To find the probability that fewer than three components fail in a burglar alarm system consisting of six independent fail-safe components, we can use the binomial probability formula.
Let \( n = 6 \) (the number of components), and the probability of success (in this case, a component failing) is \( p = 0.05 \). The probability of failure (a component working) is \( q = 1 - p = 0.95 \).
We are interested in finding the probability of fewer than three components failing, which means we want to calculate the following probabilities:
\[ P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \]
The binomial probability formula is given by:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
Where:
- \( \binom{n}{k} \) is the binomial coefficient representing the number of ways to choose \( k \) failures from \( n \) components.
- \( p \) is the probability of failure (0.05).
- \( q \) is the probability of success (0.95).
Now, we can calculate \( P(X = 0) \), \( P(X = 1) \), and \( P(X = 2) \).
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Calculate \( P(X = 0) \): \[ P(X = 0) = \binom{6}{0} (0.05)^0 (0.95)^6 = 1 \cdot 1 \cdot (0.95)^6 \approx 0.7351 \]
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Calculate \( P(X = 1) \): \[ P(X = 1) = \binom{6}{1} (0.05)^1 (0.95)^5 = 6 \cdot 0.05 \cdot (0.95)^5 \approx 6 \cdot 0.05 \cdot 0.7738 \approx 0.2335 \]
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Calculate \( P(X = 2) \): \[ P(X = 2) = \binom{6}{2} (0.05)^2 (0.95)^4 = 15 \cdot (0.05)^2 \cdot (0.95)^4 \approx 15 \cdot 0.0025 \cdot 0.8145 \approx 0.0306 \]
Now we add these probabilities together to find \( P(X < 3) \):
\[ P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \approx 0.7351 + 0.2335 + 0.0306 \approx 0.9992 \]
Now rounding to the nearest thousandth, we have:
\[ \text{Probability that fewer than three will fail} \approx 0.999 \]
Thus, the final answer is:
\[ \boxed{0.999} \]
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