Welcome to Westonci.ca, where your questions are met with accurate answers from a community of experts and enthusiasts. Discover a wealth of knowledge from experts across different disciplines on our comprehensive Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
Let's solve the problem step-by-step:
### Part (a)
Let \( Z \) denote the number of trials in which the sum of the numbers on the two dice is 3.
1. Probability of Success (p):
- When rolling two fair dice, each of which has 6 faces, there are a total of \( 6 \times 6 = 36 \) possible outcomes.
- For the sum of the numbers to be 3, the favorable outcomes are:
- (1,2) and (2,1)
- Therefore, there are 2 favorable outcomes.
- Thus, the probability \( p \) of getting a sum of 3 in one roll is:
[tex]\[ p = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{2}{36} = \frac{1}{18} \][/tex]
2. Variance of \( Z \):
- The number of trials \( n = 10 \).
- The variance of \( Z \), where \( Z \) is a binomial random variable with parameters \( n \) and \( p \), is given by:
[tex]\[ \text{Var}(Z) = n \cdot p \cdot (1 - p) \][/tex]
- Substituting the values:
[tex]\[ \text{Var}(Z) = 10 \cdot \frac{1}{18} \cdot \left(1 - \frac{1}{18}\right) = 10 \cdot \frac{1}{18} \cdot \frac{17}{18} = \frac{170}{324} \approx 0.5247 \][/tex]
So, the variance of \( Z \) is approximately \( 0.5247 \).
### Part (b)
Now, we need to find the probability that out of 10 trials, exactly 5 trials have a sum of 7.
1. Probability of Success (p) for a Sum of 7:
- We follow a similar approach as in part (a).
- The favorable outcomes for a sum of 7 are:
- (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
- Hence, there are 6 favorable outcomes.
- Thus, the probability \( p \) of getting a sum of 7 in one roll is:
[tex]\[ p = \frac{6}{36} = \frac{1}{6} \][/tex]
2. Binomial Probability:
- We need the probability of having exactly 5 successes (trials with a sum of 7) in 10 trials.
- This is a binomial probability which is given by:
[tex]\[ P(X = k) = \binom{n}{k} \cdot p^k \cdot (1 - p)^{n - k} \][/tex]
where \( \binom{n}{k} \) is the binomial coefficient, calculated as:
[tex]\[ \binom{n}{k} = \frac{n!}{k! \cdot (n - k)!} \][/tex]
and \( k = 5 \), \( n = 10 \), \( p = \frac{1}{6} \).
3. Calculating the Probability:
- Let's plug the values into the formula:
[tex]\[ P(X = 5) = \binom{10}{5} \cdot \left(\frac{1}{6}\right)^5 \cdot \left(\frac{5}{6}\right)^5 \][/tex]
- Simplifying the expression gives us:
[tex]\[ \binom{10}{5} = \frac{10!}{5! \cdot 5!} = 252 \][/tex]
[tex]\[ P(X = 5) = 252 \cdot \left(\frac{1}{6}\right)^5 \cdot \left(\frac{5}{6}\right)^5 \approx 0.0130 \][/tex]
So, the probability that exactly 5 out of 10 trials result in a sum of 7 is approximately \( 0.0130 \).
### Final Answers:
(a) The variance of \( Z \) is approximately \( 0.5247 \).
(b) The probability that there are exactly 5 trials, each of which has a sum of 7, is approximately [tex]\( 0.0130 \)[/tex].
### Part (a)
Let \( Z \) denote the number of trials in which the sum of the numbers on the two dice is 3.
1. Probability of Success (p):
- When rolling two fair dice, each of which has 6 faces, there are a total of \( 6 \times 6 = 36 \) possible outcomes.
- For the sum of the numbers to be 3, the favorable outcomes are:
- (1,2) and (2,1)
- Therefore, there are 2 favorable outcomes.
- Thus, the probability \( p \) of getting a sum of 3 in one roll is:
[tex]\[ p = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{2}{36} = \frac{1}{18} \][/tex]
2. Variance of \( Z \):
- The number of trials \( n = 10 \).
- The variance of \( Z \), where \( Z \) is a binomial random variable with parameters \( n \) and \( p \), is given by:
[tex]\[ \text{Var}(Z) = n \cdot p \cdot (1 - p) \][/tex]
- Substituting the values:
[tex]\[ \text{Var}(Z) = 10 \cdot \frac{1}{18} \cdot \left(1 - \frac{1}{18}\right) = 10 \cdot \frac{1}{18} \cdot \frac{17}{18} = \frac{170}{324} \approx 0.5247 \][/tex]
So, the variance of \( Z \) is approximately \( 0.5247 \).
### Part (b)
Now, we need to find the probability that out of 10 trials, exactly 5 trials have a sum of 7.
1. Probability of Success (p) for a Sum of 7:
- We follow a similar approach as in part (a).
- The favorable outcomes for a sum of 7 are:
- (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
- Hence, there are 6 favorable outcomes.
- Thus, the probability \( p \) of getting a sum of 7 in one roll is:
[tex]\[ p = \frac{6}{36} = \frac{1}{6} \][/tex]
2. Binomial Probability:
- We need the probability of having exactly 5 successes (trials with a sum of 7) in 10 trials.
- This is a binomial probability which is given by:
[tex]\[ P(X = k) = \binom{n}{k} \cdot p^k \cdot (1 - p)^{n - k} \][/tex]
where \( \binom{n}{k} \) is the binomial coefficient, calculated as:
[tex]\[ \binom{n}{k} = \frac{n!}{k! \cdot (n - k)!} \][/tex]
and \( k = 5 \), \( n = 10 \), \( p = \frac{1}{6} \).
3. Calculating the Probability:
- Let's plug the values into the formula:
[tex]\[ P(X = 5) = \binom{10}{5} \cdot \left(\frac{1}{6}\right)^5 \cdot \left(\frac{5}{6}\right)^5 \][/tex]
- Simplifying the expression gives us:
[tex]\[ \binom{10}{5} = \frac{10!}{5! \cdot 5!} = 252 \][/tex]
[tex]\[ P(X = 5) = 252 \cdot \left(\frac{1}{6}\right)^5 \cdot \left(\frac{5}{6}\right)^5 \approx 0.0130 \][/tex]
So, the probability that exactly 5 out of 10 trials result in a sum of 7 is approximately \( 0.0130 \).
### Final Answers:
(a) The variance of \( Z \) is approximately \( 0.5247 \).
(b) The probability that there are exactly 5 trials, each of which has a sum of 7, is approximately [tex]\( 0.0130 \)[/tex].
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.