To solve this problem, we need to use the hypergeometric distribution since the sample is drawn without replacement from a finite population.

The hypergeometric distribution describes the probability of $k$ successes (defective batteries) in a sample of size $n$, drawn from a population of size $N$, containing $K$ successes (defective batteries).

Given information:

• Total number of batteries in the lot ($N$) = 24
• Number of defective batteries in the lot ($K$) = 6
• Number of batteries inspected in the sample ($n$) = 3
• We need to find the probability that exactly 1 defective battery ($k = 1$) is in the sample.

The formula for the hypergeometric probability is:

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

Where:

• $\binom{K}{k}$ is the number of ways to choose $k$ defective batteries from $K$ defective batteries.
• $\binom{N - K}{n - k}$ is the number of ways to choose the remaining $n - k$ non-defective batteries from the $N - K$ non-defective batteries.
• $\binom{N}{n}$ is the total number of ways to choose $n$ batteries from the population of $N$.

Plugging in the values:

• $K = 6$
• $N = 24$
• $n = 3$
• $k = 1$

First, calculate the binomial coefficients:

$\binom{6}{1} = 6, \quad \binom{18}{2} = \frac{18 \times 17}{2} = 153, \quad \binom{24}{3} = \frac{24 \times 23 \times 22}{6} = 2024$

Now, use these values in the formula:

$P(X = 1) = \frac{6 \times 153}{2024} = \frac{918}{2024} \approx 0.4536$

Answer:

The probability that the sample will contain exactly one defective battery is 0.4536.

Would you like to explore more examples or details about the hypergeometric distribution? Here's a list of related questions:

1. What is the probability of getting 0 defective batteries in the sample?
2. How does the probability change if there are 8 defective batteries in the lot?
3. What is the difference between the binomial and hypergeometric distributions?
4. How do you calculate probabilities with replacement versus without replacement?
5. What are some real-world examples where the hypergeometric distribution applies?

Tip: When dealing with probabilities without replacement, always consider the hypergeometric distribution instead of binomial, as the population is finite and the events are not independent.