An ordinary (fair) die is a cube with the numbers 1 through 6 on the sides

Question

Imagine that such a die is rolled twice in succession and that the face values of the two rolls are added together. This sum is recorded as the outcome of a single trial of a random experiment.Compute the probability of each of the following events.

Event A

: The sum is greater than 9

.

Event B

: The sum is an odd number.

Answer 1

11 possible outcomes are

2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12

1/36 chance of 2 or 12

2/36 chance or 3 or 11

3/36 chance of 4 or 10

4/36 chance of 5 or 9

5/36 chance of 6 or 8

6/36 chance of 7

Probability of greater than 9 = the probability of 10, 11 or 12 = (3+2+1)/36 = 6/36 = 1/6

P(A) = P(>9) = 1/6

probability of an odd outcome = P(3,5,7,9, or 11) = (2+4+6+4+2)/36 = 18/36 = 1/2

P(odd) = 1/2

5 outcomes are odd (3,5,7,9 and 11), 6 are even (2,4,6,8,10 &12), seemingly making even outcomes more likely, but adding the probabilities of the 5 odd = the sum of the probabilities of the even outcomes

7 is most likely, an odd outcome. 2 and 12 the least likely outcomes

number of ways for each outcome:

2 1 way 1 and 1

3 2 ways, 1 and 2 or 2 and 1

4 3 ways 2 and 2, 1 and 3 or 3 and 1

5 4 ways 3 and 2, 2 and 3, 4 and 1, or 1 and 4

6 5 ways 3 and 3, 4 and 2, 2 and 4, 5 and 1 or 1 and 5

7 6 ways 4 and 3, 3 and 4, 5 and 2, 2 and 5, 6 and 1 or 1 and 6

8 5 ways (same as 6

9 4 ways (same as 5

10 3 ways (same as 4

11 2 ways (same as 3

12 1 way same as 2

total outcomes are 36 different possibilities

each probability is the specific number of ways for the outcome divided by total number of outcomes, 36