Question

You roll 2 dice. What is the probability that the sum of the dice is odd and one die shows a 5? A 6 X 6 table of dice outcomes will help you to answer this question.
is it 1/6 or 23/36?

Answer 1

@fabo can you please help me with this?

Answer 2

Have you made the table as suggested?

Answer 3

yes, but i still dont understand it

Answer 4

What does it look like, e.g. how many rows, how many columns?

Answer 5

6 rows and 6 collumns

Answer 6

Good.
The rows are labeled 1 to 6 and same for columns, right?

Answer 7

yes

Answer 8

Can you tell me the entry for the bottom right square?

Answer 9

6,6

Answer 10

Very good!

Answer 11

The question is:
"What is the probability that the sum of the dice is odd and one die shows a 5? A 6 X 6 table of dice outcomes will help you to answer this question.
"
It is a compound condition, meaning it is the intersection of two conditions, do you agree?

Answer 12

yes

Answer 13

can you name the two conditions?

Answer 14

conditional and compound

Answer 15

The conditions are:
1. The sum is odd
2. one die shows a 5
We are asked to find the probability that both of the conditions are satisfied, out of the 36 cases you enumerated. Can you count them?

Answer 16

23

Answer 17

Do you actually have 23 squares where one of the numbers is a 5 AND the sum of two numbers is odd?

Answer 18

For example, (5,5) has (at least) a 5, but the sum is not odd. So (5,5) does not count.

Answer 19

On the other hand, (5,2) has a 5, and the sum (7) is odd, so that counts as 1 successful entry.
How many successful entries are there in your table?

Answer 20

@summertimesadness are you there?

Answer 21

yes, but im more confused now

Answer 22

Did you see @satellite73 's table?
Complete all 36 squares of the table and then start counting the successful cases.

Answer 23

thank you @satellite73 for the table

Answer 24

@satellite73 Thanks, that's a neat table! :)

Answer 25

15/4?

Answer 26

Do bother with the denominator, it will be 36 (because we have 36 squares).
Count the number of squares which have a "5" AND the sum of the two numbers is odd. Review the examples I gave earlier.

Answer 27

Have you completed the table?
What is the entry of the top right square?

Answer 28

1,6

Answer 29

Does it have a "5"?

Answer 30

yes

Answer 31

I don't see a "5" in (1,6)!

Answer 32

i dont know what i am doing

Answer 33

this is my table

Answer 34

Refer to Satellite73's table, the top right square represents the outcome of throwing a one followed by a six, thus (1,6). So you didn't throw a five for that case. Is this better?

Answer 35

so it would be 17/36

Answer 36

Yes, that's the right table, except that it was mirrored left-right.
17/36 is not right.
Let look at the first row.
How many of the squares contain a 5 (that means a 5 was thrown)?

Answer 37

once

Answer 38

in the fifth column, right?

Answer 39

ohhh i get it!! thank you!

Answer 40

Is the sum odd?

Answer 41

no

Answer 42

So does that count for a success?

Answer 43

yes

Answer 44

If the square contains a 5 and does NOT have an odd sum, then it did not satisfy BOTH conditions, so it is not counted as a success. So we have no successful candidates on the first row. Does that make sense?

Answer 45

ohhh yes! thank you!!!

Answer 46

Now we move on to the second row.
How many squares contain a 5?

Answer 47

i got it! 1/6!

Answer 48

That is for the second row, because the second row contains one successful candidate, namely (2,5).
We need to do the same for all the rows and add up the total number of successful squares.