Sum      Number of ways to get this sum

2                   1

3                   2

4                   3

5                   4

6                   5

7                   6

8                   5

9                   4

10                  3

11                  2

12                  1




a)   Pr(sum is 9) = 4/36 = 1/9

b)   Pr(sum > 10) = (2+1)/36 = 3/36 = 1/12

c)   Pr(sum < 6) =  (1+2+3+4)/36 = 10/36 = 5/18



 

Answer by MathLover1(20849)   (Show Source): You can put this solution on YOUR website!
  





Two dice are rolled, find the probabilities of the following events:



a.) the sum of the dice is 

b.) the sum is  than 

c.) the sum is  than 





Solution:



The sample space of two dice is shown below.

 =  (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)  ->





a) sum equal to 



Let  be the event "sum equal to ". There are  outcomes ((3,6),(4,5),(5,4),(6,3)) which correspond to a sum equal to , hence





b) 

sum  than 

Let be the event "sum  than ". There are   outcomes  ((5,6),(6,5),(6,6)) which correspond to a sum  than , hence







c.) the sum is less than 



Let  be the event “ the sum is  than  “. There are   outcomes ((1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)) which correspond to a sum  than , hence





 

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