Question

A burglar alarm system has six fail-safe components. The probability of each failing is 0.05. Find the probability that fewer than three will fail. Round to the nearest thousandth.

Answer 1

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 \( P(X = 2) \).

1. **Calculate \( P(X = 0) \)**:
\[
P(X = 0) = \binom{6}{0} (0.05)^0 (0.95)^6 = 1 \cdot 1 \cdot (0.95)^6 \approx 0.7351
\]

2. **Calculate \( P(X = 1) \)**:
\[
P(X = 1) = \binom{6}{1} (0.05)^1 (0.95)^5 = 6 \cdot 0.05 \cdot (0.95)^5 \approx 6 \cdot 0.05 \cdot 0.7738 \approx 0.2335
\]

3. **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:

\[
\boxed{0.999}
\]