Question

In a single throw of two dice, find the probability of a total of at most 10.

Answer 1

The number of possible outcomes is 6 × 6 = 36. 

We write them as given below:  

1, 11, 21, 31, 41, 51, 6

2, 12, 22, 32, 42, 52, 6

3, 13, 23, 33, 43, 53, 6

4, 14, 24, 34, 44, 54, 6 

5, 15, 25, 35, 45, 55, 6

6, 16, 26, 36, 46, 56, 6

n(S) = 36

E = Event of getting a total of at most 10

= {(1, 1),(1, 2),(1, 3),(1, 4),(1, 5),(1, 6),(2, 1),(2, 2),(2, 3),(2, 4),(2, 5),(2, 6),(3, 1),(3, 2),(3, 3),(3, 4),(3, 5),(3, 6),(4, 1),(4, 2),(4, 3),(4, 4),(4, 5),(4, 6),(5, 1),(5, 2),(5, 3),(5, 4),(5, 5),(6, 1),(6, 2),(6, 3),(6, 4)} 

Therefore total number of favorable ways = 33 = n(E) 

Probability of getting a total of at most 10 = `(n(E))/(n(s)) = 33/36 = 11/12`