Question

A pharmaceutical company receives large shipments of ibuprofen tablets and uses an acceptance sampling plan. This plan randomly selects and tests 24 tablets, then accepts the whole batch if there is at most one that doesn't meet the required specifications. What is the probability that this whole shipment will be accepted if a particular shipment of thousands of ibuprofen tablets actually has a 5% rate of defects?

(Report answer as a decimal value accurate to four decimal places.)
P(accept shipment) =

the answer is not 0.8335

Answer 1

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.

1. **Calculate \( P(X = 0) \)**:
- \( k = 0 \)
- \( P(X = 0) = \binom{24}{0} (0.05)^0 (0.95)^{24} \)
- \( P(X = 0) = 1 \cdot 1 \cdot (0.95)^{24} \)
- \( P(X = 0) \approx 0.3769 \)

2. **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 \( (0.95)^{23} \approx 0.3960 \))
- \( P(X = 1) \approx 24 \cdot 0.05 \cdot 0.3960 \approx 0.4752 \)

3. **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:

\[
\boxed{0.8521}
\]