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 8. Event B: The sum is not divisible by 4 and not divisible by 5.

Answer 1

Event A:

Pairs that make the sum greater than 8:

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

Each of these pairs is equiprobable and which i will be using throughout the exercise:

$p(pair) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$

$p(a) = (\frac{1}{36} \times 2) \times 4 + \frac{1}{36} \times 2 = \frac{5}{18}$

The 2 (2!) is to account for permutations like ( 3 , 6 ) and ( 6 , 3 ) and the 4 are the different combinations that are composed of different numbers.

The 2 (2!) is to account for permutations like ( 3 , 6 ) and ( 6 , 3 ) and the 4 are the different combinations that are composed of different numbers.Whereas combinations like ( 5 , 5 ) cannot be switched in order, so we do not multiply by 2!

Event B

Pairs that yield a sum divisible by 4 or 5: 0,4,5,8,10,12,

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

$p(not \: b) = ( \frac{1}{36} \times 2) \times 6 + \frac{1}{36} \times 4 = \frac{4}{9}$

$p(b) = 1 - p(not \: b) = 1 - \frac{4}{9} = \frac{5}{9}$