for a parallel structure of identical components, the system can succeed if at least 1 of the components succeeds. Assume that components fail independently of each other and that each component has a 0.19 probability of failure.
A) unusual to see 1 component fail?
B) unusual to see 2 components fail?
C)
depends what you mean by "unusual"
prob that 1 will fail = C(2,1)(.19)(.81) = .3078
prob that 2 will fail = ...... you try it, let me know how you did it
To determine whether it is unusual to see a certain number of components fail in a parallel structure, we can use the concept of probability and compare it to a threshold value.
In this case, each component has a 0.19 probability of failure and we want to find out the probability of different numbers of components failing.
A) To determine whether it is unusual to see 1 component fail, we can calculate the probability of exactly one component failing. In a parallel structure, if at least 1 component succeeds, the whole system succeeds. To find the probability of exactly one component failing, we can use the binomial probability formula:
P(X = k) = nCk * p^k * (1 - p)^(n - k)
where n is the total number of components (which is not specified in the question), k is the number of components failing, and p is the probability of failure for each component.
Let's assume there are 10 components in the system:
P(1 component fails) = 10C1 * 0.19^1 * (1 - 0.19)^(10 - 1)
= 10 * 0.19 * 0.81^9
≈ 0.257
Therefore, there is approximately a 0.257 (or 25.7%) probability of seeing exactly one component fail. Since this probability is not extremely low or extremely high, it would not be considered unusual to see 1 component fail.
B) To determine whether it is unusual to see 2 components fail, we can calculate the probability of exactly two components failing:
P(2 components fail) = 10C2 * 0.19^2 * (1 - 0.19)^(10 - 2)
= 45 * 0.19^2 * 0.81^8
≈ 0.214
Therefore, there is approximately a 0.214 (or 21.4%) probability of seeing exactly two components fail. Since this probability is not extremely low or extremely high, it would not be considered unusual to see 2 components fail.
C) It seems like the question was cut off. If you have another part to the question, please provide it, and I'll be happy to assist you further.
To determine whether it is unusual to see a certain number of components fail in a parallel structure, we can use the binomial distribution. The binomial distribution calculates the probability of a certain number of successes (or failures) in a fixed number of independent trials.
In this case, the probability of failure for each component is 0.19. The components in a parallel structure are independent, meaning the failure or success of one component does not affect the others.
A) Unusual to see 1 component fail:
To find the probability of exactly one component failing, we can use the binomial distribution formula. The formula is:
P(x) = (nCx) * (p^x) * (q^(n-x))
Where:
P(x) is the probability of exactly x failures.
n is the number of trials (components in this case).
x is the number of failures.
p is the probability of failure for each component (0.19).
q is the probability of success for each component (1 - 0.19 = 0.81).
Using this formula, we can calculate:
P(1) = (1C1) * (0.19^1) * (0.81^(1-1))
P(1) = 1 * 0.19 * 1 = 0.19
Therefore, the probability of exactly one component failing is 0.19 or 19%. Since this probability is not extremely low or high, it is not considered unusual to see one component fail.
B) Unusual to see 2 components fail:
Using the same formula as above, we can calculate the probability of exactly two components failing:
P(2) = (2C2) * (0.19^2) * (0.81^(2-2))
P(2) = 1 * (0.19^2) * 1 = 0.0361
Therefore, the probability of exactly two components failing is 0.0361 or 3.61%. This probability is relatively low, so it is considered unusual to see two components fail.
C)