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 succe

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 6.
Event B: The sum is divisible by 2 or 6 (or both).
Write your answers as fractions.
(a) P (A) =
(b) P (B) =

Answer 1

P(A) = 21/36 = 7/12

P(B) = 21/36 = 7/12

Explanation:

(a) To find the probability of Event A (the sum is greater than 6), we can list all the possible outcomes of rolling two dice and adding their face values together. There are 36 possible outcomes, as each of the six sides on the first die can be combined with each of the six sides on the second die. We can then count the number of outcomes that have a sum greater than 6:

2: impossible, as the minimum sum is 1+1=2

3: 2 outcomes (1+2, 2+1)

4: 3 outcomes (1+3, 2+2, 3+1)

5: 4 outcomes (1+4, 2+3, 3+2, 4+1)

6: 5 outcomes (1+5, 2+4, 3+3, 4+2, 5+1)

7: 6 outcomes (1+6, 2+5, 3+4, 4+3, 5+2, 6+1)

8: 5 outcomes (2+6, 3+5, 4+4, 5+3, 6+2)

9: 4 outcomes (3+6, 4+5, 5+4, 6+3)

10: 3 outcomes (4+6, 5+5, 6+4)

11: 2 outcomes (5+6, 6+5)

12: 1 outcome (6+6)

Thus, there are 6+5+4+3+2+1 = 21 outcomes with a sum greater than 6. Therefore, the probability of Event A is:

P(A) = 21/36 = 7/12

(b) To find the probability of Event B (the sum is divisible by 2 or 6), we can again list all the possible outcomes and count the number of outcomes that meet the criteria:

Divisible by 2: 18 outcomes (1+1, 1+3, 1+5, 2+2, 2+4, 2+6, 3+1, 3+3, 3+5, 4+2, 4+4, 4+6, 5+1, 5+3, 5+5, 6+2, 6+4, 6+6)

Divisible by 6: 4 outcomes (1+5, 2+4, 4+2, 5+1)

Note that the outcomes that are divisible by 6 are also divisible by 2, so we need to count them only once. Thus, there are 18+4-1=21 outcomes that are divisible by 2 or 6. Therefore, the probability of Event B is:

P(B) = 21/36 = 7/12