Problem Overview

The problem is from probability and involves hypergeometric distribution. Let's break it down:

1. Situation:

   • A lot of 10 items, including 3 defective items.
   • A sample of 4 items is drawn without replacement.
   • The random variable $X$ represents the number of defective items in the sample.

2. Tasks: i. Find the probability distribution of $X$.
   ii. Calculate $P(X \leq 1)$.

Part (i): Probability Distribution of $X$

Using the hypergeometric distribution formula: $P(X = k) = \frac{\binom{D}{k} \binom{N-D}{n-k}}{\binom{N}{n}}$ Where:

• $N = 10$: Total items in the lot,
• $D = 3$: Number of defective items,
• $n = 4$: Sample size,
• $k$: Number of defective items in the sample (possible values: $k = 0, 1, 2, 3$).

Steps:

1. Calculate $\binom{N}{n} = \binom{10}{4}$, the total number of ways to choose 4 items out of 10.

2. For each $k$, calculate:

   • $\binom{D}{k} = \binom{3}{k}$, the ways to choose $k$ defective items.
   • $\binom{N-D}{n-k} = \binom{7}{4-k}$, the ways to choose $(4-k)$ non-defective items.

3. Combine these to compute $P(X = k)$ for $k = 0, 1, 2, 3$.

Calculations:

• $\binom{10}{4} = 210$
• For $k = 0, 1, 2, 3$:
  • $P(X = 0) = \frac{\binom{3}{0} \binom{7}{4}}{\binom{10}{4}} = \frac{1 \cdot 35}{210} = \frac{35}{210} = 0.1667$
  • $P(X = 1) = \frac{\binom{3}{1} \binom{7}{3}}{\binom{10}{4}} = \frac{3 \cdot 35}{210} = \frac{105}{210} = 0.5$
  • $P(X = 2) = \frac{\binom{3}{2} \binom{7}{2}}{\binom{10}{4}} = \frac{3 \cdot 21}{210} = \frac{63}{210} = 0.3$
  • $P(X = 3) = \frac{\binom{3}{3} \binom{7}{1}}{\binom{10}{4}} = \frac{1 \cdot 7}{210} = \frac{7}{210} = 0.0333$

Final Probability Distribution: $P(X = 0) = 0.1667, \quad P(X = 1) = 0.5, \quad P(X = 2) = 0.3, \quad P(X = 3) = 0.0333$

Part (ii): Calculate $P(X \leq 1)$

This is the cumulative probability: $P(X \leq 1) = P(X = 0) + P(X = 1)$ Using the values: $P(X \leq 1) = 0.1667 + 0.5 = 0.6667$

Final Answers:

1. Probability Distribution: $P(X = 0) = 0.1667, \; P(X = 1) = 0.5, \; P(X = 2) = 0.3, \; P(X = 3) = 0.0333$
2. Cumulative Probability: $P(X \leq 1) = 0.6667$

Would you like step-by-step explanations or help with additional concepts?

Related Questions:

1. How do you derive the hypergeometric distribution formula?
2. What is the difference between sampling with and without replacement?
3. Can you compare hypergeometric and binomial distributions?
4. How would the probability distribution change if the sample size increased to 5?
5. What real-world applications involve hypergeometric distributions?

Tip: When solving probability problems, always ensure whether the sampling is with or without replacement, as it significantly changes the distribution.