Question

plz help me and I kindly ask that u show work if possible? Plz don't answer if u don't know how to do the problem. PLZ help

Answer 1

8.A For this, we can start with the first roll being 1, and then writing all the possibilities for the second roll. Then do the same for 2,3,4,5&6.

(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)

8.B So lets look at the possibilities we wrote and see how many have a sum of 7.
(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)

So out of 36 possibilities, 6 of them have a sum of seven
P(sum of 7)=6/36 simplify to 1/6

Answer 2

part A is just simple multiplication
since there are 2 6 sided number cubes there are 36 possible outcomes(6 sides *6 sides)

Part B is a little trickier. First off, you must find how to get seven.
1. 1+6 and 6+1

2. 2+5 and 5+2

3.3+4 and 4+3

therefore there are six combinations

you put that number over the total combos and get 6/36 or 1/6.

Hope this helps!