Question: Suppose that a box contains $7$ cameras and that $4$ of them are defective. A sample of $2$ cameras is selected at random with replacement. Define the random variable $X$ as the number of defective cameras in the sample.
a. Write the binomial probability distribution for $X$. Round to two decimal places.
$X$ $P(X)$
$0$
$1$
$2$
b. What is the expected value of $X$? Round to two decimal places.
1. **State the problem:** We have $7$ cameras, $4$ defective and $3$ non-defective. We select $2$ cameras with replacement. $X$ is the number of defective cameras in the sample.
2. **Identify the distribution:** Since selection is with replacement and each trial is independent, $X$ follows a binomial distribution with parameters $n=2$ (number of trials) and $p=\frac{4}{7}$ (probability of defective camera).
3. **Binomial probability formula:**
$$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$
where $k=0,1,2$.
4. **Calculate probabilities:**
- For $X=0$ (no defective cameras):
$$P(0) = \binom{2}{0} \left(\frac{4}{7}\right)^0 \left(1-\frac{4}{7}\right)^2 = 1 \times 1 \times \left(\frac{3}{7}\right)^2 = \frac{9}{49} \approx 0.18$$
- For $X=1$ (one defective camera):
$$P(1) = \binom{2}{1} \left(\frac{4}{7}\right)^1 \left(\frac{3}{7}\right)^1 = 2 \times \frac{4}{7} \times \frac{3}{7} = \frac{24}{49} \approx 0.49$$
- For $X=2$ (two defective cameras):
$$P(2) = \binom{2}{2} \left(\frac{4}{7}\right)^2 \left(\frac{3}{7}\right)^0 = 1 \times \left(\frac{4}{7}\right)^2 \times 1 = \frac{16}{49} \approx 0.33$$
5. **Summarize the distribution:**
| $X$ | $P(X)$ |
|-----|--------|
| 0 | 0.18 |
| 1 | 0.49 |
| 2 | 0.33 |
6. **Expected value formula:**
$$E(X) = np$$
7. **Calculate expected value:**
$$E(X) = 2 \times \frac{4}{7} = \frac{8}{7} \approx 1.14$$
**Final answers:**
- Binomial distribution probabilities: $P(0)=0.18$, $P(1)=0.49$, $P(2)=0.33$
- Expected value: $E(X) = 1.14$