purpose and it is only necessary that one is operational. But A is distinct and must always work. Given that:

Probability (A fails) = 0.8
Probability (B fails) = 0.4
Probability (C fails) = 0.8

What is the overall probability of failure assuming these are all independent events?
Do you mean failure of circuit? Then you need A' or A B' C'. I get 0.8 + (0.2)(0.4)(0.8).

3. I found the complement, mainly because I didn't read the final question.
Let A denote A suceeds....
So in order to suceed we want.
$\displaystyle P(A(BUC))=P(A)P(BUC)=P(A)\bigl(P(B)+P(C)-P(BC)\bigr)$
$\displaystyle =(.2)(.6+.2-(.6)(.2))=.136$
So 1-.136=.864 is correct.

4. First map how can circuit fails:
If A fails $\displaystyle = 0.8$
If A works but B & C fails $\displaystyle =0.2*0.4*0.8$$\displaystyle =0.064 Probability that circuit fails \displaystyle = 0.8+0.064$$\displaystyle =0.864$