The following circuit opeprices if and just iftbelow is a path of practical devices from left to ideal.The probcapability that each device attributes is as presented.Assume that the probability that an equipment is functionaldoes not depend on whether or not various other tools arepractical.

You are watching: The following circuit operates if and only if


What is the probcapacity that device Aa) does not work-related if the device does not work? b) does not occupational if the device operates?

So I have tried working on finding the $jiyuushikan.orgbbP(A|S)$, wherein $jiyuushikan.orgbbP(S)$ is the probability that the mechanism will not work and $jiyuushikan.orgbbP(A)$ is the probcapability that gadget A will certainly not work-related. However before, I simply can not seem exactly how to uncover $jiyuushikan.orgbbP(Acap S)$.From what I computed, $jiyuushikan.orgbbP(S)=0.070625$.How have the right to I uncover $jiyuushikan.orgbbP(Acap S)$?

Edit: tbelow was a first difficulty around this one and also by solving that I currently kbrand-new what to perform through P. Btw, thank you for the edits people hehehe

probcapability conditional-probcapacity
edited Oct 8 "20 at 14:38
asked Oct 8 "20 at 10:16

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I gained a result cshed to yours as of $P(S)$.

I calculated first


Hence your $P(S)=0.070742$. Perhaps your calculation are ok but there is just a different approx.

Now to calcualte $P(overlineA cap S)=0.1cdot<1-0.95^3>=0.0142625$

(if A does not job-related, in order to have the mechanism out of job-related it is crucial that the reduced branch of the mechanism doesn"t work too)

Thus $P(overlineA|S)approx 20.16\%$

Similar arguments for the various other request

EDIT: the most basic method to calculate $P(S)=<1-0.9cdot0.8cdot0.7><1-0.95^3>=0.070742$...I did a lot of useless calculations... :(

edited Oct 8 "20 at 11:35
answered Oct 8 "20 at 10:54

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