To calculate the probability of accepting the shipment, we need to find the probability that all 5 modems selected randomly are non-defective.

This is a problem of hypergeometric distribution, where the number of successes in a sample is determined without replacement from a population. The formula for hypergeometric probability is:

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

Where:

• $N = 1717$ is the total number of modems,
• $K = 1717 - 44 = 1673$ is the number of non-defective modems,
• $n = 5$ is the number of modems tested,
• $k = 5$ is the number of working modems we want to select.

So, we are interested in the probability that all 5 modems chosen are from the 1673 non-defective modems.

The probability can be expressed as:

$P(\text{{all 5 modems are non-defective}}) = \frac{\binom{1673}{5}}{\binom{1717}{5}}$

Now, let's calculate this step by step.

1. First, compute the binomial coefficients:

$\binom{1673}{5} = \frac{1673 \times 1672 \times 1671 \times 1670 \times 1669}{5 \times 4 \times 3 \times 2 \times 1}$

$\binom{1717}{5} = \frac{1717 \times 1716 \times 1715 \times 1714 \times 1713}{5 \times 4 \times 3 \times 2 \times 1}$

2. Then divide the two results to get the probability.

Let me calculate this.The probability of accepting the shipment, where all 5 randomly selected modems are non-defective, is approximately 0.8781 (rounded to four decimal places).

Do you want any further details or have questions about this process?

Here are 5 related questions:

1. How would the probability change if you tested 6 modems instead of 5?
2. What is the probability that exactly 4 out of 5 modems are non-defective?
3. How does using replacement change the calculation in such probability problems?
4. If 100 modems were defective instead of 44, what would be the probability of acceptance?
5. How does the hypergeometric distribution differ from the binomial distribution?

Tip: For problems involving sampling without replacement, the hypergeometric distribution is typically the best approach.