Two fair dice, A and B, each with faces numbered 1 to 6, are thrown together. We need to: a) Construct a table showing all the equally likely outcomes. b) From the table, list the pairs of numbers on the two dice for which the sum is: i) 5, ii) 10, iii) more than 10, iv) at least 10. c) Find the probability that the two dice show: i) different scores, ii) same scores. d) Find the probability that the sum of the numbers on the two dice is: i) 5, ii) 10, iii) more than 10, iv) at least 10.

1. Problem Description

Two fair dice, A and B, each with faces numbered 1 to 6, are thrown together. We need to:

a) Construct a table showing all the equally likely outcomes.

b) From the table, list the pairs of numbers on the two dice for which the sum is: i) 5, ii) 10, iii) more than 10, iv) at least

1

0. c) Find the probability that the two dice show: i) different scores, ii) same scores.

d) Find the probability that the sum of the numbers on the two dice is: i) 5, ii) 10, iii) more than 10, iv) at least

1

0.

2. Solution Steps

a) Constructing the table:

The possible outcomes when throwing two dice are:

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

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

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

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

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

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

There are a total of 36 possible outcomes.

b) Listing pairs with the specified sums:

i) Sum is 5: (1,4), (2,3), (3,2), (4,1)

ii) Sum is 10: (4,6), (5,5), (6,4)

iii) Sum is more than 10: (5,6), (6,5), (6,6)

iv) Sum is at least 10: (4,6), (5,5), (6,4), (5,6), (6,5), (6,6)

c) Finding probabilities:

i) Different scores: There are 6 outcomes with the same scores (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). Therefore, there are 36 - 6 = 30 outcomes with different scores.

$P(\text{different scores}) = \frac{\text{Number of outcomes with different scores}}{\text{Total number of outcomes}} = \frac{30}{36} = \frac{5}{6}$

ii) Same scores: There are 6 outcomes with the same scores.

$P(\text{same scores}) = \frac{\text{Number of outcomes with same scores}}{\text{Total number of outcomes}} = \frac{6}{36} = \frac{1}{6}$

d) Finding probabilities of sums:

i) Sum is 5: There are 4 outcomes: (1,4), (2,3), (3,2), (4,1)

$P(\text{sum is 5}) = \frac{4}{36} = \frac{1}{9}$

ii) Sum is 10: There are 3 outcomes: (4,6), (5,5), (6,4)

$P(\text{sum is 10}) = \frac{3}{36} = \frac{1}{12}$

iii) Sum is more than 10: There are 3 outcomes: (5,6), (6,5), (6,6)

$P(\text{sum is more than 10}) = \frac{3}{36} = \frac{1}{12}$

iv) Sum is at least 10: There are 6 outcomes: (4,6), (5,5), (6,4), (5,6), (6,5), (6,6)

$P(\text{sum is at least 10}) = \frac{6}{36} = \frac{1}{6}$

3. Final Answer

a) The table is shown in the Solution Steps.

b)

i) Sum is 5: (1,4), (2,3), (3,2), (4,1)

ii) Sum is 10: (4,6), (5,5), (6,4)

iii) Sum is more than 10: (5,6), (6,5), (6,6)

iv) Sum is at least 10: (4,6), (5,5), (6,4), (5,6), (6,5), (6,6)

c)

i) P(different scores) = 5/6

ii) P(same scores) = 1/6

d)

i) P(sum is 5) = 1/9

ii) P(sum is 10) = 1/12

iii) P(sum is more than 10) = 1/12

iv) P(sum is at least 10) = 1/6