An ordinary fair die is a cube with the numbers 1 through 6 on the sides represented by painted spots imagine that such a die is rolled twice in succession and that the face values of the two roles are added together the sum is recorded as the outcome of a single trial of the random experiment. compute the probability of each of the following events: event a: the sum is greater than 6 event b: the sum is an odd number.

P(A)= 21/36

P(B)=1/2

Step-by-step explanation:

Given:

An ordinary Fair die is a cube with the numbers 1 through 6 on the sides represented by painted spots imagine that such a die is rolled twice

So the sample space of rolling two dices is:

(1,1) (2,1) (3,1) (4,1) (5,1) (6,1)

(1,2) (2,2) (3,2) (4,2) (5,2) (6,2)

(1,3) (2,3) (3,3) (4,3) (5,3) (6,3)

(1,4) (2,4) (3,4) (4,4) (5,4) (6,4)

(1,5) (2,5) (3,5) (4,5) (5,5) (6,5)

(1,6) (2,6) (3,6) (4,6) (5,6) (6,6)

Total number of sets=36

1)Now Finding probability of

Event A: The sum is greater than 6

From the above sample space sets that have the sum >6 are

(6,1)

(5,2) (6,2)

(4,3) (5,3) (6,3)

(3,4) (4,4) (5,4) (6,4)

(2,5) (3,5) (4,5) (5,5) (6,5)

(1,6) (2,6) (3,6) (4,6) (5,6) (6,6)

there are 21 sets

hence probability of Event A= 21/36

P(A)= 21/36

2) Now Finding probability of

Event B: The sum is an odd number

From the above sample space sets that have the sum an odd number are

(2,1)(4,1) (6,1)

(1,2)(3,2) (5,2)

(2,3) (4,3)(6,3)

(1,4) (3,4) (5,4)

(2,5) (4,5) (6,5)

(1,6) (3,6) (5,6)

there are 18 sets

hence the probability of Event B = 18/36

         P(B)= 1/2 !

there is no question content here, but you for the free points! : d