In a single throw of two dice, find the probability of getting
(i) A doublet
(ii) A number less than 5 on each die
(iii) Sum less than 9
Answers
Given:
In a single throw of two dice
To find:
probability of getting
- A doublet
- A number less than 5 on each die
- Sum less than 9
Solution:
The sample for the given condition is
(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 the sample space is 36
A) Doublets are:
(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)
Number of the favourable outcome is 6
P(E) = Number of the favourable outcome/ total number of the outcomes
P(E) = 6/36
P(E) = 1/6
B) A number less than 5 on each die
(1, 1)(1, 2)(1, 3)(1, 4)
(2, 1)(2, 2)(2, 3)(2, 4)
(3, 1)(3, 2)(3, 3)(3, 4)
(4, 1)(4, 2)(4, 3)(4, 4)
Number of the favourable outcome is 16
P(E) = Number of the favourable outcome/ total number of the outcomes
P(E) = 16/36
P(E) = 4/9
C) Sum less than 9
(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)(4, 4)
(5, 1)(5, 2)(5, 3)
(6, 1)(6, 2)
Number of the favourable outcome is 26
P(E) = Number of the favourable outcome/ total number of the outcomes
P(E) = 26/36