Two different dice are tossed together. Find the probability that (i) the number of each dice is
even and (ii) the sum of the numbers, appearing on the two dice is 10 (iii) the number on the
second die is twice the first die (iv) a doublet (v) the sum of the numbers appearing on the two
dice is more than 9 (vi) difference of the numbers on the two dice is 2 (vii) product of the
numbers on the dice is less than 9 (viii) product of the numbers on the dice is 12
Answers
Answer:
Step-by-step explanation:
The possible sample space 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 possible outcomes = 36
Probability = Number of favorable outcomes/ Total number of outcomes
(i) the number of each dice is even
Possible number of favorable outcomes = (2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6) = 9
Required probability = 9/36 = 1/4
(ii) the sum of numbers appearing on dice = 10
Possible number of favorable outcomes = (4,6) (5,5) (6,6) = 3
Required probability = 3/36 = 1/12
(iii) the number on second dice is twice the first dice
Possible number of favorable outcomes = (2,4) (1,2) (3,6) = 3
Required probability = 3/36 = 1/12
(iv) a doublet
Possible number of favorable outcomes = (1,1) (2,2) (3,3) (4,4),(5,5) (6,6) = 6
Required probability = 6/36 = 1/6
(v) the sum of the numbers in both dice is greater than 9
Possible number of favorable outcomes = (6,4) (6,5) (6,6) (5,5) (5,6) (4,6) = 6
Required probability = 6/36 = 1/6
(vi) difference of the numbers on two dice is 2
Possible number of favorable outcomes = (1,3) (2,4) (3,5) (4,6) (5,3) (4,6) (3,1) (4,2) = 8
Required probability = 8/36 = 2/9
(vii) product of the numbers on the dice is less than 9
Possible number of favorable outcomes = (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (3,1) (3,2) (4,1) (5,1) (6,1) = 15
Required probability = 15/36 = 5/12
(viii) product of the numbers on the dice is 12
Possible number of favorable outcomes = (2,6) (3,4) (4,3) (6,2) = 4
Required probability = 4/36 = 1/9
Answer:
Step-by-step explanation:
(i) probability that the number of each dice is even
= (3/6)(3/6)
= (1/2)(1/2)
= 1/4
ii) the sum of the numbers, appearing on the two dice is 10
3/36
= 1/12
iii) the number on the second die is twice the first die
3/36
= 1/12
(iv) a doublet = 6/36 = 1/6
v) the sum of the numbers appearing on the two
dice is more than 9
= 6/36 = 1/6
vi) difference of the numbers on the two dice is 2
8/36 = 2/9