To find the probability \( P \) that the shipment will be accepted when the defect rate is 5%, we can model this scenario using the binomial distribution.
We are interested in the situation where we select 24 tablets (the sample size \( n \)) and at most one defect is allowed for the shipment to be accepted (which translates to \( k = 0 \) or \( k = 1 \)).
The parameters are as follows:
- Probability of a defect \( p = 0.05 \) (5%)
- Number of trials (tablets tested) \( n = 24 \)
The probability mass function of the binomial distribution is given by:
\[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \]
where
- \( \binom{n}{k} \) is the binomial coefficient "n choose k".
- \( p \) is the probability of success (finding a defect).
- \( (1 - p) \) is the probability of no defect.
We need to calculate \( P(X \leq 1) = P(X = 0) + P(X = 1) \).
-
Calculate \( P(X = 0) \): \[ P(X = 0) = \binom{24}{0} (0.05)^0 (0.95)^{24} = 1 \cdot 1 \cdot (0.95)^{24} \] Calculating \( (0.95)^{24} \): \[ (0.95)^{24} \approx 0.3686 \]
-
Calculate \( P(X = 1) \): \[ P(X = 1) = \binom{24}{1} (0.05)^1 (0.95)^{23} = 24 \cdot 0.05 \cdot (0.95)^{23} \] Calculating \( (0.95)^{23} \): \[ (0.95)^{23} \approx 0.3874 \] Then, \[ P(X = 1) = 24 \cdot 0.05 \cdot 0.3874 \approx 24 \cdot 0.01937 \approx 0.4649 \]
-
Combine the two probabilities: \[ P(X \leq 1) = P(X = 0) + P(X = 1) \approx 0.3686 + 0.4649 = 0.8335 \]
Therefore, the probability that the whole shipment will be accepted is:
\[ P(accept shipment) \approx 0.8335 \]
Reporting the answer to four decimal places, we find: \[ \boxed{0.8335} \]
1 answer