Answer to Question #171453 in Statistics and Probability for Sara

Question #171453

https://www.chegg.com/homework-help/questions-and-answers/circuit-made-5-independent-components-shown-figure--chance-component-working-properly-094--q59663024


1
Expert's answer
2021-03-17T11:17:21-0400

The circuit below is made of 5 independent components as shown in the figure below. 


PC1)=P(C2)=0.94PC_1)=P(C_2)=0.94


PC3)=0.97PC_3)=0.97


P(C4)=P(C5)=085P(C_4)=P(C_5)=085

The events C1C_1 and C2C_2 are independent


P(C1C2)=P(C1)P(C2)P(C_1\cap C_2)=P(C_1)P(C_2)

The events C4C_4 and C5C_5 are independent


P(C4C5)=P(C4)P(C5)P(C_4\cap C_5)=P(C_4)P(C_5)




P(C1C2)=P(C1)+P(C2)P(C1C2)P(C_1\cup C_2)=P(C_1)+P(C_2)-P(C_1\cap C_2)

=P(C1)+P(C2)P(C1)P(C2)=P(C_1)+P(C_2)-P(C_1)P(C_2)

=0.94+0.940.94(0.94)=0.9964=0.94+0.94-0.94(0.94)=0.9964



P(C4C5)=P(C4)+P(C5)P(C4C5)P(C_4\cup C_5)=P(C_4)+P(C_5)-P(C_4\cap C_5)

=P(C4)+P(C5)P(C4)P(C5)=P(C_4)+P(C_5)-P(C_4)P(C_5)

=0.85+0.850.85(0.85)=0.9775=0.85+0.85-0.85(0.85)=0.9775



All five components are independent


P(work)=P((C1C2)C3(C4C5))P(work)=P((C_1\cup C_2)\cap C_3\cap (C_4\cup C_5))

=0.9964(0.97)(0.9775)=0.94476157=0.9964(0.97)(0.9775)=0.94476157


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