materials $1$ and $2$ are linked in parallel, so the subsystem works iff either $1$ or $2$ works;since $3$ and also $4$ are linked in series, the subsystem worksiff both 3 and 4 work. If contents work individually of the one another and $P(\textcomponent works)=0.9$, calculation $P(\textsystem works)$.

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Call $1,2$ subsystem $A$, and $3,4$ subsystem $B$ \beginalign*P<\textsystem fails> &= P + P - P<\textboth fail> \\&= (0.1)^2 + (1 - (0.9)^2) - (0.1)^2 (1-(0.9)^2) \\&= 0.1981\endalign*Thus $$P<\textsystem works> = 1 - 0.1981 = 0.8019$$  I am wondering why this doesn"t work.

Let $A$, $B$ it is in the occasions that this subsystems work, $A_i,B_i$ for $i=1,2$ it is in the event that ingredient $i$ works, and I will use a bar to show that something doesn"t work.Then, presume the device will work if at the very least one subsystem works,\beginalign*P(\textSystem works)&=P(A\cup B)\\&=P(A)+P(B)-P(A)P(B)\\&=P(A_1\cup A_2)+P(B)-P(A)P(B)\\&=P(A_1)+P(A_2)-P(A_1)P(A_2)+P(B_1)P(B_2)-P(A)P(B)\\&=.9+.9-(.9)^2+(.9)^2-P(A)P(B)\\&=0.99+ 0.81-(.99)(0.81)\\&=0.9981\endalign* You can technically go about this directly, because there space three cases toconsider: the casewhere 1 works and 2 walk not, the case where 1 does not work and 2 does, and the situation where both work. 3 and 4 must constantly work, so we account for thisprobability in all cases. The an initial case is $(.9)(1-.9)(.9)^2$, the secondcase is the same and also the third case is $(.9)(.9)(.9)^2.$ Hence, the answershould be$$2(.9)(1-.9)(.9)^2 + (.9)^4.$$ Thanks because that contributing an answer to couchsurfingcook.comematics stack Exchange!

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