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Sagot :
Let's analyze the scenario where a pair of dice is rolled and find the probabilities for various sums of the numbers on their upper faces.
We know that when rolling two 6-sided dice, there are a total of [tex]\(6 \times 6 = 36\)[/tex] possible outcomes because each die has 6 faces.
### (a) Probability of the sum being 10
The possible sums that equal 10 are:
- (4, 6)
- (5, 5)
- (6, 4)
There are 3 favorable outcomes. Therefore, the probability is:
[tex]\[ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{36} = \frac{1}{12} \approx 0.0833 \][/tex]
### (b) Probability of the sum being even
To get an even sum, we need either both numbers to be even or both to be odd. By analyzing, we find that half of the outcomes will be even sums and half will be odd sums. Therefore, the probability is:
[tex]\[ \text{Probability} = \frac{1}{2} = 0.5 \][/tex]
### (c) Probability of the sum being odd
As noted, since the sum can either be even or odd, the probability of getting an odd sum is also:
[tex]\[ \text{Probability} = \frac{1}{2} = 0.5 \][/tex]
### (d) Probability of the sum being a multiple of 3
The sums that are multiples of 3 within the range of possible sums (2 to 12) are 3, 6, 9, and 12. We include all outcomes that result in these sums:
- Sum of 3: (1, 2), (2, 1)
- Sum of 6: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1)
- Sum of 9: (3, 6), (4, 5), (5, 4), (6, 3)
- Sum of 12: (6, 6)
This gives us a total of:
[tex]\[ 2 + 5 + 4 + 1 = 12 \text{ favorable outcomes} \][/tex]
Therefore, the probability is:
[tex]\[ \text{Probability} = \frac{12}{36} = \frac{1}{3} \approx 0.3333 \][/tex]
### (e) Probability of the sum being less than 9 and even
The sums less than 9 are 2, 4, 6, and 8. We consider those sums and check which are even:
- Sum of 2: (1, 1)
- Sum of 4: (1, 3), (2, 2), (3, 1)
- Sum of 6: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1)
- Sum of 8: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2)
This gives us a total of:
[tex]\[ 1 + 3 + 5 + 5 = 14 \text{ favorable outcomes} \][/tex]
Therefore, the probability is:
[tex]\[ \text{Probability} = \frac{14}{36} \approx 0.3889 \][/tex]
### 1) Probability of the sum being greater than 12
The highest possible sum when rolling two 6-sided dice is 12 (6 + 6), so it's impossible to have a sum greater than 12. Thus, the probability is:
[tex]\[ \text{Probability} = 0 \][/tex]
In summary:
(a) The probability of the sum being 10 is approximately [tex]\(0.0833\)[/tex].
(b) The probability of the sum being even is [tex]\(0.5\)[/tex].
(c) The probability of the sum being odd is [tex]\(0.5\)[/tex].
(d) The probability of the sum being a multiple of 3 is approximately [tex]\(0.3333\)[/tex].
(e) The probability of the sum being less than 9 and even is approximately [tex]\(0.3889\)[/tex].
(1) The probability of the sum being greater than 12 is [tex]\(0\)[/tex].
We know that when rolling two 6-sided dice, there are a total of [tex]\(6 \times 6 = 36\)[/tex] possible outcomes because each die has 6 faces.
### (a) Probability of the sum being 10
The possible sums that equal 10 are:
- (4, 6)
- (5, 5)
- (6, 4)
There are 3 favorable outcomes. Therefore, the probability is:
[tex]\[ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{36} = \frac{1}{12} \approx 0.0833 \][/tex]
### (b) Probability of the sum being even
To get an even sum, we need either both numbers to be even or both to be odd. By analyzing, we find that half of the outcomes will be even sums and half will be odd sums. Therefore, the probability is:
[tex]\[ \text{Probability} = \frac{1}{2} = 0.5 \][/tex]
### (c) Probability of the sum being odd
As noted, since the sum can either be even or odd, the probability of getting an odd sum is also:
[tex]\[ \text{Probability} = \frac{1}{2} = 0.5 \][/tex]
### (d) Probability of the sum being a multiple of 3
The sums that are multiples of 3 within the range of possible sums (2 to 12) are 3, 6, 9, and 12. We include all outcomes that result in these sums:
- Sum of 3: (1, 2), (2, 1)
- Sum of 6: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1)
- Sum of 9: (3, 6), (4, 5), (5, 4), (6, 3)
- Sum of 12: (6, 6)
This gives us a total of:
[tex]\[ 2 + 5 + 4 + 1 = 12 \text{ favorable outcomes} \][/tex]
Therefore, the probability is:
[tex]\[ \text{Probability} = \frac{12}{36} = \frac{1}{3} \approx 0.3333 \][/tex]
### (e) Probability of the sum being less than 9 and even
The sums less than 9 are 2, 4, 6, and 8. We consider those sums and check which are even:
- Sum of 2: (1, 1)
- Sum of 4: (1, 3), (2, 2), (3, 1)
- Sum of 6: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1)
- Sum of 8: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2)
This gives us a total of:
[tex]\[ 1 + 3 + 5 + 5 = 14 \text{ favorable outcomes} \][/tex]
Therefore, the probability is:
[tex]\[ \text{Probability} = \frac{14}{36} \approx 0.3889 \][/tex]
### 1) Probability of the sum being greater than 12
The highest possible sum when rolling two 6-sided dice is 12 (6 + 6), so it's impossible to have a sum greater than 12. Thus, the probability is:
[tex]\[ \text{Probability} = 0 \][/tex]
In summary:
(a) The probability of the sum being 10 is approximately [tex]\(0.0833\)[/tex].
(b) The probability of the sum being even is [tex]\(0.5\)[/tex].
(c) The probability of the sum being odd is [tex]\(0.5\)[/tex].
(d) The probability of the sum being a multiple of 3 is approximately [tex]\(0.3333\)[/tex].
(e) The probability of the sum being less than 9 and even is approximately [tex]\(0.3889\)[/tex].
(1) The probability of the sum being greater than 12 is [tex]\(0\)[/tex].
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