To solve this problem, we can model the failure of the components using the binomial probability formula, since we have a fixed number of independent trials (the six components) and each trial has two possible outcomes (failure or success). The binomial probability formula is given by:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
Where:
- \( n \) is the total number of trials (in this case, 6 components),
- \( k \) is the number of successful trials we are interested in (in this case, 3 failures),
- \( p \) is the probability of success on an individual trial (in this case, the probability of a component failing, which is 0.05),
- \( (1-p) \) is the probability of failure on an individual trial (in this case, the probability of a component not failing, which is 0.95).
Substituting the values into the formula:
- \( n = 6 \)
- \( k = 3 \)
- \( p = 0.05 \)
- \( 1-p = 0.95 \)
First, we calculate the binomial coefficient \( \binom{n}{k} \):
\[ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]
Now we can calculate the probability:
\[ P(X = 3) = \binom{6}{3} (0.05)^3 (0.95)^{6-3} \] \[ = 20 \times (0.05)^3 \times (0.95)^3 \] \[ = 20 \times (0.000125) \times (0.857375) \] \[ = 20 \times 0.0001076298828125 \] \[ \approx 0.0021525984 \]
Rounding this to the nearest thousandth, we get:
\[ \approx 0.002 \]
Thus, the probability that exactly three components will fail is approximately 0.002.
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