Question

Two dice are thrown simultaneously. Find the probability of getting:
a total of at least 10

Answer 1

Elementary events associated to the random experiment of throwing two dice are:
(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), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
$\therefore$ Total number of elementary events = 6 X 6 = 36
Let A be the event of getting a total of at least 10 i.e. 10, 11, 12. Then, the elementary events favourable to A are:
(6, 4), (4, 6), (5, 5), (6, 5), (5, 6) and (6, 6)
$\therefore$ Favourable number of elementary events = 6
Hence, required probability = $\frac{6}{36}=\frac{1}{6}$