Question

a pair of dice is rolled. find the probability of rolling a) a sum not more than 6, b) a sum not less than 4, c) a sum between 3 and 8 (exclusive)

Explanation:

Step1: Calculate total outcomes

When rolling two dice, each die has 6 possible outcomes. So the total number of outcomes when rolling two dice is $6\times6 = 36$.

Step2: Find outcomes for sum not more than 6

The possible sums not more than 6 are 2, 3, 4, 5, 6.
For sum = 2: (1,1) - 1 outcome.
For sum = 3: (1,2), (2,1) - 2 outcomes.
For sum = 4: (1,3), (2,2), (3,1) - 3 outcomes.
For sum = 5: (1,4), (2,3), (3,2), (4,1) - 4 outcomes.
For sum = 6: (1,5), (2,4), (3,3), (4,2), (5,1) - 5 outcomes.
Total number of outcomes with sum not more than 6 is $1 + 2+3 + 4+5=15$.
Probability = $\frac{15}{36}=\frac{5}{12}$.

Step3: Find outcomes for sum not less than 4

The opposite of sum not less than 4 is sum equal to 2 or 3.
Sum = 2 has 1 outcome and sum = 3 has 2 outcomes. So there are $1 + 2=3$ outcomes with sum 2 or 3.
Number of outcomes with sum not less than 4 is $36-(1 + 2)=33$.
Probability = $\frac{33}{36}=\frac{11}{12}$.

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

The possible sums are 4, 5, 6, 7.
Sum = 4 has 3 outcomes, sum = 5 has 4 outcomes, sum = 6 has 5 outcomes, sum = 7 has 6 outcomes.
Total number of outcomes is $3+4 + 5+6=18$.
Probability = $\frac{18}{36}=\frac{1}{2}$.

Answer:

a) $\frac{5}{12}$
b) $\frac{11}{12}$
c) $\frac{1}{2}$