Answer:
(a)2,3,4,5,6,7,8,9,10,11 and 12.
(b)(1,4), (2,3), (3,2) and (4,1)
(c) [tex]\dfrac{1}{9}$ or 11.11\%[/tex]
Step-by-step explanation:
When two six-sided dice are rolled, the possible outcomes are:
[tex](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)[/tex]
If Dominic can roll a total of five using the two dice, he will win the game.
(a) The possible sums Dominic can roll using the pair of dice is the sum
x+y where the for each outcome (x,y). These are:
[tex]2,3,4,5,6,7\\3,4,5,6,7,8\\4,5,6,7,8,9\\5,6,7,8,9,10\\6,7,8,9,10,11\\7,8,9,10,11,12[/tex]
The possible sums are 2,3,4,5,6,7,8,9,10,11 and 12.
(b) The outcomes that could win the game for Dominic are the outcomes which adds up to 5. These are:
(1,4), (2,3), (3,2) and (4,1)
(c)Probability that Dominic will win the game on this roll
Total Possible sums =36
Number of sums that add up to 5 =4
Therefore:
The probability that Dominic will win the game on this roll
[tex]=\dfrac{4}{36}\\=\dfrac{1}{9}\\ $Expressed as a percentage, we have$:\\\dfrac{1}{9} \times 100 =11.11\%[/tex]