from a lot of 10 items containing 3 defectives, a sample of 4 items is drawn at random. Let the random variable X denote the no. of defective items in the sample is drawn without replacement, find the probability distribution of X

(i) probability distribution of X

(ii) P(X is less than and equal to 1)

(iii) P(X is less than 1)

(iv) P( x is less than 2 but greater than 1)

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.

Explore the probability distribution of defective items in a sample of 4 items drawn from a lot of 10 items containing 3 defectives. Calculate probabilities for different scenarios using the hypergeometric distribution formula.

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