The reliability of a parallel system is given by $1 - (1 - R_1)(1 - R_2)$, where $R_1$ and $R_2$ are the reliabilities of the individual components. In this case, $R_1 = 0.89$ and $R_2 = 0.87$.
$R_{parallel\_left} = 1 - (1 - 0.89)(1 - 0.87) = 1 - (0.11)(0.13) = 1
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