To calculate the probability that the whole shipment will be accepted under this acceptance sampling plan, we can use the binomial probability formula.
In this case:
- The number of trials (n) = 24 (the number of tablets tested),
- The number of successes (k) = 0 or 1 (we accept the shipment if there is at most one defective tablet),
- The probability of success (p) = 0.05 (the probability of a defective tablet),
- The probability of failure (q) = 1 - p = 0.95 (the probability of a non-defective tablet).
We can break this down into two parts: the probability of finding exactly 0 defective tablets and the probability of finding exactly 1 defective tablet. Then we will sum these probabilities to find the total probability of acceptance.
The binomial probability formula is given by:
\[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \]
Where \( \binom{n}{k} \) is the number of combinations of n items taken k at a time.
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Calculate \( P(X = 0) \):
- \( k = 0 \)
- \( P(X = 0) = \binom{24}{0} (0.05)^0 (0.95)^{24} \)
- \( P(X = 0) = 1 \cdot 1 \cdot (0.95)^{24} \)
- \( P(X = 0) \approx 0.3769 \)
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Calculate \( P(X = 1) \):
- \( k = 1 \)
- \( P(X = 1) = \binom{24}{1} (0.05)^1 (0.95)^{23} \)
- \( P(X = 1) = 24 \cdot 0.05 \cdot (0.95)^{23} \)
- \( P(X = 1) \approx 24 \cdot 0.05 \cdot 0.3960 \) (using \( (0.95)^{23} \approx 0.3960 \))
- \( P(X = 1) \approx 24 \cdot 0.05 \cdot 0.3960 \approx 0.4752 \)
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Total Acceptance Probability:
Combining both probabilities: \[ P(\text{accept shipment}) = P(X = 0) + P(X = 1) \] \[ P(\text{accept shipment}) \approx 0.3769 + 0.4752 = 0.8521 \]
Thus, the final probability that the whole shipment will be accepted is:
\[ \boxed{0.8521} \]
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