Given that components 1, 2, and 3 each have reliabilities of 0.95, and assuming the system configuration as depicted in the image (series-parallel configuration):
* Reliability of the parallel components (2 and 3): Rparallel=1[(1R2)×(1R3)]R_{\text{parallel}} = 1 - [(1
- R_2) \times (1 - R_3)]Rparallel=1[(1R2)×(1R3)] Rparallel=1[(10.95)×(10.95)]R_{\text{parallel}} = 1
- [(1 - 0.95) \times (1 - 0.95)]Rparallel=1[(10.95)×(10.95)] Rparallel=1[0.05×0.05]R_{\text{parallel}} =
1 - [0.05 \times 0.05]Rparallel=1[0.05×0.05] Rparallel=10.0025R_{\text{parallel}} = 1 -
0.0025Rparallel=10.0025 Rparallel=0.9975R_{\text{parallel}} = 0.9975Rparallel=0.9975
* Reliability of the entire system (series of component 1 and the parallel system):
Rsystem=R1×RparallelR_{\text{system}} = R_1 \times R_{\text{parallel}}Rsystem=R1×Rparallel Rsystem=0.95×0.9975R_{\text{system}} = 0.95 \times 0.9975Rsystem=0.95×0.9975 Rsystem=0.947625R_{\text{system}} = 0.947625Rsystem=0.947625 Rounded to three decimal places, the system reliability is 0.948.