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A system contains two components, C and D, connected in parallel as shown in the diagram Assume C and D function independently. For the system to function, either C or D must function.

1.  If the probability that C fails is 0.1 and the probability that D fails is 0.12, find the probability that the system functions. Round the answer to four decimal places.

2. If both C and D have probability p of failing, what must the value of p be so that the probability that the system functions is 0.98?

3. If three components are connected in parallel, function independently, and each has probability p of failing, what must the value of p be so that the probability that the system functions is 0.98?

4. If components function independently, and each component has probability 0.5 of failing, what is the minimum number of components that must be connected in parallel so that the probability that the system functions is at least 0.98?

math   Statistics-And-Probability   
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Answer #1

Solcetion: Suntem has o compon ents Cand D Connected in paxatlet Now,we haue to fnd out: probabiltly that the Stem tunction iVen, ealready know the pobabity of Sntem finction is, PC system function) 亡-p ( System Riu) teye Pytemnctnfatr) p (system function) ะ-plenoc ) 디-(001a) F0.988 Probobilihy o Syte function 08e, both c and D hove olobiity f foitnq -p Probobility of Sytem inchion Now, coe findout: thaf must the value ol p- To findoeul the p value , xe murt bte the ptsyxlem fnction) hle knoto), Syslem fiu, cai D S IP - p(c) p(oc) cand D of fai ing pobability叩 So, plec) and p (D%p Nouw, p( Sytem function) pSem falu) 2 but in given, pitm finchon) D8 0.98-P . . The value of P & o.luw,3 3) NoL,Ce indout that the value of p iven, three componentA ove connected in pe each component ha proboliliy of faitin p lProbobi btliti^ of SuxHemunctione 018 Km faits) each Component failtrng Paobabilt P(system foin) FiPXpp) Su^lem function) I- l- P )-o.98 t) Now.uwe linout lhat ix the Mirimum numbei of Componentx ihot murt be Connected in pavattel To findout Minimum number of componentk oth the heip of robaby oftem function. rrh componentof hobabiy of fai ling 050 no 1- (0.5) n MiMimum pCSyAlem faiK) em touK)ech component lodex piObabi (ench componet faiex piobability) Nout p(sHs ten-function) -0-5 0.98-o 5 . 1-05098 NO, 0.5 21-0.98 Nouo, Apply Jnonbo th Kidea n. An(0.5)n02) in (o-5) メ1.6931 n5644 the minimum number of components6

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