Question

Two number cubes are rolled. What is the probability of rolling a sum of 9 or a sum that is even?
I get confused because I don't know if each would be a separate probability, like independent or dependent and if I multiply them or if they are together, what would I do?

Answer 1

can the sum be 9 and even at the same time?

Answer 2

btw this has nothing whatsoever to do with "independent" or "dependent" because you are not multiplying
because the sum cannot be both 9 and even, those events are called "disjoint"
to find the probability of one or the other ADD

Answer 3

Okay so, I would take the total number of outcomes, and then take the amount of outcomes that can be a sum of 9 or 7, which would be 2/6 or 1/3?

Answer 4

for one dice

Answer 5

it says "sum is 9 OR even" right?

Answer 6

Wow I misread that, I thought it said 7, sorry. Long day

Answer 7

what does it mean by sum though? You should know english is not my first language so I struggle a bit with word problems.

Answer 8

yeah must have been
sum is 7 or 9 would be \[\frac{6}{36}+\frac{4}{36}\]

Answer 9

why the 36?

Answer 10

it means "roll two dice, record the total" which is usually what you do when you roll two dice (although not always)

Answer 11

Ah okay, like monopoly.

Answer 12

the 36 because there are 36 elements in the sample space (set of all possible outcomes) when you roll two dice
6 outcomes possible for the first die, another 6 for the second \(6\times 6=36\) total

Answer 13

yeah, or craps

Answer 14

or even my favorite candy land

Answer 15

Oh okay, so now I understand 6 / 26 and 4/ 36, so I would need to add them?

Answer 16

IF the question was "what is the probability you roll at 7 or a 9" yes, but it is not

Answer 17

Yeah yeah I understand that. So, since it isn't. It's asking for even numbers, which correct me if I'm wrong, this is not my first language but even means 2,4,6, . . etc right?

Answer 18

right, it means the total (when you add) is 2,4,6,8,10 or 12

Answer 19

don't think too hard about that probability, half the numbers are even

Answer 20

I'm afraid my brain is shutting down or something. Let me just re read the problem. For some reason today has not been a very good math day. So, two dices, each one has 6, right? So, half the numbers are even, 2,4 and 6. So there are two, there should be 6/36 that can be even? (I'm sorry if I'm wrong)

Answer 21

lets go slow

Answer 22

i am going to attach a table hold on

Answer 23

Okay, I'm just trying to visualize this

Answer 24

The only possible thing that comes on my head is 6/12 for even numbers and 1/12 for hte nine. Oh wait let me look at that

Answer 25

that table is supposed to represent the 36 possible outcomes when you roll two dice
for example \((3,2)\) means the left die is a 6 and the top die is a 2, for a total of 5

Answer 26

we can list the number of ways to roll a 9 on two dice, there are only 4 ways to do it \[(5,4),(4,5),(6,3),(3,6)\]

Answer 27

since we believe the dice are fair, so all 36 outcomes are equally likely, we would say the probability you roll a total of 9 is \(\frac{4}{36}\)

Answer 28

Okay. . . I sort of understand that

Answer 29

ok good

Answer 30

now you want to add that the probability you roll an even number (the total ls even)

Answer 31

that probability you can compute by counting, but it is easier just to realize that half the numbers are even, so it must be \(\frac{1}{2}\)

Answer 32

Okay, so, that would be me taking the 4/36 living it alone right? Or is the 1/2 a total separate number? It's just you said "you want to add" and I don't know whether you mean adding the numbers or as in "you want to count this factor in your thought process"

Answer 33

leaving*

Answer 34

mean you add the probabilities \[\frac{1}{2}+\frac{4}{36}\]

Answer 35

okay good! Phew I thought I was lost again.

Answer 36

This gets me 11/18

Answer 37

yes looks good

Answer 38

Phew, wow,I'm sorry for all the trouble. It has been a very long day. I am truly grateful for your help.

Answer 39

@moonrose just a side note. in this problem, the two probabilities are disjoint — the number rolled is either odd (2,4,6,8) or a 9, but it is not both. Some problems like this you may have a situation where the two cases overlap. For example, what is the probability that the sum is even, or there is a 1 showing on one or both of the dice? There your work would be a bit more complicated because you can't just add the two probabilities together — doing that would count some rolls twice. 1 1 would be counted both as being an even sum AND as a roll where where a 1 is showing on one or both of the dice. Think about it later, when your brain has recovered :-)