This problem is a case of binomial probability calculation since we're looking at two possible outcomes for an event - success or failure of a component.
First, we establish a couple of elements:
1. The total number of components in the subsystem is 4.
2. Each component has a 0.2 probability of failing in less than 1000 hours. Hence, the probability of a component successfully operating for more than 1000 hours is 1 - 0.2 = 0.8.
This is because the sum of the probabilities of all possible outcomes for an event always equals 1.
Now, we know that the subsystem will continue to operate if at least 2 out of 4 components are functioning. This includes scenarios where 2, 3 or all 4 components are working. Hence, our task is to calculate these three probabilities and then add them to get the total probability of the subsystem working.
1. The probability that exactly 2 components are working can be calculated as the number of ways to choose 2 out of 4 components (which is given by the binomial coefficient 4 choose 2) multiplied by the success probability (0.8) to the power of the number of successes (2) and then multiplied by the failure probability (0.2) to the power of the number of failures (2). The calculated value is approximately 0.1536.
2. For exactly 3 working components, the method is largely the same. The binomial coefficient is 4 choose 3, the number of successes increases to 3, and the number of failures reduces to 1. The probability is approximately 0.4096.
3. For all (or 4) components working, all must be successes, hence the calculation is just the success probability to the power of 4. This gives an approximate value of 0.4096.
These three situations combined will cover all cases when the subsystem will keep working. So, to find the total probability, we add all three probabilities:
0.1536 (2 components working) + 0.4096 (3 components working) + 0.4096 (4 components working) = 0.9728
Hence, the probability that the subsystem operates longer than 1000 hours, regardless of failures, is about 0.9728 or 97.28%.