(Probability) Problem

[Narmer]

There's a problem in my statistics book in which the answer is already given. However, I cannot reproduce the this answer. Is the book correct?
|----A(0.7)-------B(0.7)|
- -------------------------- -
|---C(0.8)--D(0.8)---E(0.8)|
(Please disregard the lines in the middle, I'm trying to make it look like two connected parallel systems forming a single circuit.)

Where A and B are parallel and C,D, and E are parallel.
A circuit system is given in Figure 2.11. Assume the components fail independently.
a) What is the probability that the entire system works? (Answer is 0.75112. I was able to get this one correct.)
b) Given that the system works, what is the probability that the component A is not working? (Answer is 0.2045. I cannot reproduce this answer.)

This is NOT A Homework problem. But it's related to another problem I have to solve.

 

[her209]

Your circuit is all screwed up...

 

[Narmer]

Originally posted by: her209
Your circuit is all screwed up...

I know. I can't draw.
I'm trying to make it look like this with A,B on the top and C,D,E on the bottom.

 

[reverend boltron]

It would probably make it look a bit nicer if you copied it from the book and posted a pic of it on pics.bbzzdd.com

 

[dogooder]

P(A doesn't work given that the circuit works) = P(A doesn't work and the circuit works) / P(the circuit works) = .3 * .8^3 / .75112

 

[Narmer]

Originally posted by: dogooder
P(A doesn't work given that the circuit works) = P(A doesn't work and the circuit works) / P(the circuit works) = .3 * .8^3 / .75112

Yeah, that's it. Damn and it's so simple too. Is there a reason why it isn't .8^3*.7*.3/.75112?

 

[Born2bwire]

Originally posted by: Narmer

Originally posted by: dogooder
P(A doesn't work given that the circuit works) = P(A doesn't work and the circuit works) / P(the circuit works) = .3 * .8^3 / .75112

Yeah, that's it. Damn and it's so simple too. Is there a reason why it isn't .8^3*.7*.3/.75112?

Because B can be working or broken.

 

[chuckywang]

I have no idea what the circuit looks like, otherwise I could help.

 

[Narmer]

Originally posted by: Born2bwire

Originally posted by: Narmer

Originally posted by: dogooder
P(A doesn't work given that the circuit works) = P(A doesn't work and the circuit works) / P(the circuit works) = .3 * .8^3 / .75112

Yeah, that's it. Damn and it's so simple too. Is there a reason why it isn't .8^3*.7*.3/.75112?

Because B can be working or broken.

If B can be working or broken, then why was it taken out of the (solution) equation altogether?

 

[RapidSnail]

If I knew what you were talking about I would help .

 

[Born2bwire]

Originally posted by: Narmer

Originally posted by: Born2bwire

Originally posted by: Narmer

Originally posted by: dogooder
P(A doesn't work given that the circuit works) = P(A doesn't work and the circuit works) / P(the circuit works) = .3 * .8^3 / .75112

Yeah, that's it. Damn and it's so simple too. Is there a reason why it isn't .8^3*.7*.3/.75112?

Because B can be working or broken.

If B can be working or broken, then why was it taken out of the (solution) equation altogether?

Because the probability of it working or not working is 1.