The outcomes when two dice are thrown together are:
(1, 1), (1, 2), (2, 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)
- Let A be the event of getting the number 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 outcones = 15
$\therefore\ \text{P(A)}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$
$=\frac{15}{36}$
$=\frac{5}{12}$
- Let B be the event of getting the number 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).
- Let C be the event of getting the number 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
$\therefore\ \text{P(A)}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$
$=\frac{3}{36}$
$=\frac{1}{12}$