Question

Two dice are thrown simultaneously. Find the probability of getting:
same number on both dice i.e. a doublet.

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 the same number on both dice. Then, elementary events favourable to event A are:
(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), and (6, 6).
$\therefore$ favourable number of elementary events = 6
Hence, required probability = $\frac{6}{36}=\frac{1}{6}$.