Answer :
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].
