Question

1 2 3 4 5 6
first number cube
1 1,1 1,2 1,3 1,4 1,5 1,6
2 2,1 2,2 2,3 2,4 2,5 2,6
3 3,1 3,2 3,3 3,4 3,5 3,6
4 4,1 4,2 4,3 4,4 4,5 4,6
5 5,1 5,2 5,3 5,4 5,5 5,6
6 6,1 6,2 6,3 6,4 6,5 6,6
which outcome would allow asher to have a higher probability of winning the game? select three options.
☐ rolling a sum of 4
☐ rolling a sum of 9
☐ rolling a sum that is less than 5
☐ rolling a sum that is greater than 5 but less than 7
☐ rolling a sum that is greater than 9 but less than 11

Explanation:

Step1: Count total outcomes

Total possible outcomes when rolling two dice: $6 \times 6 = 36$

Step2: Calculate prob of sum=4

Find pairs that sum to 4: (1,3), (2,2), (3,1). Number of outcomes = 3.
Probability: $\frac{3}{36} = \frac{1}{12}$

Step3: Calculate prob of sum=9

Find pairs that sum to 9: (3,6), (4,5), (5,4), (6,3). Number of outcomes = 4.
Probability: $\frac{4}{36} = \frac{1}{9}$

Step4: Calculate prob of sum<5

Sums are 2,3,4. Pairs: (1,1), (1,2), (2,1), (1,3), (2,2), (3,1). Number of outcomes = 6.
Probability: $\frac{6}{36} = \frac{1}{6}$

Step5: Calculate prob of 5<sum<7

Only sum=6. Pairs: (1,5), (2,4), (3,3), (4,2), (5,1). Number of outcomes = 5.
Probability: $\frac{5}{36}$

Step6: Calculate prob of 9<sum<11

Only sum=10. Pairs: (4,6), (5,5), (6,4). Number of outcomes = 3.
Probability: $\frac{3}{36} = \frac{1}{12}$

Step7: Compare probabilities

Order from highest to lowest: $\frac{1}{6} > \frac{1}{9} > \frac{5}{36} > \frac{1}{12} = \frac{1}{12}$

Answer:

• rolling a sum of 9
• rolling a sum that is less than 5
• rolling a sum that is greater than 5 but less than 7