Suppose that a box contains 6 cameras and that 5 of them are defective. A sample of 2 cameras is selected at random, with replacement. Define the random variable  as the number of defective cameras in the sample.

Write the probability distribution for .
P()
	
	
	


What is the expected value of ?    
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To define the random variable \( X \) as the number of defective cameras in a sample of 2 selected at random with replacement from a box containing 6 cameras (5 defective and 1 non-defective), we can first determine the probabilities associated with each possible outcome.

### Probability Distribution

The possible values of \( X \) are 0, 1, or 2, representing the number of defective cameras in the sample. Since the selection is done with replacement, the probabilities for each sample can be calculated as follows:

1. **Probability of selecting a defective camera (D)**: 
   \[
   P(D) = \frac{5}{6}
   \]

2. **Probability of selecting a non-defective camera (N)**: 
   \[
   P(N) = \frac{1}{6}
   \]

Now, we calculate the probabilities for each possible value of \( X \):

- \( P(X = 0) \): Both selected cameras are non-defective.
  \[
  P(X = 0) = P(N) \times P(N) = \left(\frac{1}{6}\right) \times \left(\frac{1}{6}\right) = \frac{1}{36}
  \]

- \( P(X = 1) \): One camera is defective and one is non-defective. This can happen in two ways: either the first camera is defective and the second is non-defective, or vice versa.
  \[
  P(X = 1) = P(D) \times P(N) + P(N) \times P(D) = \left(\frac{5}{6}\right) \times \left(\frac{1}{6}\right) + \left(\frac{1}{6}\right) \times \left(\frac{5}{6}\right) = 2 \times \left(\frac{5}{36}\right) = \frac{10}{36} = \frac{5}{18}
  \]

- \( P(X = 2) \): Both selected cameras are defective.
  \[
  P(X = 2) = P(D) \times P(D) = \left(\frac{5}{6}\right) \times \left(\frac{5}{6}\right) = \frac{25}{36}
  \]

### Summary of the Probability Distribution
- \( P(X = 0) = \frac{1}{36} \)
- \( P(X = 1) = \frac{5}{18} \)
- \( P(X = 2) = \frac{25}{36} \)

### Expected Value of \( X \)

The expected value \( E(X) \) is calculated using the formula:
\[
E(X) = \sum (x \cdot P(X = x))
\]
Calculating it step-by-step:
\[
E(X) = 0 \cdot P(X = 0) + 1 \cdot P(X = 1) + 2 \cdot P(X = 2)
\]
\[
E(X) = 0 \cdot \frac{1}{36} + 1 \cdot \frac{5}{18} + 2 \cdot \frac{25}{36}
\]
\[
E(X) = 0 + \frac{5}{18} + \frac{50}{36}
\]
Convert \( \frac{5}{18} \) to have a common denominator with \( \frac{50}{36} \):
\[
\frac{5}{18} = \frac{10}{36}
\]
So,
\[
E(X) = \frac{10}{36} + \frac{50}{36} = \frac{60}{36} = \frac{5}{3}
\]

### Final Results
- **Probability Distribution**:
  - \( P(X = 0) = \frac{1}{36} \)
  - \( P(X = 1) = \frac{5}{18} \)
  - \( P(X = 2) = \frac{25}{36} \)

- **Expected Value**: 
  \[
  E(X) = \frac{5}{3}
  \]

If you have any further questions or need more details, feel free to ask! Here are some related questions to consider:

1. How would the probabilities change if there were 4 defective cameras?
2. What is the variance of the random variable \( X \)?
3. How would the expected value change if the sample size increased to 3 cameras?
4. What are the implications of sampling without replacement?
5. How does the probability distribution affect decision-making in quality control? 

**Tip:** Always check if the selection is with or without replacement, as it significantly impacts the probabilities and calculations!

Suppose that a box contains 6 cameras and that 5 of them are defective. A sample of 2 cameras is selected at random, with replacement. Define the random variable as the number of defective cameras in the sample. Write the probability distribution for P(). What is the expected value of ?

Explore the probability distribution and expected value for a scenario where a box contains 6 cameras, with 5 being defective. This problem involves random sampling with replacement, detailing the calculations of P(X) and E(X). Suitable for high school students.

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