Two different dice are thrown together Find the probability of getting a number greater than 3

SOLUTION:

The outcomes when two dice are thrown together 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)

Total number of outcomes = 36 (i) Let A be the event of getting the numbers whose sum is less than 7.

The outcomes in favour of event A are (1, 1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (4,1), (4,2) and (5,1).

Number of favourable outcomes = 15

∴ P(A ) = Number of favourable outcomesTotal number of outcomes= $\frac{15}{36}$ = $\frac{5}{12}$

(ii) Let B be the event of getting the numbers whose product is less than 16. The outcomes in favour of event B 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), (4,1), (4,2), (4,3), (5,1), (5,2), (5,3), (6,1) and (6,2).

Number of favourable outcomes = 25

∴ P(B ) = Number of favourable outcomes/Total number of outcomes= $\frac{25}{36}$ (iii) Let C be the event of getting the numbers which are doublets of odd numbers. The outcomes in favour of event C are (1,1), (3,3) and (5,5). Number of favourable outcomes = 3

∴ P(C ) = Number of favourable outcomes/Total number of outcomes= $\frac{3}{36}$ = $\frac{1}{12}$

When two dice are thrown simultaneously, the possible outcomes can be listed as follows:

Outcome	1	2	3	4	5	6
1	(1, 1)	(1, 2)	(1, 3)	(1, 4)	(1, 5)	(1, 6)
2	(2, 1)	(2, 2)	(2, 3)	(2, 4)	(2, 5)	(2, 6)
3	(3, 1)	(3, 2)	(3, 3)	(3, 4)	(3, 5)	(3, 6)
4	(4, 1)	(4, 2)	(4, 3)	(4, 4)	(4, 5)	(4, 6)
5	(5, 1)	(5, 2)	(5, 3)	(5, 4)	(5, 5)	(5, 6)
6	(6, 1)	(6, 2)	(6, 3)	(6, 4)	(6, 5)	(6, 6)