Question

a pair of dice is rolled. find the probability of rolling (a) a sum not more than 3, (b) a sum not less than 6, (c) a sum between 3 and 8 (exclusive). (a) how many possible outcomes are there from rolling two dice? (simplify your answer.)

Explanation:

Step1: Calculate total outcomes

Each die has 6 faces. When rolling two dice, by the multiplication - principle, the total number of possible outcomes is $6\times6 = 36$.

Step2: Find outcomes for sum not more than 3

The possible sums not more than 3 are 2 and 3. For a sum of 2, the outcome is (1,1). For a sum of 3, the outcomes are (1,2) and (2,1). So there are $1 + 2=3$ favorable outcomes. The probability $P_1=\frac{3}{36}=\frac{1}{12}$.

Step3: Find outcomes for sum not less than 6

The possible sums not less than 6 are 6, 7, 8, 9, 10, 11, 12.

• Sum of 6: (1,5), (2,4), (3,3), (4,2), (5,1) - 5 outcomes.
• Sum of 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) - 6 outcomes.
• Sum of 8: (2,6), (3,5), (4,4), (5,3), (6,2) - 5 outcomes.
• Sum of 9: (3,6), (4,5), (5,4), (6,3) - 4 outcomes.
• Sum of 10: (4,6), (5,5), (6,4) - 3 outcomes.
• Sum of 11: (5,6), (6,5) - 2 outcomes.
• Sum of 12: (6,6) - 1 outcome.

The total number of favorable outcomes is $5 + 6+5 + 4+3 + 2+1 = 26$. The probability $P_2=\frac{26}{36}=\frac{13}{18}$.

Step4: Find outcomes for sum between 3 and 8 (exclusive)

The possible sums are 4, 5, 6, 7.

• Sum of 4: (1,3), (2,2), (3,1) - 3 outcomes.
• Sum of 5: (1,4), (2,3), (3,2), (4,1) - 4 outcomes.
• Sum of 6: 5 outcomes.
• Sum of 7: 6 outcomes.

The total number of favorable outcomes is $3 + 4+5 + 6 = 18$. The probability $P_3=\frac{18}{36}=\frac{1}{2}$.

Answer:

(a) $\frac{1}{12}$
(b) $\frac{13}{18}$
(c) $\frac{1}{2}$