This is a problem of selecting components from a total population, where some are defective, and others function properly. Since the selection is random and without replacement, we can solve this using the concepts of combinatorics and hypergeometric distribution.

Given:

• Total number of components $N = 15$
• Number of defective components $D = 4$
• Number of functional components $F = 15 - 4 = 11$
• Number of components selected $n = 8$

We'll use the hypergeometric probability formula for our calculations:

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

where:

• $X$ is the number of defective components in the selection,
• $\binom{a}{b}$ is the binomial coefficient (combinations) "a choose b."

Let's solve each part.

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Part (i): The probability that all selected components function properly.

This means none of the selected components are defective, i.e., $k = 0$ defective components.

$P(X = 0) = \frac{\binom{4}{0} \binom{11}{8}}{\binom{15}{8}}$

We can now calculate this.

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Part (ii): The probability that exactly 3 components are defective.

Here, we need to calculate the probability that exactly 3 defective components are selected, i.e., $k = 3$.

$P(X = 3) = \frac{\binom{4}{3} \binom{11}{5}}{\binom{15}{8}}$

We will compute this next.

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Part (iii): The probability that at least 1 component is defective.

The probability of at least one defective component is the complement of the event where all components function properly. So:

$P(\text{at least 1 defective}) = 1 - P(X = 0)$

We'll now calculate all the values using combinations.### The results are:

(i) The probability that all selected components function properly: $P(X = 0) = 0.0256 \text{ (approximately)}$

(ii) The probability that exactly 3 components are defective: $P(X = 3) = 0.2872 \text{ (approximately)}$

(iii) The probability that at least 1 component is defective: $P(\text{at least 1 defective}) = 0.9744 \text{ (approximately)}$

Let me know if you need any further details or explanations.

Relative questions:

1. How would the probabilities change if the total number of components were 20?
2. What would be the probability of selecting exactly 2 defective components?
3. How can the hypergeometric distribution be applied in quality control processes?
4. How would the probabilities change if you selected only 5 components instead of 8?
5. Can the binomial distribution be used here, and under what conditions?

Tip:

When selecting without replacement, the hypergeometric distribution is ideal for calculating probabilities, as it considers the changing population size after each selection.