Answer :
Sure! Let's solve this step-by-step:
We define the random variable [tex]\(X\)[/tex] as the number of defective cameras in a sample of 2 cameras selected at random with replacement from a box of 7 cameras, where 3 cameras are defective.
### a. Binomial Probability Distribution for [tex]\(X\)[/tex]
The scenario described follows a binomial distribution, since each camera selection is independent of the other, and each selection can either result in a defective camera (success) or a non-defective camera (failure).
1. Parameters of the Binomial Distribution:
- Number of trials ([tex]\(n\)[/tex]): 2
- Probability of success ([tex]\(p\)[/tex]):
[tex]\[ p = \frac{\text{number of defective cameras}}{\text{total number of cameras}} = \frac{3}{7} \][/tex]
2. Binomial Probability Formula:
The probability mass function (PMF) of a binomial distribution is given by:
[tex]\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \][/tex]
where [tex]\(\binom{n}{k}\)[/tex] is the binomial coefficient, which represents the number of ways to choose [tex]\(k\)[/tex] successes from [tex]\(n\)[/tex] trials.
3. Calculating the Probabilities:
- For [tex]\(k = 0\)[/tex] (no defective cameras):
[tex]\[ P(X = 0) = \binom{2}{0} p^0 (1-p)^2 = (1 - p)^2 \][/tex]
- For [tex]\(k = 1\)[/tex] (one defective camera):
[tex]\[ P(X = 1) = \binom{2}{1} p^1 (1-p)^1 = 2p(1 - p) \][/tex]
- For [tex]\(k = 2\)[/tex] (two defective cameras):
[tex]\[ P(X = 2) = \binom{2}{2} p^2 (1 - p)^0 = p^2 \][/tex]
4. Substituting [tex]\(p = \frac{3}{7}\)[/tex]:
[tex]\[ P(X = 0) = \left(1 - \frac{3}{7}\right)^2 = \left(\frac{4}{7}\right)^2 \approx 0.33 \][/tex]
[tex]\[ P(X = 1) = 2 \cdot \frac{3}{7} \cdot \frac{4}{7} = 2 \cdot \frac{12}{49} \approx 0.49 \][/tex]
[tex]\[ P(X = 2) = \left(\frac{3}{7}\right)^2 = \frac{9}{49} \approx 0.18 \][/tex]
Thus, the binomial probability distribution for [tex]\(X\)[/tex] is:
[tex]\[ P(X = 0) \approx 0.33, \quad P(X = 1) \approx 0.49, \quad P(X = 2) \approx 0.18 \][/tex]
### b. Expected Value of [tex]\(X\)[/tex]
The expected value (mean) of a binomial random variable [tex]\(X\)[/tex] is given by:
[tex]\[ E(X) = n \cdot p \][/tex]
Substituting the values [tex]\(n = 2\)[/tex] and [tex]\(p = \frac{3}{7}\)[/tex], we get:
[tex]\[ E(X) = 2 \cdot \frac{3}{7} \approx 0.86 \][/tex]
#### Summary:
- The binomial probability distribution for [tex]\(X\)[/tex] is approximately: [tex]\([0.33, 0.49, 0.18]\)[/tex].
- The expected value of [tex]\(X\)[/tex] is approximately: [tex]\(0.86\)[/tex].
We define the random variable [tex]\(X\)[/tex] as the number of defective cameras in a sample of 2 cameras selected at random with replacement from a box of 7 cameras, where 3 cameras are defective.
### a. Binomial Probability Distribution for [tex]\(X\)[/tex]
The scenario described follows a binomial distribution, since each camera selection is independent of the other, and each selection can either result in a defective camera (success) or a non-defective camera (failure).
1. Parameters of the Binomial Distribution:
- Number of trials ([tex]\(n\)[/tex]): 2
- Probability of success ([tex]\(p\)[/tex]):
[tex]\[ p = \frac{\text{number of defective cameras}}{\text{total number of cameras}} = \frac{3}{7} \][/tex]
2. Binomial Probability Formula:
The probability mass function (PMF) of a binomial distribution is given by:
[tex]\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \][/tex]
where [tex]\(\binom{n}{k}\)[/tex] is the binomial coefficient, which represents the number of ways to choose [tex]\(k\)[/tex] successes from [tex]\(n\)[/tex] trials.
3. Calculating the Probabilities:
- For [tex]\(k = 0\)[/tex] (no defective cameras):
[tex]\[ P(X = 0) = \binom{2}{0} p^0 (1-p)^2 = (1 - p)^2 \][/tex]
- For [tex]\(k = 1\)[/tex] (one defective camera):
[tex]\[ P(X = 1) = \binom{2}{1} p^1 (1-p)^1 = 2p(1 - p) \][/tex]
- For [tex]\(k = 2\)[/tex] (two defective cameras):
[tex]\[ P(X = 2) = \binom{2}{2} p^2 (1 - p)^0 = p^2 \][/tex]
4. Substituting [tex]\(p = \frac{3}{7}\)[/tex]:
[tex]\[ P(X = 0) = \left(1 - \frac{3}{7}\right)^2 = \left(\frac{4}{7}\right)^2 \approx 0.33 \][/tex]
[tex]\[ P(X = 1) = 2 \cdot \frac{3}{7} \cdot \frac{4}{7} = 2 \cdot \frac{12}{49} \approx 0.49 \][/tex]
[tex]\[ P(X = 2) = \left(\frac{3}{7}\right)^2 = \frac{9}{49} \approx 0.18 \][/tex]
Thus, the binomial probability distribution for [tex]\(X\)[/tex] is:
[tex]\[ P(X = 0) \approx 0.33, \quad P(X = 1) \approx 0.49, \quad P(X = 2) \approx 0.18 \][/tex]
### b. Expected Value of [tex]\(X\)[/tex]
The expected value (mean) of a binomial random variable [tex]\(X\)[/tex] is given by:
[tex]\[ E(X) = n \cdot p \][/tex]
Substituting the values [tex]\(n = 2\)[/tex] and [tex]\(p = \frac{3}{7}\)[/tex], we get:
[tex]\[ E(X) = 2 \cdot \frac{3}{7} \approx 0.86 \][/tex]
#### Summary:
- The binomial probability distribution for [tex]\(X\)[/tex] is approximately: [tex]\([0.33, 0.49, 0.18]\)[/tex].
- The expected value of [tex]\(X\)[/tex] is approximately: [tex]\(0.86\)[/tex].
