To solve this problem, we will use the binomial probability formula since we are dealing with a fixed number of trials (in this case, 10 tablets), and each trial can either result in a defective or non-defective tablet.

The binomial probability formula is:

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

Where:

• $P(X = k)$ is the probability of exactly $k$ successes (in this case, defective tablets).
• $\binom{n}{k}$ is the binomial coefficient (the number of ways to choose $k$ successes from $n$ trials).
• $n$ is the total number of trials (in this case, 10 tablets).
• $k$ is the number of successes (in this case, 8 defective tablets).
• $p$ is the probability of success on a single trial (in this case, 0.12, the defect rate).
• $1 - p$ is the probability of failure on a single trial (in this case, 0.88, the non-defective rate).

Let's plug in the values:

• $n = 10$
• $k = 8$
• $p = 0.12$

Now calculate the probability.The probability that exactly 8 out of 10 tablets produced will be defective, given a defect rate of 12%, is approximately $1.50 \times 10^{-6}$, or 0.0000015.

Would you like more details on the calculation? Here are some related questions:

1. What is the probability that 0 tablets are defective?
2. What is the probability that at least 2 tablets are defective?
3. What is the expected number of defective tablets in a batch of 10?
4. How does the defect rate change if the probability of defect increases to 20%?
5. What is the cumulative probability of producing 7 or fewer defective tablets?

Tip: For small probabilities like this, increasing the sample size might yield more meaningful insights.