Question

two 6 - sided dice are rolled. what is the probability that the sum is greater than 6?

Explanation:

Step1: Find total number of outcomes

Each die has 6 sides. When two dice are rolled, the total number of outcomes is $6\times6 = 36$ according to the multiplication - principle.

Step2: Find number of favorable outcomes

The possible sums of two dice range from 2 (1 + 1) to 12 (6+6). The sums less than or equal to 6 are:
Sum = 2: (1,1) - 1 way
Sum = 3: (1,2), (2,1) - 2 ways
Sum = 4: (1,3), (3,1), (2,2) - 3 ways
Sum = 5: (1,4), (4,1), (2,3), (3,2) - 4 ways
Sum = 6: (1,5), (5,1), (2,4), (4,2), (3,3) - 5 ways
The number of outcomes with sum less than or equal to 6 is $1 + 2+3 + 4+5=\frac{5\times(5 + 1)}{2}=15$
The number of outcomes with sum greater than 6 is $36-15 = 21$

Step3: Calculate probability

The probability $P$ that the sum is greater than 6 is $P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}=\frac{21}{36}=\frac{7}{12}$

Answer:

$\frac{7}{12}$