Given a lot of 10 items containing 3 defective items, a sample of 4 items is drawn at random without replacement. Let the random variable $X$ denote the number of defective items in the sample.

(i) Probability Distribution of $X$

Since the sample is drawn without replacement, $X$ can take values $0$, $1$, $2$, or $3$ because there can be 0, 1, 2, or 3 defective items in the sample. The probability distribution of $X$ is calculated using the hypergeometric distribution formula:

$P(X = x) = \frac{\binom{3}{x} \binom{7}{4-x}}{\binom{10}{4}}$

Where:

• $\binom{3}{x}$ is the number of ways to choose $x$ defective items from 3.
• $\binom{7}{4-x}$ is the number of ways to choose $4-x$ non-defective items from 7.
• $\binom{10}{4}$ is the total number of ways to choose 4 items from 10.

Let's compute these probabilities:

1. $P(X = 0)$: $P(X = 0) = \frac{\binom{3}{0} \binom{7}{4}}{\binom{10}{4}} = \frac{1 \times 35}{210} = \frac{35}{210} = \frac{1}{6} \approx 0.1667$

2. $P(X = 1)$: $P(X = 1) = \frac{\binom{3}{1} \binom{7}{3}}{\binom{10}{4}} = \frac{3 \times 35}{210} = \frac{105}{210} = \frac{1}{2} = 0.5$

3. $P(X = 2)$: $P(X = 2) = \frac{\binom{3}{2} \binom{7}{2}}{\binom{10}{4}} = \frac{3 \times 21}{210} = \frac{63}{210} = \frac{3}{10} = 0.3$

4. $P(X = 3)$: $P(X = 3) = \frac{\binom{3}{3} \binom{7}{1}}{\binom{10}{4}} = \frac{1 \times 7}{210} = \frac{7}{210} \approx 0.0333$

So, the probability distribution of $X$ is: $P(X = 0) = 0.1667, \quad P(X = 1) = 0.5, \quad P(X = 2) = 0.3, \quad P(X = 3) = 0.0333$

(ii) $P(X \leq 1)$

This is the probability that $X$ is less than or equal to 1. It is the sum of $P(X = 0)$ and $P(X = 1)$: $P(X \leq 1) = P(X = 0) + P(X = 1) = 0.1667 + 0.5 = 0.6667$

(iii) $P(X < 1)$

This is the probability that $X$ is less than 1, which is the probability of $X = 0$: $P(X < 1) = P(X = 0) = 0.1667$

(iv) $P(1 < X < 2)$

This is the probability that $X$ is greater than 1 but less than 2. However, since $X$ is a discrete variable, $P(1 < X < 2)$ is the probability that $X = 2$: $P(1 < X < 2) = P(X = 2) = 0.3$

Summary:

1. Probability distribution of $X$: $P(X = 0) = 0.1667$, $P(X = 1) = 0.5$, $P(X = 2) = 0.3$, $P(X = 3) = 0.0333$.
2. $P(X \leq 1) = 0.6667$
3. $P(X < 1) = 0.1667$
4. $P(1 < X < 2) = 0.3$

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Would you like any further explanation or details on this? Here are five related questions:

1. How does the hypergeometric distribution differ from the binomial distribution?
2. What is the expected value $E(X)$ for the given distribution?
3. How would the distribution change if the sample size was increased to 5?
4. Can the variance of $X$ be computed, and how would it be used?
5. What would the probability distribution look like if the sample was drawn with replacement?

Tip: Understanding the hypergeometric distribution is essential for analyzing scenarios involving sampling without replacement, often used in quality control and other real-world applications.