Question

1. calculate each probability when rolling 2 number cubes and summing the resulting numbers. explain your calculations.

a. p(prime number)
b. p(greater than 7)
c. p(1)

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

Explanation:

When rolling two number cubes, there are a total of $6 \times 6 = 36$ possible outcomes.

Step1: Identify prime sum outcomes

Prime sums are 2,3,5,7,11.
Count valid pairs:

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

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

Step2: Identify sums >7 outcomes

Sums >7 are 8,9,10,11,12.
Count valid pairs:

• Sum=8: (2,6),(3,5),(4,4),(5,3),(6,2) → 5 outcomes
• Sum=9: (3,6),(4,5),(5,4),(6,3) → 4 outcomes
• Sum=10: (4,6),(5,5),(6,4) → 3 outcomes
• Sum=11: (5,6),(6,5) → 2 outcomes
• Sum=12: (6,6) → 1 outcome

Total outcomes >7: $5+4+3+2+1=15$
Probability: $\frac{15}{36} = \frac{5}{12}$

Step3: Identify sum=1 outcomes

The minimum sum of two cubes is 2, so there are 0 outcomes where the sum is 1.
Probability: $\frac{0}{36} = 0$

Step4: Predict occurrences for 180 tosses

For each sum category, multiply probability by 180:

• Prime number: $180 \times \frac{5}{12} = 75$
• Greater than 7: $180 \times \frac{5}{12} = 75$
• Sum=1: $180 \times 0 = 0$

Answer:

11.
a. $P(\text{prime number}) = \frac{5}{12}$
b. $P(\text{greater than 7}) = \frac{5}{12}$
c. $P(1) = 0$

12.

• Prime number sum: 75 times
• Sum greater than 7: 75 times
• Sum of 1: 0 times