Question

You roll 2 dice. What is the probability that the sum of the dice is odd or 1 die shows a 4? A 6 X 6 table of dice outcomes will help you to answer this question.

Write fractions using the slash ( / ) key. Reduce fractions to their lowest terms

Answer 1

Ok let's do this in 3 parts, for the 3 questions.
You can start enumerating the possibilities for the sum is odd:
(1,2), (1,4), (1,6)
(2, 1), (2, 3), (2, 5)
(3, 2), ...
You should see a pattern that will allow you the count the number of outcomes for which the sum of the 2 dice is odd

Answer 2

2 parts* for the 2 questions* oops typo lol

Answer 3

For the other part, you can make use of the addition rule. You basically want \(P(1st~die=4)\) OR \(P(2nd~die=4)\)

Answer 4

Ok it is in fact 1 question, then you can just add the probabilities for the 1st and 2nd part together I believe

Answer 5

im so confused

Answer 6

There are 2 "parts" to what appears to be 1 question that have to be brought together. You already know that there are 36 (6 x 6) total possible rolls, so your denominator for the probability will be 36 (at least to start, before possible reducing).

For part 1, the "addition to odd" part, you can either inspect the table (easiest way, but time-consuming) or realize that you can get this "addition to odd" by having either the first die odd with the second even or the first die even with the second die odd. If you take all those into consideration, you will see 18 rolls. But we're not done yet . . .

For part 2, you will have ALREADY taken into account a lot of rolls with a 4 on one of the dice. So we look at the remainder of the rolls containing a 4 ON ONLY ONE OF THE DIE as that appears to be the consideration of the problem if read closely. This gives the remaining rolls, with a 4 on only one of the dice, and where the sum is now positive. This gives only the additional rolls of

42, 24, 46, and 64

That's 4 additional rolls to the 18 in part 1

P = (18 + 4)/36 = 22/36 = 11/18

Answer 7

That part 2 above would seem to exclude the roll of 44. It would seem we have to exclude that roll, because if we included it, we wouldn't have ". . . or 1 die shows a 4".

All good now, @neon.fire ?

Answer 8

im having a mind blank.

Answer 9

Well, take your time. If you are having trouble doing it in your head, don't worry, you are not expected to do it that way. But then you will definitely have to go through the whole table. First cross off all dice combinations that sum to odd. That will be 18 rolls. Look at the remainder of the rolls and include the ones where you see only one 4. That will be 4 additional rolls. That's the super-easy way and requires no further explanation. You just have to do the crossing off manually.

The final probability is: (# of rolls fulfilling requirements)/ (total # of rolls)

That's all there is to it.

Answer 10

so 4/36?

Answer 11

I mentioned a part 1 and a part 2. I mentioned a number of rolls you get from part 1 and a number of rolls you get from part 2. The answers are already worked out and printed. You just have to read them. I actually listed the answer twice.

Answer 12

My first post actually gives the answer explicitly, just read it. My third post gives the answer implicitly. Just read it.

Answer 13

ohp got it now. sorry!

Answer 14

np! You might want to go over this one a couple of times to consolidate your understanding, though, @neon.fire

Answer 15

i made the table and reread those instructions. before i was trying to understand it without the table.

Answer 16

It is a good thing to work it out with the table and then try it without the table in your head if you can. If you can't, don't worry about it at all. If you were to do problems like this a lot, it would become second nature.

Answer 17

I used a table and marked each outcome that meets the conditions. I counted 23 qualifying outcomes.

Answer 18

@kropot72 you have to exclude the roll of 44.

Answer 19

@tcarroll010 Sorry. You are correct.

Answer 20

@kropot72 np.

Answer 21

Not to confuse you, and you can feel free to disregard this post if it gets confusing, but you can do this problem another way by combining part 1 and part 2 by taking the negation of the 2-part requirements. You could, at one time, exclude all rolls that are positive with no "4" or 2 "4"'s in them. That can be confusing for a lot of people, so feel very free to disregard this.