Suppose you roll two fair dice. Create (or reference) a sample space to represent the situation then determine
each of these probabilities.

a. Getting an even sum OR a sum of 3

b. Getting doubles AND an even sum of at least 8.

Rolling two dice can output any number between 2 and 12. Here are the probabilities (given by how many combinations would yield that sum):

You get 2 if you roll (1, 1)

You get 3 if you roll (1, 2), (2, 1)

You get 4 if you roll (1, 3), (2, 2), (3, 1)

You get 5 if you roll (1, 4), (2, 3), (3, 2), (4, 1)

You get 6 if you roll (1, 5), (2, 4), (3, 3), (4, 2), (5, 1)

You get 7 if you roll (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

You get 8 if you roll (2, 6), (3, 5), (4, 4), (5, 3), (6, 2)

You get 9 if you roll (3, 6), (4, 5), (5, 4), (6, 3)

You get 10 if you roll (4, 6), (5, 5), (6, 4)

You get 11 if you roll (5, 6), (6, 5)

You get 12 if you roll (6, 6)

So, you get an even sum OR a 3 if you roll a total of 2, 3, 4, 6, 8, 10 or 12. The sum of their probabilities is

You get doubles and at least 8 if you get (4,4), (5,5) or (6,6). Each of these pairs happen with probability 1/36, for a total of

mr. flanders’ class sold 47 doughnuts and   mr. rodriquez’s class sold 38 cartons of milk

step-by-step explanation:

let x be the number of doughnuts sold by mr. flanders’ class and y be the number of cartons sold by mr. rodriquez’s class.

1. together, the classes sold 85 items, then

x+y=85.

2. both classes earn earned $104.49 for their school. if doughnuts were sold for $1.35 each, then x doughnuts costed $1.35x. if cartons of milk were sold for $1.08 each, then y cartons costed $1.08y. hence,

1.35x+1.08y=104.49.

3. solve the system of two equations: