To determine the probability of accepting the shipment, we need to calculate the probability that all 44 modems selected from the shipment are not defective. This involves the hypergeometric distribution since we're drawing without replacement.

Hypergeometric Distribution Formula:

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

Where:

• $N$ = Total number of items = 2727
• $D$ = Number of defective items = 22
• $n$ = Number of selected items = 44
• $k$ = Number of defective items in the sample = 0 (since we want all modems to work)

The probability of selecting 44 working modems (i.e., 0 defective modems) is given by the ratio of the number of ways to choose 44 non-defective modems from 2705 (non-defective modems) to the total number of ways to choose 44 modems from 2727.

Steps:

1. The total number of ways to select 44 modems from 2727 is $\binom{2727}{44}$.
2. The number of ways to select 44 non-defective modems from the 2705 non-defective modems is $\binom{2705}{44}$.

Thus, the probability of accepting the shipment is:

$P(\text{accept}) = \frac{\binom{2705}{44}}{\binom{2727}{44}}$

This is quite difficult to compute manually, but it can be solved using hypergeometric probability tools or approximations. I will now compute the value.The probability of accepting the shipment is approximately 0.6982 (rounded to four decimal places).

Would you like further details or have any questions?

Here are some related questions you might find useful:

1. What is the probability of rejecting the shipment?
2. How would the probability change if 50 modems were tested instead of 44?
3. What is the probability of finding exactly 1 defective modem in the 44 tested?
4. How does the hypergeometric distribution differ from the binomial distribution?
5. How would increasing the number of defective modems affect the probability of acceptance?

Tip: The hypergeometric distribution is ideal when sampling without replacement, while the binomial distribution assumes sampling with replacement.