This is a problem involving binomial probability, where we want to find the probability that a shipment is accepted based on the acceptance sampling plan.

Given:

• The total number of tablets in the shipment is 6000, but only 51 tablets are randomly selected for testing.
• The defect rate in the shipment is 4% (or 0.04).
• The acceptance criterion is that the shipment is accepted if 0 or 1 tablet in the sample fails to meet the specifications.

The process can be modeled as a binomial distribution because:

• Each tablet is either defective or not defective (two outcomes).
• The probability of selecting a defective tablet is 0.04, and the probability of selecting a non-defective tablet is 0.96.
• The number of trials (tablets selected) is 51.

Let $X$ be the random variable representing the number of defective tablets in the sample. We want to find $P(X \leq 1)$, the probability that 0 or 1 defective tablet is found in the sample of 51.

The binomial distribution is defined as: $P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$ where:

• $n = 51$ (number of trials, or tablets selected),
• $p = 0.04$ (probability of a tablet being defective),
• $k$ is the number of defective tablets (which we want to be 0 or 1),
• $\binom{n}{k}$ is the binomial coefficient.

Calculate the probability:

We need to calculate:

• $P(X = 0)$ — the probability that no tablets are defective.
• $P(X = 1)$ — the probability that exactly one tablet is defective.
• Then, we sum these probabilities to get $P(X \leq 1) = P(X = 0) + P(X = 1)$.

Step 1: Compute $P(X = 0)$

$P(X = 0) = \binom{51}{0} (0.04)^0 (0.96)^{51} = 1 \cdot 1 \cdot (0.96)^{51}$ $P(X = 0) \approx (0.96)^{51} \approx 0.1297$

Step 2: Compute $P(X = 1)$

$P(X = 1) = \binom{51}{1} (0.04)^1 (0.96)^{50} = 51 \cdot 0.04 \cdot (0.96)^{50}$ $P(X = 1) \approx 51 \cdot 0.04 \cdot 0.1358 \approx 0.2767$

Step 3: Calculate $P(X \leq 1)$

Now, we can find the total probability: $P(X \leq 1) = P(X = 0) + P(X = 1)$ $P(X \leq 1) = 0.1297 + 0.2767 = 0.4064$

Final answer:

The probability that the whole shipment will be accepted is approximately 0.4064 or 40.64%.

Will almost all such shipments be accepted, or will many be rejected?

Since the probability of acceptance is 40.64%, this means that only a little less than half of the shipments will be accepted. Therefore, many shipments will be rejected because the acceptance probability is less than 50%.

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Would you like me to explain any part of this process in more detail? Here are some additional related questions:

1. How would the acceptance probability change if the defect rate were higher or lower?
2. What other types of probability distributions could model a similar situation?
3. How can the acceptance sampling plan be adjusted to increase the probability of acceptance?
4. What is the binomial coefficient, and how do you compute it manually?
5. How can you use a normal approximation for large binomial distributions?

Tip: When dealing with large binomial problems, you can sometimes use a normal approximation if $n$ is large and $p$ is not too close to 0 or 1, which simplifies the calculations.