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Sagot :
Certainly, let’s go through the solution step-by-step to determine the probability that the sum of numbers rolled on two fair dice is a multiple of 5.
### Step 1: Understanding the Possibility Space
Each die has 6 faces, numbered from 1 to 6. When throwing two dice, each face of the first die can pair with each face of the second die. Hence, the total number of possible outcomes can be calculated as:
[tex]\[ \text{Total Outcomes} = 6 \times 6 = 36 \][/tex]
### Step 2: Listing the Possible Outcomes
All possible outcomes when throwing two dice can be written as ordered pairs [tex]\((a, b)\)[/tex], where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] range from 1 to 6. Here is a part of the list:
[tex]\[ (a, b) = (1, 1), (1, 2), (1, 3), \ldots, (6, 6) \][/tex]
### Step 3: Finding the Favorable Outcomes
We need to find the outcomes where the sum [tex]\(a + b\)[/tex] is a multiple of 5. The possible sums that are multiples of 5 between 2 (minimum sum when both dice show 1) and 12 (maximum sum when both dice show 6) are 5, 10.
Let's list the orders and sums:
- Sum = 5:
- (1, 4), (2, 3), (3, 2), (4, 1)
- Sum = 10:
- (4, 6), (5, 5), (6, 4)
By counting these, we get:
- There are 4 outcomes that result in a sum of 5.
- There are 3 outcomes that result in a sum of 10.
So, the total number of favorable outcomes is:
[tex]\[ \text{Favorable Outcomes} = 4 + 3 = 7 \][/tex]
### Step 4: Calculating the Probability
The probability is the ratio of the number of favorable outcomes to the total number of outcomes. Thus,
[tex]\[ \text{Probability} = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{7}{36} \][/tex]
Expressing this as a decimal, we get approximately:
[tex]\[ \text{Probability} = 0.1944 \][/tex]
### Conclusion
The probability that the sum of the numbers rolled on two fair dice is a multiple of 5 is approximately [tex]\( \frac{7}{36} \)[/tex] or 0.1944 (19.44%).
### Step 1: Understanding the Possibility Space
Each die has 6 faces, numbered from 1 to 6. When throwing two dice, each face of the first die can pair with each face of the second die. Hence, the total number of possible outcomes can be calculated as:
[tex]\[ \text{Total Outcomes} = 6 \times 6 = 36 \][/tex]
### Step 2: Listing the Possible Outcomes
All possible outcomes when throwing two dice can be written as ordered pairs [tex]\((a, b)\)[/tex], where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] range from 1 to 6. Here is a part of the list:
[tex]\[ (a, b) = (1, 1), (1, 2), (1, 3), \ldots, (6, 6) \][/tex]
### Step 3: Finding the Favorable Outcomes
We need to find the outcomes where the sum [tex]\(a + b\)[/tex] is a multiple of 5. The possible sums that are multiples of 5 between 2 (minimum sum when both dice show 1) and 12 (maximum sum when both dice show 6) are 5, 10.
Let's list the orders and sums:
- Sum = 5:
- (1, 4), (2, 3), (3, 2), (4, 1)
- Sum = 10:
- (4, 6), (5, 5), (6, 4)
By counting these, we get:
- There are 4 outcomes that result in a sum of 5.
- There are 3 outcomes that result in a sum of 10.
So, the total number of favorable outcomes is:
[tex]\[ \text{Favorable Outcomes} = 4 + 3 = 7 \][/tex]
### Step 4: Calculating the Probability
The probability is the ratio of the number of favorable outcomes to the total number of outcomes. Thus,
[tex]\[ \text{Probability} = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{7}{36} \][/tex]
Expressing this as a decimal, we get approximately:
[tex]\[ \text{Probability} = 0.1944 \][/tex]
### Conclusion
The probability that the sum of the numbers rolled on two fair dice is a multiple of 5 is approximately [tex]\( \frac{7}{36} \)[/tex] or 0.1944 (19.44%).
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