To calculate the probability that the whole shipment will be accepted under the given conditions, we can use the binomial distribution. The acceptance sampling plan states that we accept the shipment if there are at most one defective tablet among the 24 tested tablets.
Let \( p \) be the probability of a tablet being defective, which is given as 5% or 0.05. The probability of a tablet being non-defective is then \( 1 - p = 1 - 0.05 = 0.95 \).
We need to find the probability of having at most one defective tablet (0 or 1 defective tablet) in the sample of 24.
This can be calculated as follows:
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The probability of having exactly \( k \) defective tablets in a sample of \( n \) tablets follows the binomial distribution:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
where \( n = 24 \), \( p = 0.05 \), and \( k \) can be 0 or 1.
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Calculate \( P(X = 0) \):
\[ P(X = 0) = \binom{24}{0} (0.05)^0 (0.95)^{24} = 1 \cdot 1 \cdot (0.95)^{24} \approx 0.2885 \]
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Calculate \( P(X = 1) \):
\[ P(X = 1) = \binom{24}{1} (0.05)^1 (0.95)^{23} = 24 \cdot 0.05 \cdot (0.95)^{23} \approx 24 \cdot 0.05 \cdot 0.3037 \approx 0.3637 \]
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Now, add the probabilities of having 0 or 1 defective tablet:
\[ P(\text{accept shipment}) = P(X = 0) + P(X = 1) \approx 0.2885 + 0.3637 \approx 0.6522 \]
Thus, the probability that the shipment will be accepted is approximately:
\[ \boxed{0.6522} \]
This value is accurate to four decimal places.
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