A shipment of 20 similar laptop computers to a retail outlet contains 3 that are defective. If a school makes
a random purchase of 2 of these computers, find the probability distribution for the number of defectives.

To find the probability distribution for the number of defective computers in the random purchase of 2 computers out of the 20 available, we can follow these steps:

### Step 1: Define the Random Variable
Let \( X \) represent the number of defective computers in the purchase of 2 computers. \( X \) can take the values 0, 1, or 2.

### Step 2: Calculate the Probabilities
We will use the hypergeometric distribution because the sampling is done without replacement. The probability mass function for a hypergeometric distribution is given by:

\[
P(X = k) = \frac{\binom{D}{k} \binom{N-D}{n-k}}{\binom{N}{n}}
\]

Where:
- \( N \) is the total number of items (in this case, 20 computers).
- \( D \) is the number of defective items (in this case, 3 defective computers).
- \( n \) is the number of items drawn (in this case, 2 computers).
- \( k \) is the number of defective items drawn, which can be 0, 1, or 2.

### Step 3: Calculate Each Probability
1. **Probability \( P(X = 0) \):**
   \[
   P(X = 0) = \frac{\binom{3}{0} \binom{17}{2}}{\binom{20}{2}} = \frac{1 \times \binom{17}{2}}{\binom{20}{2}}
   \]
   \[
   P(X = 0) = \frac{1 \times \frac{17 \times 16}{2}}{\frac{20 \times 19}{2}} = \frac{1 \times 136}{190} = \frac{136}{190} \approx 0.7158
   \]

2. **Probability \( P(X = 1) \):**
   \[
   P(X = 1) = \frac{\binom{3}{1} \binom{17}{1}}{\binom{20}{2}} = \frac{3 \times 17}{190} = \frac{51}{190} \approx 0.2684
   \]

3. **Probability \( P(X = 2) \):**
   \[
   P(X = 2) = \frac{\binom{3}{2} \binom{17}{0}}{\binom{20}{2}} = \frac{3 \times 1}{190} = \frac{3}{190} \approx 0.0158
   \]

### Step 4: Summary of the Probability Distribution
\[
P(X = 0) \approx 0.7158
\]
\[
P(X = 1) \approx 0.2684
\]
\[
P(X = 2) \approx 0.0158
\]

So, the probability distribution for the number of defective computers in the purchase is:

\[
\begin{array}{c|c}
\text{Number of Defective Computers (X)} & \text{Probability} \\
\hline
0 & 0.7158 \\
1 & 0.2684 \\
2 & 0.0158 \\
\end{array}
\]

This distribution shows that there is a high probability of getting no defective computers, a smaller probability of getting one defective computer, and a very low probability of getting two defective computers.

Would you like details on the calculations, or do you have any questions? Here are five related questions:

1. What is the expected number of defective computers in the purchase?
2. How does the probability distribution change if the number of defective computers increases?
3. What is the probability of getting at least one defective computer?
4. How would the distribution change if the school purchases 3 computers instead of 2?
5. What assumptions are made in this probability model?

**Tip:** The hypergeometric distribution is ideal for problems involving sampling without replacement, where the probability of success changes with each draw.

Explore the probability distribution for the number of defective computers in a random purchase of 2 out of 20 available laptops. Learn how to calculate probabilities using the hypergeometric distribution, including detailed steps and examples.

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