Answer:
[tex]\frac{7}{36}[/tex]
Step-by-step explanation:
Two dices are rolled so total number of sample space =6×6=36
sample space
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (22) (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)
The event which has sum 6, (1,5) (2,4) (3,3) (4,2) (5,1)
So the required probability [tex]=\frac{6}{36}=\frac{1}{6}[/tex]
The event which has sum 12 , (6,6)
So the required probability = [tex]\frac{1}{36}[/tex]
So the total probability = [tex]\frac{1}{36}+\frac{1}{6}=\frac{7}{6}[/tex]
