Data Analysis and Probabilit

Question

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 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 divisible by 3

 or 6

 (or both).

Write your answers as exact fractions.

Answer 1

For the first die there a 6 outcomes. There are also 6 outcomes for the second die giving 6 x 6 = 36 total outcomes.

Event A: The sum is greater than 9:

6 + 4 = 10

5 + 5 = 10

4 + 6 = 10

5 + 6 = 11

6 + 5 = 11

6 + 6 = 12

6/36 = 1/6

Event B: The sum is divisible by 3 or 6. There are 12 possible outcomes where the sum of the two dice are divisible by 3 or 6.

12/36 = 1/3

Answer 2

There are 36 possible results of this experiment ... considering the results of the roll of the first die and the role of the second die.

First roll Second role

1 1

1 2

1 3

1 4

1 5

1 6

2 1

2 2 and so on from there

Event B - sum divisible by 3, 6 or both

The possible totals of the two dice that would be divisible by 3,6, or both would be

3, 6, 9 , 12

# of ways to sum 3 = 2 (1,2) (2,1)

# of ways to sum 6 = 5 (1,5) (5,1) (2,4) (4,2) (3,3)

# of ways to sum 9 = 2 (5,4) (4,5) (3,6) (6,3)

# of ways to sum 12 = 1 (6,6)

Total events that are divisible by 3,6, or both = 12 /36 total possible event results

Answer is 1/3

Event A - sum greater than 9

Results that would create this 6 on first roll 4 on second (6,4)

also (6,5), (6.6)

Five on first roll sum greater than 9 (5,5) (5,6)

Four on first roll sum greater than 9 (4,6)

# of successful rolls 6/36 total rolls = 1/6

Answer 3

The total number of possible sums of the two die are 6*6 = 36

For event A, we need to how many cases are where the two die sum to a value greater than 9.

Since the largest number on a die is 6, the lowest number on one of the two die has to be 4 and the largest sum is when both die are 6.

The sum of the values must be either 10, 11 or 12.

Here are the values of the die that add to sum greater than 9 (10,11,12):

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

Since there are 6 instances where the sum exceeds 9, the probability = 6/total possible sums = 6/36=1/6

For event B, sum must be divisible by 3, 6 or both. Those happen when the sum of the two die is 3, 6, 9 or 12:

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

Since there are 12 instances where the sums are divisible by 3 or 6 or both, the probability = 10/36 = 5/18