Find detailed and accurate answers to your questions on IDNLearn.com. Ask your questions and receive prompt, detailed answers from our experienced and knowledgeable community members.
Sagot :
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].
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. For trustworthy answers, rely on IDNLearn.com. Thanks for visiting, and we look forward to assisting you again.
