answer:To find the probability that at least one component will fail, it's easier to first calculate the probability that none of the components fail and then subtract this value from 1. The probability that all components function properly (do not fail) is the product of their individual probabilities of functioning properly: P(function properly) = P(component 1 functions properly) * P(component 2 functions properly) * ... * P(component 5 functions properly) = (1 - 0.02) * (1 - 0.03) * (1 - 0.01) * (1 - 0.02) * (1 - 0.04) = 0.98 * 0.97 * 0.99 * 0.98 * 0.96 ≈ 0.904952 Now, we can find the probability that at least one component will fail: P(at least one fails) = 1 - P(all function properly) = 1 - 0.904952 ≈ 0.095048 So, there is approximately a 9.50% chance that at least one component will fail.,To find the probability that at least one component will fail, we can first find the probability that none of the components fail and then subtract that from 1. The probability that a component does not fail is equal to 1 minus its failure probability. So, for each component, we have: Component 1: 1 - 0.02 = 0.98 Component 2: 1 - 0.03 = 0.97 Component 3: 1 - 0.01 = 0.99 Component 4: 1 - 0.02 = 0.98 Component 5: 1 - 0.04 = 0.96 Now, we find the probability that none of the components fail by multiplying the probabilities of each component not failing: P(None fail) = 0.98 * 0.97 * 0.99 * 0.98 * 0.96 ≈ 0.8853 Finally, we find the probability that at least one component will fail by subtracting the probability that none of them fail from 1: P(At least one fails) = 1 - P(None fail) = 1 - 0.8853 ≈ 0.1147 So, the probability that at least one component will fail is approximately 0.1147 or 11.47%.,To find the probability that at least one component will fail, we can calculate the probability that none of the components fail and then subtract this probability from 1. The probability that each individual component does not fail is given by 1 - its failure probability: 1 - 0.02 = 0.98 (component 1) 1 - 0.03 = 0.97 (component 2) 1 - 0.01 = 0.99 (component 3) 1 - 0.02 = 0.98 (component 4) 1 - 0.04 = 0.96 (component 5) Since the components function independently, the probability that none of them fail is the product of their individual probabilities of not failing: (0.98 * 0.97 * 0.99 * 0.98 * 0.96) Now, we find the complementary probability, which is the probability that at least one component will fail: 1 - (0.98 * 0.97 * 0.99 * 0.98 * 0.96) ≈ 0.177 or 17.7% So, the probability that at least one component will fail is approximately 17.7%.

answer:To find the probability that at least one backup system will be operational, we can first find the probability that all backup systems fail and then subtract that from 1. The probability of a system failure is 0.1, so the probability of a system being operational is 1 - 0.1 = 0.9. Since the backup systems operate independently, the probability of all 5 backup systems failing is (0.1)^5 = 0.00001. Now, we can find the probability that at least one backup system will be operational by subtracting the probability of all systems failing from 1: 1 - 0.00001 = 0.99999 So, the probability that at least one backup system will be operational if the spacecraft is launched is approximately 0.99999 or 99.999%., To find the probability that at least one backup system will be operational, it's easier to first find the probability that all systems fail and then subtract this value from 1. The probability that all five systems fail is (0.1)^5 since they operate independently. Now, we can calculate the probability that at least one system is operational: P(at least one system operational) = 1 - P(all systems fail) = 1 - (0.1)^5 = 1 - 0.000001 = 0.999999 So, the probability that at least one backup system will be operational if the spacecraft is launched is approximately 0.999999 or 99.9999%.,To find the probability that at least one backup system will be operational, we can first find the probability that none of the backup systems will be operational and then subtract that from 1. The probability of a system failure for each backup system is 0.1. Since the systems operate independently, we can multiply their probabilities to find the probability that all of them fail: 0.1 * 0.1 * 0.1 * 0.1 * 0.1 = 0.000001 Now, subtract this probability from 1 to find the probability that at least one backup system will be operational: 1 - 0.000001 = 0.999999 So, there is a 99.9999% chance that at least one backup system will be operational when the spacecraft is launched.

answer:To find the probability of a complete system failure, we can first determine the probability of the system not failing and then subtract that from 1. The probability of a single component not failing is 1 - 0.1 = 0.9. Since there are four components, the probability of none of them failing is (0.9)^4 = 0.6561. Now, to find the probability of at least one component failing, we can subtract the probability of none of them failing from 1: 1 - 0.6561 = 0.3439 So, the probability of a complete system failure due to at least one component failing is 0.3439 or 34.39%., To solve this problem, we'll use the principle of inclusion-exclusion for probabilities. Let A1, A2, A3, and A4 be the events that each of the four components fail, respectively. We want to find the probability P(A1 ∪ A2 ∪ A3 ∪ A4), which is the probability that at least one component fails. First, let's calculate the individual probabilities: P(A1) = P(A2) = P(A3) = P(A4) = 0.1 Now, we'll calculate the probabilities of each pair of events: P(A1 ∩ A2) = P(A1) * P(A2) = 0.1 * 0.1 = 0.01 Similarly, P(A1 ∩ A3) = P(A1 ∩ A4) = P(A2 ∩ A3) = P(A2 ∩ A4) = P(A3 ∩ A4) = 0.01 Next, we'll calculate the probabilities of each triplet of events: P(A1 ∩ A2 ∩ A3) = P(A1) * P(A2) * P(A3) = 0.1^3 = 0.001 Similarly, P(A1 ∩ A2 ∩ A4) = P(A1 ∩ A3 ∩ A4) = P(A2 ∩ A3 ∩ A4) = 0.001 Finally, we'll calculate the probability of all four events happening together: P(A1 ∩ A2 ∩ A3 ∩ A4) = P(A1) * P(A2) * P(A3) * P(A4) = 0.1^4 = 0.0001 Now, we can apply the principle of inclusion-exclusion: P(A1 ∪ A2 ∪ A3 ∪ A4) = P(A1) + P(A2) + P(A3) + P(A4) - P(A1 ∩ A2) - P(A1 ∩ A3) - P(A1 ∩ A4) - P(A2 ∩ A3) - P(A2 ∩ A4) - P(A3 ∩ A4) + P(A1 ∩ A2 ∩ A3) + P(A1 ∩ A2 ∩ A4) + P(A1 ∩ A3 ∩ A4) - P(A1 ∩ A2 ∩ A3 ∩ A4) Plugging in the values, we get: P(A1 ∪ A2 ∪ A3 ∪ A4) = 0.1 + 0.1 + 0.1 + 0.1 - 4 * 0.01 + 3 * 0.001 - 0.0001 P(A1 ∪ A2 ∪ A3 ∪ A4) ≈ 0.392 So, the probability that a complete system failure will occur due to at least one component failing is approximately 0.392.,To find the probability of at least one component failing, we can first find the probability of none of the components failing and then subtract that from 1. The probability of a component not failing is 1 - 0.1 = 0.9. Since there are four components, the probability of all four components not failing is (0.9)^4 = 0.6561. Now, to find the probability of at least one component failing, we subtract the probability of none of them failing from 1: 1 - 0.6561 = 0.3439. So, the probability of a complete system failure due to at least one component failing is 0.3439 or 34.39%.

answer:To find the probability that at least 2 components will fail, we can first find the probability that 0 or 1 component will fail and then subtract that from 1. Let p = 0.02 be the probability of failure for a single component, and q = 1 - p = 0.98 be the probability of success (no failure) for a single component. The probability that 0 components will fail is simply q^100, since all 100 components must succeed. The probability that exactly 1 component will fail is given by the binomial probability formula: P(X = 1) = C(n, k) * p^k * q^(n-k) where n = 100 (number of components), k = 1 (number of failures), and C(n, k) is the number of combinations of n items taken k at a time. C(n, k) = n! / (k!(n-k)!) C(100, 1) = 100! / (1!(100-1)!) = 100 P(X = 1) = 100 * 0.02^1 * 0.98^99 ≈ 0.2707 Now, we can find the probability that at least 2 components will fail: P(X ≥ 2) = 1 - P(X = 0) - P(X = 1) P(X ≥ 2) = 1 - 0.98^100 - 0.2707 ≈ 0.9323 So, the probability that at least 2 components will fail due to this type of failure is approximately 0.9323 or 93.23%., To solve this problem, it's easier to first calculate the probability that no components or only one component fails, and then subtract that value from 1 to find the probability that at least two components fail. The probability of a single component not failing is 1 - 0.02 = 0.98. Since the components are independent, the probability of all components not failing is (0.98)^100 ≈ 0.3697. Now, we need to calculate the probability that only one component fails while the others do not. There are 100 ways for one component to fail and the rest to succeed: P(exactly one failure) = C(100, 1) * (0.02)^1 * (0.98)^99 ≈ 100 * 0.02 * 0.3697^99 ≈ 0.3704 Finally, to find the probability that at least two components fail, we subtract the probabilities of no failures and exactly one failure from 1: P(at least two failures) = 1 - P(no failures) - P(exactly one failure) = 1 - 0.3697 - 0.3704 ≈ 0.2599 So, the probability that at least two components will fail due to this type of failure is approximately 0.2599 or 25.99%.,To find the probability that at least 2 components will fail, we can first find the probability that 0 or 1 component will fail and then subtract this value from 1. Let X be the number of failed components. We can model X as a binomial random variable with n = 100 trials (components) and a probability of success (failure) p = 0.02. p(X = k) = C(n, k) * p^k * (1-p)^(n-k) where C(n, k) is the number of combinations of n items taken k at a time. p(X = 0) = C(100, 0) * 0.02^0 * (1-0.02)^(100-0) = 1 * 1 * (0.98)^100 ≈ 0.13206 p(X = 1) = C(100, 1) * 0.02^1 * (1-0.02)^(100-1) ≈ 0.31109 Now find the complementary probability, which is the probability of 0 or 1 component failing: p(X = 0 or X = 1) = p(X = 0) + p(X = 1) ≈ 0.13206 + 0.31109 ≈ 0.44315 Finally, subtract this value from 1 to find the probability of at least 2 components failing: p(X ≥ 2) = 1 - p(X = 0 or X = 1) ≈ 1 - 0.44315 ≈ 0.55685 So, the probability that at least 2 components will fail due to this type of failure is approximately 0.55685.