We are given that a box contains 8 cameras, 4 of which are defective. A sample of 2 cameras is selected at random with replacement, and we are asked to define the random variable $X$ as the number of defective cameras in the sample. Since the selection is with replacement, the number of defective cameras in the sample follows a binomial distribution.

Binomial Probability Distribution:

The binomial probability distribution can be written as:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$ where:

• $n$ is the number of trials (in this case, 2),
• $k$ is the number of defective cameras (values of $k$ can be 0, 1, or 2),
• $p$ is the probability of selecting a defective camera,
• $1-p$ is the probability of selecting a non-defective camera.

The probability of selecting a defective camera is: $p = \frac{4}{8} = 0.5$

So, the binomial distribution is:

$P(X = k) = \binom{2}{k} (0.5)^k (0.5)^{2-k}$

For $k = 0, 1, 2$:

• $P(X = 0) = \binom{2}{0} (0.5)^0 (0.5)^2 = 1 \times 1 \times 0.25 = 0.25$
• $P(X = 1) = \binom{2}{1} (0.5)^1 (0.5)^1 = 2 \times 0.5 \times 0.5 = 0.50$
• $P(X = 2) = \binom{2}{2} (0.5)^2 (0.5)^0 = 1 \times 0.25 \times 1 = 0.25$

Thus, the binomial probability distribution for $X$ is:

$P(X = 0) = 0.25, \quad P(X = 1) = 0.50, \quad P(X = 2) = 0.25$

Expected Value of $X$:

The expected value $E(X)$ of a binomial distribution is given by:

$E(X) = n \times p$ where:

• $n = 2$ is the number of trials,
• $p = 0.5$ is the probability of selecting a defective camera.

Thus, the expected value is:

$E(X) = 2 \times 0.5 = 1.00$

Final Answer:

• The binomial probability distribution for $X$ is:

  $P(X = 0) = 0.25, \quad P(X = 1) = 0.50, \quad P(X = 2) = 0.25$

• The expected value $E(X) = 1.00$.

Would you like more details or have any questions?

Here are 5 related questions:

1. What is the variance of this binomial distribution?
2. How would the distribution change if the sampling was done without replacement?
3. What is the probability of selecting exactly 1 defective camera in 3 trials with replacement?
4. How does the binomial distribution relate to the normal distribution for large sample sizes?
5. How would you calculate the cumulative distribution function (CDF) for this problem?

Tip: In binomial distributions, the variance is calculated as $\sigma^2 = n \times p \times (1-p)$.