Question

1. if the number cubes are tossed 180 times, how many times do you predict the following sums would occur?

a. 1
b. 4
c. 9
d. 10
e. 12
f. prime number

Explanation:

Step1: Find total cube outcomes

When tossing 2 number cubes, total possible outcomes: $6 \times 6 = 36$

Step2: Calculate probability for sum=1

No outcomes give sum=1. Probability: $\frac{0}{36}=0$
Expected occurrences: $180 \times 0 = 0$

Step3: Calculate probability for sum=4

Outcomes for sum=4: (1,3),(2,2),(3,1) → 3 outcomes.
Probability: $\frac{3}{36}=\frac{1}{12}$
Expected occurrences: $180 \times \frac{1}{12} = 15$

Step4: Calculate probability for sum=9

Outcomes for sum=9: (3,6),(4,5),(5,4),(6,3) → 4 outcomes.
Probability: $\frac{4}{36}=\frac{1}{9}$
Expected occurrences: $180 \times \frac{1}{9} = 20$

Step5: Calculate probability for sum=10

Outcomes for sum=10: (4,6),(5,5),(6,4) → 3 outcomes.
Probability: $\frac{3}{36}=\frac{1}{12}$
Expected occurrences: $180 \times \frac{1}{12} = 15$

Step6: Calculate probability for sum=12

Outcome for sum=12: (6,6) → 1 outcome.
Probability: $\frac{1}{36}$
Expected occurrences: $180 \times \frac{1}{36} = 5$

Step7: Calculate probability for prime sum

Prime sums (2,3,5,7,11):

• Sum=2: (1,1) →1
• Sum=3: (1,2),(2,1) →2
• Sum=5: (1,4),(2,3),(3,2),(4,1) →4
• Sum=7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) →6
• Sum=11: (5,6),(6,5) →2

Total prime outcomes: $1+2+4+6+2=15$
Probability: $\frac{15}{36}=\frac{5}{12}$
Expected occurrences: $180 \times \frac{5}{12} = 75$

Answer:

a. 0
b. 15
c. 20
d. 15
e. 5
f. 75