Among 15 electrical components exactly 4 are known not to function properly. If 8 components are randomly selected, find the following probabilities:
(i) The probability that all selected components function properly.
(ii) The probability that exactly 3 are defective.
(iii) The probability that at least 1 component is defective.

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.

---

### 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.

---

### 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.

---

### 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.

Among 15 electrical components exactly 4 are known not to function properly. If 8 components are randomly selected, find the following probabilities: (i) The probability that all selected components function properly. (ii) The probability that exactly 3 are defective. (iii) The probability that at least 1 component is defective.

Solve a probability problem involving 15 electrical components, 4 of which are defective. Learn how to calculate the probability of selecting only functional components, exactly 3 defective ones, and at least 1 defective component using hypergeometric distribution. Suitable for college-level probability or advanced high school students.

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