Answer :
Let's break down each scenario step by step to determine the correct probabilities.
### Scenario 1: Probability that the sum of the two numbers rolled is greater than 3, given that the sum of the numbers is not greater than 5.
First, list all possible outcomes when rolling two six-sided dice:
- (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
- (2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
- (3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
- (4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
- (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
- (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
We need to consider only those combinations where the sum is not greater than 5. Those combinations are:
- (1,1) = 2
- (1,2) = 3
- (2,1) = 3
- (1,3) = 4
- (3,1) = 4
- (2,2) = 4
- (1,4) = 5
- (4,1) = 5
- (2,3) = 5
- (3,2) = 5
So, the total number of valid combinations where the sum is not greater than 5 is 10.
Next, we count the number of cases where the sum is greater than 3 but still not greater than 5:
- (1,3), (3,1) = 4
- (2,2) = 4
- (1,4), (4,1) = 5
- (2,3), (3,2) = 5
So, there are 7 combinations where the sum is greater than 3 but not greater than 5.
Thus, the probability is:
[tex]\[ P(\text{sum > 3 | sum ≤ 5}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{7}{10} = 0.7 \][/tex]
### Scenario 2: Probability that the sum of the two numbers rolled is an odd number, given that one of the numbers is a 6.
List all possible outcomes where one of the numbers is 6:
- (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
- (1,6), (2,6), (3,6), (4,6), (5,6), (6,6)
There are a total of 11 unique outcomes in this scenario (we count (6,6) only once here to avoid repetition).
Next, we count the number of cases where the sum is odd. The combinations that give an odd sum when one of the numbers is a 6 are:
- (6,1) = 7 (odd)
- (6,3) = 9 (odd)
- (6,5) = 11 (odd)
- (1,6) = 7 (odd)
- (3,6) = 9 (odd)
- (5,6) = 11 (odd)
So, there are 6 such outcomes.
Thus, the probability is:
[tex]\[ P(\text{sum is odd | one number is 6}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{6}{11} \][/tex]
### Summary:
1. The probability that the sum of the two numbers rolled is greater than 3, given that the sum of the numbers is not greater than 5, is [tex]\( 0.7 \)[/tex].
2. The probability that the sum of the two numbers rolled is an odd number, given that one of the numbers is 6, is [tex]\( \frac{6}{11} \approx 0.545 \)[/tex].
### Scenario 1: Probability that the sum of the two numbers rolled is greater than 3, given that the sum of the numbers is not greater than 5.
First, list all possible outcomes when rolling two six-sided dice:
- (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
- (2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
- (3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
- (4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
- (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
- (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
We need to consider only those combinations where the sum is not greater than 5. Those combinations are:
- (1,1) = 2
- (1,2) = 3
- (2,1) = 3
- (1,3) = 4
- (3,1) = 4
- (2,2) = 4
- (1,4) = 5
- (4,1) = 5
- (2,3) = 5
- (3,2) = 5
So, the total number of valid combinations where the sum is not greater than 5 is 10.
Next, we count the number of cases where the sum is greater than 3 but still not greater than 5:
- (1,3), (3,1) = 4
- (2,2) = 4
- (1,4), (4,1) = 5
- (2,3), (3,2) = 5
So, there are 7 combinations where the sum is greater than 3 but not greater than 5.
Thus, the probability is:
[tex]\[ P(\text{sum > 3 | sum ≤ 5}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{7}{10} = 0.7 \][/tex]
### Scenario 2: Probability that the sum of the two numbers rolled is an odd number, given that one of the numbers is a 6.
List all possible outcomes where one of the numbers is 6:
- (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
- (1,6), (2,6), (3,6), (4,6), (5,6), (6,6)
There are a total of 11 unique outcomes in this scenario (we count (6,6) only once here to avoid repetition).
Next, we count the number of cases where the sum is odd. The combinations that give an odd sum when one of the numbers is a 6 are:
- (6,1) = 7 (odd)
- (6,3) = 9 (odd)
- (6,5) = 11 (odd)
- (1,6) = 7 (odd)
- (3,6) = 9 (odd)
- (5,6) = 11 (odd)
So, there are 6 such outcomes.
Thus, the probability is:
[tex]\[ P(\text{sum is odd | one number is 6}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{6}{11} \][/tex]
### Summary:
1. The probability that the sum of the two numbers rolled is greater than 3, given that the sum of the numbers is not greater than 5, is [tex]\( 0.7 \)[/tex].
2. The probability that the sum of the two numbers rolled is an odd number, given that one of the numbers is 6, is [tex]\( \frac{6}{11} \approx 0.545 \)[/tex].
