Suppose components A1, A2, B1, B2 and C1 operate independently in the system shown below. Assume that all the components in the system start operating at the same time and a component does not work again once it fails. The probability of functioning for each component is 0.92. The entire system works if A1 or A2 works, and B1 or B2 works, and C1 works. Find the probability that the entire system works. Show your reasoning.

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∻Solution by Steps∻ ‖ step 1 ⋮ The probability that component A1 or A2 works is given by the formula: $ P(A1 \cup A2) = 1 - P(A1^c) \cdot P(A2^c) $. Since the probability of functioning for each component is 0.92, the probability of failure for each component is $ P(A1^c) = P(A2^c) = 1 - 0.92 = 0.08 $. Thus, we have: ‖ ‖ step 2 ⋮ Calculating the probability that at least one of A1 or A2 works: $$ P(A1 \cup A2) = 1 - (0.08 \cdot 0.08) = 1 - 0.0064 = 0.9936 $$ ‖ ‖ step 3 ⋮ Similarly, the probability that component B1 or B2 works is calculated in the same way: $$ P(B1 \cup B2) = 1 - P(B1^c) \cdot P(B2^c) = 1 - (0.08 \cdot 0.08) = 0.9936 $$ ‖ ‖ step 4 ⋮ The probability that component C1 works is simply $ P(C1) = 0.92 $. ‖ ‖ step 5 ⋮ The entire system works if A1 or A2 works, and B1 or B2 works, and C1 works. Therefore, the overall probability is: $$ P(\text{System works}) = P(A1 \cup A2) \cdot P(B1 \cup B2) \cdot P(C1) = 0.9936 \cdot 0.9936 \cdot 0.92 $$ ‖ ‖ step 6 ⋮ Calculating the final probability: $$ P(\text{System works}) = 0.9936 \cdot 0.9936 \cdot 0.92 \approx 0.846 $$ ‖ ∻Answer∻ ⚹ The probability that the entire system works is approximately 0.846 ⚹ ∻Key Concept∻ ⚹ The concept of independent events and their probabilities in a system ⚹ ∻Explanation∻ ⚹ The answer is derived from calculating the probabilities of independent components working together to ensure the entire system functions. Each component's failure probability affects the overall system's reliability. ⚹◊What is the probability that at least one of the components A1 or A2 functions?⍭ Generate me a similar question◊

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