two fair six-sided dice are rolled and the sum of the dots on the top faces is recorded.
a) complete the table, showing the number of ways each sum can occur
sum: 1 2 3 4 5 6 7 8 9 10 11 12
ways:1 2 3
b)use the table to find the probability of the following events.
A:the sum is prime
B:the sum is a divisor of 12
C:the sum is a power of 2
D:the sum is greater than 3
First of all, they cannot sum to 1. What table are you completing?
the one that says sum: and ways: that is what the book says!
Two fair six-sided dice are rolled and the sum of the dots on the top faces is recorded. Use the table, showing the number of ways each sum can occur, to find the probability of the following events.
SUM
2
3
4
5
6
7
8
9
10
11
12
WAYS
1
2
3
4
5
6
5
4
3
2
1
Give exact answers in fraction form.
A: The sum is prime.
Entry field with incorrect answer now contains modified data
B: The sum is a divisor of 12.
Entry field with incorrect answer
C: The sum is a power of 2.
Entry field with incorrect answer
D: The sum is greater than 3.
Entry field with incorrect answer
Two fair six-sided dice are rolled and the sum of the dots on the top faces is recorded. Use the table, showing the number of ways each sum can occur, to find the probability of the following events.
SUM
2
3
4
5
6
7
8
9
10
11
12
WAYS
1
2
3
4
5
6
5
4
3
2
1
Give exact answers in fraction form.
A: The sum is prime.
Entry field with incorrect answer now contains modified data
B: The sum is a divisor of 12.
Entry field with incorrect answer
C: The sum is a power of 2.
Entry field with incorrect answer
D: The sum is greater than 3.
Entry field with incorrect answer
To complete the table, we need to calculate the number of ways each sum can occur when two fair six-sided dice are rolled.
a) The easiest way to determine the number of ways a sum can occur is by listing all the possible combinations of rolls that add up to that sum. Here is a breakdown of the possible combinations for each sum:
Sum: 1
Ways: (1, 1)
Sum: 2
Ways: (1, 2), (2, 1)
Sum: 3
Ways: (1, 3), (2, 2), (3, 1)
Sum: 4
Ways: (1, 4), (2, 3), (3, 2), (4, 1)
Sum: 5
Ways: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1)
Sum: 6
Ways: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)
Sum: 7
Ways: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2)
Sum: 8
Ways: (3, 6), (4, 5), (5, 4), (6, 3)
Sum: 9
Ways: (4, 6), (5, 5), (6, 4)
Sum: 10
Ways: (5, 6), (6, 5)
Sum: 11
Ways: (6, 6)
Sum: 12
Ways: (6, 6)
b) Now let's use the table to find the probability of the following events:
A: The sum is prime. There are 4 prime numbers (2, 3, 5, 7) in the range of possible sums (2 to 12). Looking at the table, we can see that there are 6 ways to get a sum of 2 or 3. Therefore, the probability of the sum being prime is (6/36) or (1/6).
B: The sum is a divisor of 12. Divisors of 12 are 1, 2, 3, 4, 6, and 12. From the table, we can see that there are 6 ways to get a sum of 2 or 12, 8 ways to get a sum of 3 or 9, and 10 ways to get a sum of 4, 6, or 8. Therefore, the total number of ways to get a sum that is a divisor of 12 is (6 + 8 + 10 = 24). The probability of the sum being a divisor of 12 is (24/36) or (2/3).
C: The sum is a power of 2. The power of 2 values in the range of possible sums (2 to 12) are 2, 4, 8, and 12. From the table, we can see that there are 1 way to get a sum of 2, 3 ways to get a sum of 4, 5 ways to get a sum of 8, and 2 ways to get a sum of 12. Therefore, the total number of ways to get a sum that is a power of 2 is (1 + 3 + 5 + 2 = 11). The probability of the sum being a power of 2 is (11/36).
D: The sum is greater than 3. To find this probability, we need to calculate the probability of all sums less than or equal to 3 and subtract it from 1. From the table, we can see that there are 1 + 2 + 3 = 6 ways to get sums of 1, 2, or 3. Therefore, the probability of the sum being greater than 3 is (1 - 6/36) or (5/6).