Let's solve the problem step by step: --- ## **Step 1: Initial System Reliability** The system is two components in series: - \( R_1 = .7 \) - \( R_2 = .95 \) **Reliability of series system:** \[ R_{\text{series}} = R_1 \times R_2 = .7 \times .95 = .665 \] --- ## **Step 2: Option (a) - Replace Component 1 by Three in Parallel** Replace component 1 with three in parallel (each has reliability .7), and keep component 2 as is. **Reliability of three parallel components (all have same reliability \( R \)):** \[ R_{\text{parallel}} = 1 - (1 - R)^n \] where \( n = 3, R = .7 \): \[ R_{1,\text{parallel}} = 1 - (1 - .7)^3 = 1 - (.3)^3 = 1 - .027 = .973 \] **System reliability:** \[ R_{\text{system, a}} = R_{1,\text{parallel}} \times R_2 = .973 \times .95 = .924 \] --- ## **Step 3: Option (b) - Replace Both by Two in Parallel Each** Both components are replaced by two parallel subsystems. **Reliability for two parallel components:** \[ R_{\text{parallel,2}} = 1 - (1 - R)^2 \] - For component 1: \( R_1 = .7 \) \[ R_{1,\text{parallel,2}} = 1 - (1 - .7)^2 = 1 - (.3)^2 = 1 - .09 = .91 \] - For component 2: \( R_2 = .95 \) \[ R_{2,\text{parallel,2}} = 1 - (1 - .95)^2 = 1 - (.05)^2 = 1 - .0025 = .9975 \] **System reliability:** \[ R_{\text{system, b}} = R_{1,\text{parallel,2}} \times R_{2,\text{parallel,2}} = .91 \times .9975 = .9077 \] --- ## **Step 4: Conclusion** - **Option (a):** \( R_{\text{system, a}} = .924 \) - **Option (b):** \( R_{\text{system, b}} = .9077 \) ### **Final Answer:** > **It is preferable to replace component 1 by three components in parallel (Option a), as this yields a higher system reliability of .924 compared to .9077 for Option b.**