To solve this problem, we need to compute the probability that the shipment is rejected, which happens if 1 or more defective components are found in a random sample of 44 components out of the total 170 components.

Key Information:

• Total components: $170$
• Defective components: $44$
• Number of components sampled: $44$
• The shipment is rejected if 1 or more defective components are found in the sample.

This is a hypergeometric probability problem because we are selecting items without replacement from a finite population (170 components, with 44 being defective).

The probability of rejecting the shipment is the complement of the probability that none of the components in the sample are defective. In other words, we first calculate the probability of selecting 0 defective components in the sample of 44, then subtract this from 1.

Step 1: Probability of selecting 0 defective components

To compute this, we use the hypergeometric probability formula:

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

Where:

• $N = 170$ is the total number of components,
• $D = 44$ is the number of defective components,
• $n = 44$ is the number of components sampled,
• $k = 0$ is the number of defective components selected (since we are calculating the probability of 0 defects),
• $\binom{a}{b}$ is the combination function (number of ways to choose $b$ objects from $a$).

So, we need:

$P(0 \text{ defects}) = \frac{\binom{44}{0} \binom{170 - 44}{44}}{\binom{170}{44}}$

Step 2: Subtract from 1

The probability of rejecting the shipment (1 or more defective components) is:

$P(\text{rejected}) = 1 - P(0 \text{ defects})$

Step 3: Calculation

Let me calculate this.The probability that the shipment is rejected, which occurs if 1 or more defective components are found, is approximately $0.9999998$.

This means that there is an almost certain chance that the shipment will be rejected under these conditions.

Do you need further explanation or details?

Here are five related questions you might find interesting:

1. How would the probability change if fewer components were defective?
2. What is the probability of exactly 1 defective component being in the sample?
3. How does the hypergeometric distribution differ from a binomial distribution?
4. How would the result differ if we sampled only 20 components?
5. What is the expected number of defective components in the sample?

Tip: The hypergeometric distribution is useful when sampling without replacement, while the binomial distribution applies to independent trials with replacement.