Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Discover comprehensive answers to your questions from knowledgeable professionals on our user-friendly platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
To find the probabilities of the given events when rolling a fair die twice, we will follow a sequence of logical steps:
### Step 1: List all possible outcomes
When you roll a die twice, each roll has 6 possible outcomes. Therefore, the total number of outcomes for rolling the die twice is:
[tex]\[ 6 \times 6 = 36. \][/tex]
The possible sums (2 to 12) and their frequencies are counted as below:
- Sum = 2 : (1,1)
- Sum = 3 : (1,2), (2,1)
- Sum = 4 : (1,3), (2,2), (3,1)
- Sum = 5 : (1,4), (2,3), (3,2), (4,1)
- Sum = 6 : (1,5), (2,4), (3,3), (4,2), (5,1)
- Sum = 7 : (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
- Sum = 8 : (2,6), (3,5), (4,4), (5,3), (6,2)
- Sum = 9 : (3,6), (4,5), (5,4), (6,3)
- Sum = 10 : (4,6), (5,5), (6,4)
- Sum = 11 : (5,6), (6,5)
- Sum = 12 : (6,6)
### Step 2: Compute [tex]$P(A)$[/tex], where Event [tex]$A$[/tex] is that the sum is greater than 7
Firstly, identify which sums are greater than 7: 8, 9, 10, 11, and 12.
- Number of ways to get each sum:
- Sum = 8 : 5 ways
- Sum = 9 : 4 ways
- Sum = 10 : 3 ways
- Sum = 11 : 2 ways
- Sum = 12 : 1 way
Total favorable outcomes for Event [tex]$A$[/tex] = 5 (8) + 4 (9) + 3 (10) + 2 (11) + 1 (12) = 15.
Probability of Event [tex]$A$[/tex]:
[tex]\[ P(A) = \frac{\text{Number of favorable outcomes for Event } A}{\text{Total number of outcomes}} = \frac{15}{36} \approx 0.42 \][/tex]
Thus:
[tex]\[ P(A) = 0.42 \][/tex]
### Step 3: Compute [tex]$P(B)$[/tex], where Event [tex]$B$[/tex] is that the sum is not divisible by 4 and not divisible by 5
Identify the sums that are not divisible by 4 and not by 5:
- Exclude sums divisible by 4: 4, 8, 12
- Exclude sums divisible by 5: 5, 10
Therefore, sums that are not divisible by 4 or 5: 2, 3, 6, 7, 9, 11
Number of ways to get each sum:
- Sum = 2 : 1 way
- Sum = 3 : 2 ways
- Sum = 6 : 5 ways
- Sum = 7 : 6 ways
- Sum = 9 : 4 ways
- Sum = 11 : 2 ways
Total favorable outcomes for Event [tex]$B$[/tex] = 1 (2) + 2 (3) + 5 (6) + 6 (7) + 4 (9) + 2 (11) = 20.
Probability of Event [tex]$B$[/tex]:
[tex]\[ P(B) = \frac{\text{Number of favorable outcomes for Event } B}{\text{Total number of outcomes}} = \frac{20}{36} \approx 0.56 \][/tex]
Thus:
[tex]\[ P(B) = 0.56 \][/tex]
### Summary of Results:
(a) \( P(A) = 0.42 \)
(b) [tex]\( P(B) = 0.56 \)[/tex]
### Step 1: List all possible outcomes
When you roll a die twice, each roll has 6 possible outcomes. Therefore, the total number of outcomes for rolling the die twice is:
[tex]\[ 6 \times 6 = 36. \][/tex]
The possible sums (2 to 12) and their frequencies are counted as below:
- Sum = 2 : (1,1)
- Sum = 3 : (1,2), (2,1)
- Sum = 4 : (1,3), (2,2), (3,1)
- Sum = 5 : (1,4), (2,3), (3,2), (4,1)
- Sum = 6 : (1,5), (2,4), (3,3), (4,2), (5,1)
- Sum = 7 : (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
- Sum = 8 : (2,6), (3,5), (4,4), (5,3), (6,2)
- Sum = 9 : (3,6), (4,5), (5,4), (6,3)
- Sum = 10 : (4,6), (5,5), (6,4)
- Sum = 11 : (5,6), (6,5)
- Sum = 12 : (6,6)
### Step 2: Compute [tex]$P(A)$[/tex], where Event [tex]$A$[/tex] is that the sum is greater than 7
Firstly, identify which sums are greater than 7: 8, 9, 10, 11, and 12.
- Number of ways to get each sum:
- Sum = 8 : 5 ways
- Sum = 9 : 4 ways
- Sum = 10 : 3 ways
- Sum = 11 : 2 ways
- Sum = 12 : 1 way
Total favorable outcomes for Event [tex]$A$[/tex] = 5 (8) + 4 (9) + 3 (10) + 2 (11) + 1 (12) = 15.
Probability of Event [tex]$A$[/tex]:
[tex]\[ P(A) = \frac{\text{Number of favorable outcomes for Event } A}{\text{Total number of outcomes}} = \frac{15}{36} \approx 0.42 \][/tex]
Thus:
[tex]\[ P(A) = 0.42 \][/tex]
### Step 3: Compute [tex]$P(B)$[/tex], where Event [tex]$B$[/tex] is that the sum is not divisible by 4 and not divisible by 5
Identify the sums that are not divisible by 4 and not by 5:
- Exclude sums divisible by 4: 4, 8, 12
- Exclude sums divisible by 5: 5, 10
Therefore, sums that are not divisible by 4 or 5: 2, 3, 6, 7, 9, 11
Number of ways to get each sum:
- Sum = 2 : 1 way
- Sum = 3 : 2 ways
- Sum = 6 : 5 ways
- Sum = 7 : 6 ways
- Sum = 9 : 4 ways
- Sum = 11 : 2 ways
Total favorable outcomes for Event [tex]$B$[/tex] = 1 (2) + 2 (3) + 5 (6) + 6 (7) + 4 (9) + 2 (11) = 20.
Probability of Event [tex]$B$[/tex]:
[tex]\[ P(B) = \frac{\text{Number of favorable outcomes for Event } B}{\text{Total number of outcomes}} = \frac{20}{36} \approx 0.56 \][/tex]
Thus:
[tex]\[ P(B) = 0.56 \][/tex]
### Summary of Results:
(a) \( P(A) = 0.42 \)
(b) [tex]\( P(B) = 0.56 \)[/tex]
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.