a missile guidance system has five fail-safe components. the probability of each failing is 0.05. find these probabilities.
a) Exactly 2 will fail
b) More than 2 will fail
c) All will fail.
d) Compare the answer for parts a, b, c, and explain why these results are reasonable
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Answer
To find the probabilities for each scenario, we can use the binomial probability formula, which is given by:
P(X=k) = nCk * p^k * (1-p)^(n-k)
Where:
P(X=k) is the probability of exactly k components failing
n is the total number of components (5 in this case)
k is the number of components failing
p is the probability of an individual component failing (0.05 in this case)
(1-p) is the probability of an individual component not failing (0.95 in this case)
nCk is the combination formula, which represents the number of ways to choose k elements from n elements.
Let's calculate the probabilities for each scenario:
a) Exactly 2 will fail:
P(X=2) = 5C2 * 0.05^2 * (1-0.05)^(5-2)
= 10 * 0.0025 * 0.857375
≈ 0.021
b) More than 2 will fail:
P(X > 2) = P(X=3) + P(X=4) + P(X=5)
To calculate each term, we use the same formula as in part a.
P(X=3) = 5C3 * 0.05^3 * (1-0.05)^(5-3)
= 10 * 0.000125 * 0.816375
≈ 0.00122
P(X=4) = 5C4 * 0.05^4 * (1-0.05)^(5-4)
= 5 * 0.00000625 * 0.7785
≈ 0.000024
P(X=5) = 5C5 * 0.05^5 * (1-0.05)^(5-5)
= 1 * 0.0000003125 * 0.7371
≈ 0.00000023
P(X > 2) ≈ P(X=3) + P(X=4) + P(X=5) ≈ 0.00122 + 0.000024 + 0.00000023 ≈ 0.00127
c) All will fail:
P(X=5) = 5C5 * 0.05^5 * (1-0.05)^(5-5)
= 1 * 0.0000003125 * 1
≈ 0.0000003125
d) Comparing the answers:
The probability for exactly 2 failing is higher than the probability for more than 2 failing, which is reasonable since fewer components failing is generally more likely.
Similarly, the probability for all components failing is extremely low compared to the other scenarios, indicating that having all components fail is highly unlikely.
These results are reasonable because the probability of failure for each component is relatively low (0.05), and the more components that need to fail for a specific scenario, the lower the probability becomes. Additionally, the probabilities obtained satisfy the condition that the sum of all probabilities for each possible outcome is equal to 1.
To find the probabilities for the given situations, we will use the binomial probability formula:
P(X = k) = nCk * p^k * (1 - p)^(n - k)
Where:
- P(X = k) is the probability of X (the number of failures) being equal to k.
- nCk represents the number of ways to choose k failures out of n components.
- p is the probability of a single component failing.
- (1 - p) represents the probability of a single component not failing.
- n is the total number of components (in this case, 5).
Let's solve each part of the question step by step:
a) Exactly 2 will fail:
Using the formula, we have:
P(X = 2) = 5C2 * 0.05^2 * (1 - 0.05)^(5 - 2)
= (10) * (0.0025) * (0.95)^3
≈ 0.0706
Therefore, the probability that exactly two components will fail is approximately 0.0706.
b) More than 2 will fail:
To find this probability, we need to calculate the probabilities of 3, 4, and 5 failures and sum them up:
P(X > 2) = P(X = 3) + P(X = 4) + P(X = 5)
Using the formula for each of these situations, we get:
P(X = 3) = 5C3 * 0.05^3 * (1 - 0.05)^(5 - 3)
= (10) * (0.000125) * (0.95)^2
≈ 0.002375
P(X = 4) = 5C4 * 0.05^4 * (1 - 0.05)^(5 - 4)
= (5) * (6.25e-06) * (0.95)^1
≈ 1.486e-05
P(X = 5) = 5C5 * 0.05^5 * (1 - 0.05)^(5 - 5)
= (1) * (3.125e-07) * (1)
≈ 3.125e-07
P(X > 2) ≈ 0.002375 + 1.486e-05 + 3.125e-07
≈ 0.002418
Therefore, the probability of more than two components failing is approximately 0.002418.
c) All will fail:
P(X = 5) = 5C5 * 0.05^5 * (1 - 0.05)^(5 - 5)
= (1) * (3.125e-07) * (1)
≈ 3.125e-07
Therefore, the probability that all five components will fail is approximately 3.125e-07.
d) Comparing the results:
The reasonableness of the results can be understood by considering the probabilities assigned to different outcomes. In this case, the probabilities decrease as we move from more extreme outcomes (e.g., all fail or none fail) to less extreme outcomes (e.g., exactly two failing or more than two failing).
This is reasonable because the probability of a single component failing is relatively small (0.05), indicating that it is unlikely for all components to fail. Likewise, it is more likely for a few components to fail compared to none or all, which is reflected in the probabilities calculated. Therefore, the results align with our expectations based on the given probability of each component failing.