Question

2.Suppose you roll an eight-sided die two times hoping to get two numbers whose sum is even. What is the sample space? How many favorable outcomes are there?

Answer 1

i dont understand this part
- two dimensions, each ranging from 1 to 8

Answer 2

A sample space is a set consisting of all the possible outcomes of an event.
We use brackets curly brackets to show the items in a set.

Answer 3

i'm ok with your sample space, but maybe also specify that \(x,y\in \mathbb{Z}\)

Answer 4

I don't know my brother helped me.

Answer 5

guess he's wrong

Answer 6

32 sounds right for the favourable outcomes

Answer 7

when rolling an eight sided die it ranges {1 2 3 4 5 6 7 8 }

Answer 8

{\((x,y)|1\le x,y\le8, x,y\in \mathbb{Z}\)}

Answer 9

why don't we just ignore that and start over. I'll give you some formulas.

Answer 10

its the same sample space you provided but with an extra condition... this is to make sure you use the values 1,2,3,...,8 only and not a continuous range (1,8)

Answer 11

Sample space = 8x8 = 64
Favorable outcomes = 32

Answer 12

kitby i think that in ur sample space there are sums of number whose result is even and odd too

Answer 13

the same space is all possibilities you can encounter. The favourable outcomes are only ones for which the two numbers added together are even

Answer 14

oh! i got it it is the sample space 64

Answer 15

i was thinking in the favorable outcomes

Answer 16

Yeah there are 64 possibilities. But we usually enumerate all poissiblties when they ask for "the sample space", which is what @Martinmelissa was trying to do writing it in set notation

Answer 17

if there is 32 favorable outcomes how can you put it in a sample space? he is asking for two numbers whose sum is even.

Answer 18

We want the number of favourable outcomes. So this ok the pairs you get for which their sum is even:

(1,1), (1,3), (1,5), (1,7)
(2,2), (2, 4), (2, 6), (2, 8)
....

Answer 19

ok so the sample space would be what?

Answer 20

since the two numbers are even.

Answer 21

yeah, with all the pairs enumerated

Answer 22

well the sum of the 2 numbers is even

Answer 23

yeah

Answer 24

I really need help. I don't get this. Sample space is hard.

Answer 25

sum of two odd numbers is also even, don't forget

Answer 26

it's 1 even + 1 odd that gives you an odd number. it may be helpful to count the odd outcomes and subtract them from the total outcomes to get the even outcomes, rather than trying to determine directly...

Answer 27

The sample space is the list of \(all\) possible outcomes when you roll the two dice, regardless of what the sum of the 2 numbers is.

So, we get that:
S={\((1,1),(1,2),(1,3),...,(1,8),\)
\((2,1),(2,1),(2,3),...,(2,8),\)
\(...\)
\((8,1),(8,2),(8,3),...,(8,8)\)}

Answer 28

32- the odd outcomes = ....

Answer 29

this will give 8 rows, with 8 pairs (x,y) in each row... which gives 8*8=64 outcomes in the sample space

Now you can list the favourable outcomes in a systematic way. Use 1 row to count all possibilities for the 1st die:

(1,1), (1,3), (1,5), (1,7)
(2,2), (2, 4), (2, 6), (2, 8)
(3,1), (3, 3), (3, 5), (3, 7)
...
(8,2) ,(8,4), (8,6), (8,8)

You can see that when he 1st die is odd, there are 4 possibilities . So, when the first die is 1,3,5, or 7, we get 4 possibilities for each. So 4*4=16

We get the same thing when the 1st die is even: when the 1st die is 2,4,6,8 we still get 4 valid pairs for each number. So 4*4 = 16

So in total there are 16+16 = 32 favourable outcomes

Answer 30

You don't have to actually list all the favourable pairs. You should notice a clear pattern that allows you to deduce the information above^^

Answer 31

okay thx

Answer 32

is this right? it is another question. just making sure
Pablo rolled a standard die sixty times. He got a 1 twelve times. How does this result compare to the expected results? Explain.

Since a die has 6 sides, one expects each number to be rolled 1/6 of the time. For 60 rolls, one would expect 60(1/6) = 10 rolls to in 1.12 rolls of 1 is 2 more than expected.

Answer 33

i'm not sure I get the 10 rolls in 1.12 rolls part. But Other than that it seems good. He would expect a "1" 10 times, but he got it in 12 times, which is 2 more rolls than what's expected

Answer 34

thanks

Answer 35

your welcome