Question

Two dice are thrown simultaneously. Given that sum of the numbers is NOT more than 5, what is the probability that sum is more than 3?

7

10


3

10


5

6


3

4


9

10

Answer 1

@perl

Answer 2

n(S)= 36
Let A denotes the event of getting even number
S THE SUM
Ii.e, 2,4,6,8,10 12

Hence A={( 1,1 ),(1,3 ) (1,5 ) (2,2 ) , (2,4 ) , (2,6 ), ( 4,4 ), (4,6 ), ( 4,2 ),
(5,3 ),( 3,5 ),( 6,6 ), (6,4 ),.( 5,1 ), (3,1 ) ,( 6,2 ),( 3,3 ), (5,5 ) },
n(A)= 18

thus,required probability= P(A)= n(A)/n(S)= 18/36= 1/2 =50%

Answer 3

Hope you understand all dat