Answer :
Certainly! Let’s solve the problem step by step.
Given:
- We roll a fair die twice and sum the face values.
- An ordinary fair die has 6 faces, numbered from 1 to 6.
### Step 1: Total Number of Possible Outcomes
When rolling two dice, each die has 6 faces. Therefore, the total number of possible outcomes when rolling two dice is:
[tex]\[ 6 \times 6 = 36 \][/tex]
### Step 2: Calculate the Probability of Event [tex]$A$[/tex]
Event [tex]$A$[/tex]: The sum is greater than 8.
Let us count the favorable outcomes where the sum of the two dice is greater than 8:
- Possible sums greater than 8 are: 9, 10, 11, and 12.
Possible combinations for each sum:
- Sum = 9: (3,6), (4,5), (5,4), (6,3) → 4 outcomes
- Sum = 10: (4,6), (5,5), (6,4) → 3 outcomes
- Sum = 11: (5,6), (6,5) → 2 outcomes
- Sum = 12: (6,6) → 1 outcome
So, the total number of favorable outcomes for event [tex]$A$[/tex] is:
[tex]\[ 4 + 3 + 2 + 1 = 10 \][/tex]
The probability of event [tex]$A$[/tex] is:
[tex]\[ P(A) = \frac{\text{Number of favorable outcomes for } A}{\text{Total number of possible outcomes}} = \frac{10}{36} \][/tex]
Rounding to two decimal places:
[tex]\[ P(A) \approx 0.28 \][/tex]
### Step 3: Calculate the Probability of Event [tex]$B$[/tex]
Event [tex]$B$[/tex]: The sum is not divisible by 4.
Let us count the favorable outcomes where the sum of the two dice is not divisible by 4:
- Possible sums: 2 through 12
- Sums that are divisible by 4 are: 4, 8, 12
Possible combinations for each sum:
- Sum = 4: (1,3), (2,2), (3,1) → 3 outcomes
- Sum = 8: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 outcomes
- Sum = 12: (6,6) → 1 outcome
So, the total number of outcomes where the sum is divisible by 4:
[tex]\[ 3 + 5 + 1 = 9 \][/tex]
Thus, the number of favorable outcomes for event [tex]$B$[/tex] (not divisible by 4) is:
[tex]\[ 36 - 9 = 27 \][/tex]
The probability of event [tex]$B$[/tex] is:
[tex]\[ P(B) = \frac{\text{Number of favorable outcomes for } B}{\text{Total number of possible outcomes}} = \frac{27}{36} \][/tex]
Rounding to two decimal places:
[tex]\[ P(B) \approx 0.75 \][/tex]
### Answer Summary:
(a) \( P(A) = 0.28 \)
(b) [tex]\( P(B) = 0.75 \)[/tex]
Given:
- We roll a fair die twice and sum the face values.
- An ordinary fair die has 6 faces, numbered from 1 to 6.
### Step 1: Total Number of Possible Outcomes
When rolling two dice, each die has 6 faces. Therefore, the total number of possible outcomes when rolling two dice is:
[tex]\[ 6 \times 6 = 36 \][/tex]
### Step 2: Calculate the Probability of Event [tex]$A$[/tex]
Event [tex]$A$[/tex]: The sum is greater than 8.
Let us count the favorable outcomes where the sum of the two dice is greater than 8:
- Possible sums greater than 8 are: 9, 10, 11, and 12.
Possible combinations for each sum:
- Sum = 9: (3,6), (4,5), (5,4), (6,3) → 4 outcomes
- Sum = 10: (4,6), (5,5), (6,4) → 3 outcomes
- Sum = 11: (5,6), (6,5) → 2 outcomes
- Sum = 12: (6,6) → 1 outcome
So, the total number of favorable outcomes for event [tex]$A$[/tex] is:
[tex]\[ 4 + 3 + 2 + 1 = 10 \][/tex]
The probability of event [tex]$A$[/tex] is:
[tex]\[ P(A) = \frac{\text{Number of favorable outcomes for } A}{\text{Total number of possible outcomes}} = \frac{10}{36} \][/tex]
Rounding to two decimal places:
[tex]\[ P(A) \approx 0.28 \][/tex]
### Step 3: Calculate the Probability of Event [tex]$B$[/tex]
Event [tex]$B$[/tex]: The sum is not divisible by 4.
Let us count the favorable outcomes where the sum of the two dice is not divisible by 4:
- Possible sums: 2 through 12
- Sums that are divisible by 4 are: 4, 8, 12
Possible combinations for each sum:
- Sum = 4: (1,3), (2,2), (3,1) → 3 outcomes
- Sum = 8: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 outcomes
- Sum = 12: (6,6) → 1 outcome
So, the total number of outcomes where the sum is divisible by 4:
[tex]\[ 3 + 5 + 1 = 9 \][/tex]
Thus, the number of favorable outcomes for event [tex]$B$[/tex] (not divisible by 4) is:
[tex]\[ 36 - 9 = 27 \][/tex]
The probability of event [tex]$B$[/tex] is:
[tex]\[ P(B) = \frac{\text{Number of favorable outcomes for } B}{\text{Total number of possible outcomes}} = \frac{27}{36} \][/tex]
Rounding to two decimal places:
[tex]\[ P(B) \approx 0.75 \][/tex]
### Answer Summary:
(a) \( P(A) = 0.28 \)
(b) [tex]\( P(B) = 0.75 \)[/tex]
