Let's analyze the question shown in the image:

The problem states:

• There are 8 cameras in a box, 4 of which are defective.
• A sample of 2 cameras is drawn with replacement.
• $X$ represents the number of defective cameras in the sample.

Step 1: Probability Distribution

Since we're drawing with replacement, the probabilities for defective and non-defective cameras remain constant. We need to calculate the probability for each possible outcome, which corresponds to how many defective cameras are selected (0, 1, or 2 defective cameras).

• $P(X = 0)$: This is the probability of selecting 0 defective cameras (i.e., both are non-defective). The probability of choosing a non-defective camera is $P(\text{non-defective}) = \frac{4}{8} = 0.5$. For 2 cameras, this becomes $P(X = 0) = 0.5 \times 0.5 = 0.25$.

• $P(X = 1)$: This is the probability of selecting 1 defective and 1 non-defective camera. Since the cameras are drawn with replacement, there are two possible ways this can happen: first defective, then non-defective, or first non-defective, then defective. The probability for either of these cases is $0.5 \times 0.5 = 0.25$. Thus, $P(X = 1) = 2 \times 0.25 = 0.5$.

• $P(X = 2)$: This is the probability of selecting 2 defective cameras. The probability of choosing a defective camera is $P(\text{defective}) = \frac{4}{8} = 0.5$, so for two defective cameras, $P(X = 2) = 0.5 \times 0.5 = 0.25$.

Step 2: Expected Value

The expected value of $X$ (denoted $E(X)$) is calculated using the formula:

$E(X) = \sum_{k} k \cdot P(X = k)$

Substitute the values for $k = 0, 1, 2$ and their corresponding probabilities:

$E(X) = 0 \times P(X = 0) + 1 \times P(X = 1) + 2 \times P(X = 2)$ $E(X) = 0 \times 0.25 + 1 \times 0.5 + 2 \times 0.25 = 0 + 0.5 + 0.5 = 1$

Answers:

1. $P(X = 0) = 0.25$
2. $P(X = 1) = 0.5$
3. $P(X = 2) = 0.25$
4. Expected value $E(X) = 1$

Would you like further details or have any additional questions?

Here are some related questions for further understanding:

1. How does the probability distribution change if we sample without replacement?
2. What if the number of defective cameras changes to 2 instead of 4?
3. How do you calculate the variance of the distribution in this scenario?
4. What is the significance of the expected value in real-world scenarios?
5. How does the sampling size impact the probability distribution in this case?

Tip: The expected value represents the average number of defective cameras you can expect in the sample after many trials.