This problem can be modeled using the binomial distribution, which is used to calculate probabilities for experiments with two possible outcomes: success or failure. In this case, the outcomes are "defective tablet" or "non-defective tablet."

Given:

• $n = 25$ (the number of tablets tested),
• $p = 0.12$ (the probability of a defective tablet),
• The shipment is accepted if there are at most 1 defective tablet, i.e., 0 or 1 defective tablets are acceptable.

Binomial Distribution Formula:

The probability of exactly $k$ defective tablets in $n$ trials is given by the binomial probability formula:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

Where:

• $X$ is the number of defective tablets,
• $\binom{n}{k}$ is the binomial coefficient $\frac{n!}{k!(n-k)!}$,
• $p$ is the probability of a defective tablet.

We need to find $P(X \leq 1)$, which is the probability of having 0 or 1 defective tablets. This can be expressed as:

$P(X \leq 1) = P(X = 0) + P(X = 1)$

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

$P(X = 0) = \binom{25}{0} (0.12)^0 (0.88)^{25} = (1)(0.88)^{25}$

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

$P(X = 1) = \binom{25}{1} (0.12)^1 (0.88)^{24} = 25(0.12)(0.88)^{24}$

Now let's compute these values.The probability that the whole shipment will be accepted is approximately $0.1805$ (accurate to four decimal places).

Do you have any questions or need further details?

Here are five related questions for deeper understanding:

1. What is the probability that exactly 2 tablets are defective in this batch?
2. How would the probability change if the defective rate were 15% instead of 12%?
3. What is the expected number of defective tablets in the sample of 25?
4. What is the probability that none of the 25 tablets are defective?
5. How would the acceptance probability change if the sample size were 30 tablets instead of 25?

Tip: In binomial distributions, as the sample size increases, the distribution tends to a normal distribution under certain conditions (Central Limit Theorem).