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Sagot :
Alright, let's work through this problem step-by-step.
We'll start by considering all possible outcomes when rolling two dice. Each die has 6 faces, so when rolling two dice, there are a total of [tex]\(6 \times 6 = 36\)[/tex] possible outcomes.
### (a) Calculating Probability [tex]\(P(A)\)[/tex]
Event [tex]\(A\)[/tex] is the event where the sum of the two dice is greater than 9. We need to determine the successful outcomes for this event.
The possible sums greater than 9 are: [tex]\(10, 11, \)[/tex] and [tex]\(12\)[/tex].
Let's list the combinations for each sum:
- Sum of 10: (4,6), (5,5), (6,4) → 3 outcomes
- Sum of 11: (5,6), (6,5) → 2 outcomes
- Sum of 12: (6,6) → 1 outcome
Adding these together, we get:
[tex]\[ 3 + 2 + 1 = 6 \text{ successful outcomes} \][/tex]
To find the probability, we divide the number of successful outcomes by the total number of outcomes:
[tex]\[ P(A) = \frac{6}{36} = \frac{1}{6} \approx 0.17 \][/tex]
So,
[tex]\[ P(A) \approx 0.17 \][/tex]
### (b) Calculating Probability [tex]\(P(B)\)[/tex]
Event [tex]\(B\)[/tex] is the event where the sum is neither divisible by 4 nor by 5.
First, let's find the totals that are either divisible by 4 or by 5.
- Sums divisible by 4: [tex]\(4, 8, 12\)[/tex]
- Combinations for 4: (1,3), (2,2), (3,1) → 3 outcomes
- Combinations for 8: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 outcomes
- Combinations for 12: (6,6) → 1 outcome
Adding these together, we get:
[tex]\[ 3 + 5 + 1 = 9 \text{ outcomes where sum is divisible by 4} \][/tex]
- Sums divisible by 5: [tex]\(5, 10\)[/tex]
- Combinations for 5: (1,4), (2,3), (3,2), (4,1) → 4 outcomes
- Combinations for 10: (4,6), (5,5), (6,4) → 3 outcomes
Adding these together, we get:
[tex]\[ 4 + 3 = 7 \text{ outcomes where sum is divisible by 5} \][/tex]
Now, let's count the overlap (sums divisible by both 4 and 5, i.e., [tex]\(20\)[/tex]), but as neither 4 nor 5 can appear as results of two dice roll, there are no overlaps seemingly.
However, note that sum [tex]\(20\)[/tex] can't be twice of sum of a 6-sided die.
Thus, we have identified:
- Total outcomes divisible by 4: 9
- Total outcomes divisible by 5: 7
Since these sets are mutually exclusive (since their overlap is 0):
[tex]\[ \text {Total outcomes divisible by 4 or 5} = 9 + 7 = 16 \][/tex]
To find sums not divisible by 4 or 5, we subtract these from the total outcomes:
[tex]\[ \text{Total outcomes not divisible by 4 or 5} = 36 - 16 = 20 \][/tex]
Finally, the probability for Event [tex]\(B\)[/tex] is:
[tex]\[ P(B) = \frac{20}{36} \approx 0.56 \][/tex]
Therefore, rounding to two decimal places:
[tex]\[ (a) \, P(A) \approx 0.17 \][/tex]
[tex]\[ (b) \, P(B) \approx 0.56 \][/tex]
We'll start by considering all possible outcomes when rolling two dice. Each die has 6 faces, so when rolling two dice, there are a total of [tex]\(6 \times 6 = 36\)[/tex] possible outcomes.
### (a) Calculating Probability [tex]\(P(A)\)[/tex]
Event [tex]\(A\)[/tex] is the event where the sum of the two dice is greater than 9. We need to determine the successful outcomes for this event.
The possible sums greater than 9 are: [tex]\(10, 11, \)[/tex] and [tex]\(12\)[/tex].
Let's list the combinations for each sum:
- Sum of 10: (4,6), (5,5), (6,4) → 3 outcomes
- Sum of 11: (5,6), (6,5) → 2 outcomes
- Sum of 12: (6,6) → 1 outcome
Adding these together, we get:
[tex]\[ 3 + 2 + 1 = 6 \text{ successful outcomes} \][/tex]
To find the probability, we divide the number of successful outcomes by the total number of outcomes:
[tex]\[ P(A) = \frac{6}{36} = \frac{1}{6} \approx 0.17 \][/tex]
So,
[tex]\[ P(A) \approx 0.17 \][/tex]
### (b) Calculating Probability [tex]\(P(B)\)[/tex]
Event [tex]\(B\)[/tex] is the event where the sum is neither divisible by 4 nor by 5.
First, let's find the totals that are either divisible by 4 or by 5.
- Sums divisible by 4: [tex]\(4, 8, 12\)[/tex]
- Combinations for 4: (1,3), (2,2), (3,1) → 3 outcomes
- Combinations for 8: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 outcomes
- Combinations for 12: (6,6) → 1 outcome
Adding these together, we get:
[tex]\[ 3 + 5 + 1 = 9 \text{ outcomes where sum is divisible by 4} \][/tex]
- Sums divisible by 5: [tex]\(5, 10\)[/tex]
- Combinations for 5: (1,4), (2,3), (3,2), (4,1) → 4 outcomes
- Combinations for 10: (4,6), (5,5), (6,4) → 3 outcomes
Adding these together, we get:
[tex]\[ 4 + 3 = 7 \text{ outcomes where sum is divisible by 5} \][/tex]
Now, let's count the overlap (sums divisible by both 4 and 5, i.e., [tex]\(20\)[/tex]), but as neither 4 nor 5 can appear as results of two dice roll, there are no overlaps seemingly.
However, note that sum [tex]\(20\)[/tex] can't be twice of sum of a 6-sided die.
Thus, we have identified:
- Total outcomes divisible by 4: 9
- Total outcomes divisible by 5: 7
Since these sets are mutually exclusive (since their overlap is 0):
[tex]\[ \text {Total outcomes divisible by 4 or 5} = 9 + 7 = 16 \][/tex]
To find sums not divisible by 4 or 5, we subtract these from the total outcomes:
[tex]\[ \text{Total outcomes not divisible by 4 or 5} = 36 - 16 = 20 \][/tex]
Finally, the probability for Event [tex]\(B\)[/tex] is:
[tex]\[ P(B) = \frac{20}{36} \approx 0.56 \][/tex]
Therefore, rounding to two decimal places:
[tex]\[ (a) \, P(A) \approx 0.17 \][/tex]
[tex]\[ (b) \, P(B) \approx 0.56 \][/tex]
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