To find the probabilities of the given events when rolling a fair die twice, we'll follow a structured approach.
### Step 1: Total Possible Outcomes
When a die is rolled twice, each of the first six outcomes (1 to 6) can pair with any of the six outcomes on the second roll. Therefore, the total number of possible outcomes is:
[tex]\[ 6 \times 6 = 36 \][/tex]
### Step 2: Event A - The Sum is Greater than 6
We need to find the pairs [tex]\((i, j)\)[/tex] where [tex]\(i + j > 6\)[/tex].
Here are the valid pairs:
- For [tex]\(i = 1\)[/tex]: (None, as 1 + any number from 1 to 6 is always [tex]\(≤ 6\)[/tex])
- For [tex]\(i = 2\)[/tex]: (5,6)
- For [tex]\(i = 3\)[/tex]: (4,5,6)
- For [tex]\(i = 4\)[/tex]: (3,4,5,6)
- For [tex]\(i = 5\)[/tex]: (2,3,4,5,6)
- For [tex]\(i = 6\)[/tex]: (1,2,3,4,5,6)
Counting these pairs:
- 0 pairs for [tex]\(i = 1\)[/tex]
- 2 pairs for [tex]\(i = 2\)[/tex]
- 3 pairs for [tex]\(i = 3\)[/tex]
- 4 pairs for [tex]\(i = 4\)[/tex]
- 5 pairs for [tex]\(i = 5\)[/tex]
- 6 pairs for [tex]\(i = 6\)[/tex]
Adding these together:
[tex]\[ 0 + 2 + 3 + 4 + 5 + 6 = 20 \][/tex]
So, there are 20 favorable outcomes for Event A.
#### Probability P(A):
[tex]\[ P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} = \frac{20}{36} = \frac{5}{9} \approx 0.58 \][/tex]
(Rounded to two decimal places)
### Step 3: Event B - The Sum is Divisible by 5
We need to find the pairs [tex]\((i, j)\)[/tex] where [tex]\( (i + j) \mod 5 = 0 \)[/tex]. In other words, the sums need to be 5 or 10 (as these are the sums that are divisible by 5 within our range).
Pairs that sum to 5:
- (1,4), (2,3), (3,2), (4,1)
Pairs that sum to 10:
- (4,6), (5,5), (6,4)
Counting these pairs:
- 4 pairs (sum to 5)
- 3 pairs (sum to 10)
Adding these together:
[tex]\[ 4 + 3 = 7 \][/tex]
So, there are 7 favorable outcomes for Event B.
#### Probability P(B):
[tex]\[ P(B) = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} = \frac{7}{36} \approx 0.19 \][/tex]
(Rounded to two decimal places)
### Final Answers:
(a) [tex]\( P(A) = 0.58 \)[/tex]
(b) [tex]\( P(B) = 0.19 \)[/tex]
