This problem involves a hypergeometric distribution because we are selecting a sample without replacement from a finite population.

The hypergeometric probability formula is:

$P(X = k) = \frac{\binom{D}{k} \binom{N-D}{n-k}}{\binom{N}{n}}$

Where:

• $N$ is the population size (2299 calculators),
• $D$ is the number of defective calculators (10),
• $n$ is the sample size (125 calculators),
• $k$ is the number of defective calculators in the sample.

We are looking for the probability of having at most 2 defective calculators, so we need to calculate:

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

Let's calculate this step by step.

The probability of having at most 2 defective calculators in a sample of 125, selected from a batch of 2299 calculators where there are 10 defective ones, is approximately 0.9858.

Would you like further details or clarifications?

Here are 5 related questions to consider:

1. What if the sample size was larger than 125?
2. How does the probability change if the number of defective calculators increases?
3. Can we use a different distribution for this problem? Why or why not?
4. What is the probability of having exactly 1 defective unit in the sample?
5. How would the solution change if we were sampling with replacement?

Tip: Hypergeometric distributions are ideal for sampling without replacement when the population size is not too large compared to the sample size.