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If n components, functioning independently, are connected in series, and if the ith i^{th}ith​ component has reliability Ri(t)R_i(t)Ri​(t)​, then
Question

If n components, functioning independently, are connected in series, and if the ith i^{th}​ component has reliability Ri(t)R_i(t)​, then the reliability of the entire system R(t) is given by

A.

R(t)=R1(t)+R2(t)++Rn(t)R(t)=R_1(t)+R_2(t)+\cdots+R_n(t)​​

B.

R(t)=R1(t)R2(t)Rn(t) R(t)=R_1(t)R_2(t)\cdots R_n(t)​​

C.

R(t)=R1(t)R2(t)+R3(t)R4(t)++Rn1(t)Rn(t)R(t)=R_1(t)R_2(t)+R_3(t)R_4(t)+\cdots+R_{n-1}(t)R_n(t)​​

D.

R(t)=R1(t)R2(t)+R3(t)R4(t)++Rn1(t)Rn(t) R(t)=R_1(t)-R_2(t)+R_3(t)-R_4(t)+\cdots+R_{n-1}(t)-R_n(t)​​

Correct option is B

Given:
The components are connected in series and function independently.
Formula Used:
For a series system,
R(t)=i=1nRi(t).R(t)=\prod_{i=1}^{n}R_i(t).​​
Solution:
A series system functions only if every component functions.
Since the components are independent,
the reliability of the entire system is the product of the reliabilities of all components.
Hence,
R(t)=R1(t)R2(t)Rn(t).R(t)=R_1(t)R_2(t)\cdots R_n(t).​​
The correct answer is (b) R(t)=R1(t)R2(t)Rn(t).R(t)=R_1(t)R_2(t)\cdots R_n(t).​​

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