Question

Suppose you roll a six-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

not neccisarly what?

Answer 2

......somebody with sense explain it properly plese? and jack it seems as if you don't...

Answer 3

your sample space is essentially pairs of rolls \((a,b)\) where \(a,b\) are numbers that you can get from rolling a die so an integer from \(1\) to \(6\). the number of favorable outcomes is the number of pairs \((a,b)\) that when you add \(a,b\) you get an even number. recall that adding two even or two odd numbers always yields an even number so \(a,b\) must both either be even or odd to give even \(a+b\)

Answer 4

Well the first time your roll a die we have six possible outomes..so if we throw it 2 times it will be 6*6=36.I guess.

Answer 5

lol good job

Answer 6

to count the number of favorable outcomes, just count the number of pairs \((a,b)\) where either \(a,b\) are both even or \(a,b\) are both odd. We have \(6\) possibilities for \(a\) and so this determines \(3\) possibilities for \(b\) (if \(a\) is even we only have \(3\) even numbers possible for \(b\) and vice versa for odds) so the total number of possibilities is \(3\times6=18\)

Answer 7

There are 18 different scenarios where the sum is even.

Answer 8

@oldrin.bataku My brain compasity got smaller readin that, i'm so confuzzled

Answer 9

There are 3 ways to roll even and 3 ways to roll odd , and 3 ways to pair another 3 even numbers up with the even ones and another 3 odd with the odd ones $$3\times3+3\times3=18$$

Answer 10

alternatively you could just realize that \(a+b\) will always either be even or odd and both happen equally so \(a+b\) will be *even* for half of the possible pairs \((a,b)\) and *odd* for the other half of the pairs \((a,b)\). it makes sense that the number of favorable outcomes is just \(36/2=18\)

Answer 11

both happen equally as often*

Answer 12

Ye 36/2 makes sense.

Answer 13

whats the sample space?

Answer 14

The sample space has 36 possible combinations of numbers. These can be set out in column form as follows:
6,6 5,6 4,6 3,6 2,6 1,6
6,5 5,5 4,5 3,5 2,5 1,5
6,4 5,4 4,4 3,4 2,4 1,4
6,3 5,3 4,3 3,3 2,3 1,3
6,2 5,2 4,2 3,2 2,2 1,2
6,1 5,1 4,1 3,1 2,1 1,1

Answer 15

@Ryleighblue the 'space' of all possible outcomes from rolling two dice

Answer 16

assign a two tuple to both dice rolls*

Answer 17

Okay y'all @oldrin.bataku has helped me to the fullest Thanks to all ho tried :D

Answer 18

Note that your sample space consists of 36 pairs total outcomes since there is 6 outcomes for the first die and 6 for the second hence 6 x 6 = 36 outcomes. The only "Favourable outcomes" are the ones that lead to an even sum, i.e. if 'x' is the number I first rolled and 'y' is the second number I roll then \(\bf x+y=2k, \ 0 < k \le 6\). The is only possible 'x' and 'y' are both odd integers. Since 1, 3, 5 are the only odd numbers in then 'x' can take on 3 odd numbers and 'y'can take on 3 odd numbers, which is 3 x 3 = 9 possible outcomes. But we can also get an even sum if 'x' and 'y' are both even. Hence in this case each 'x' and 'y' can take on 2,4,6 so we once again get 3 x 3 = 9 possible outcomes. Hence there is a total of 18 favourable outcomes out of a sample space of the sample space (shown below) of 36 outcomes:Fill in the table to get your sample space.

@Ryleighblue

Answer 19

$$(a,b)\in \mathbb{N_6}^2$$

Answer 20

Just count up the number of pairs in the table that I posted where the sum of the pair of numbers is even.

Answer 21

@kropot72 Put in a lot of effort typing that up lol =D

Answer 22

I like how this question has been closed.

Answer 23

@Jack111 are you sure that notation is a good idea? i'd rather say the state space is equivalent to \(\mathbb{Z}_6\,^2\) since \(\mathbb{Z}_k\) is a more standard notation than \(\mathbb{N}_k\)

Answer 24

@genius12 Copy and paste :0

Answer 25

@oldrin.bataku I might think your refering to the 6 adic numbers

Answer 26

@genius12 My original typing tho!

Answer 27

@uri I'm pretty sure the OP doesn't require more help..there is like 2 million solutions posted above.

@kropot72 Either way, you did put in effort at some point lols...

Answer 28

integers*

Answer 29

though your prolly right lol

Answer 30

@Jack111 well k-adic numbers are usually given by \(\mathbb{Q}_k\)? :-p

Answer 31

thats why I specified integers* ;[ afterword

Answer 32

anyway maybe somthing like $$\mathbb{N_{n\leq 6}}$$

Answer 33

Using you should say 1 <= n <= 6 instead of n <= 6

Answer 34

reason I suggest \(\mathbb{Z}_k\) is since it refers to set of least residues \(\mod k\) :-p one problem with that notation @Jack111 is that it's still ambiguous whether \(0\) is included and if it is then your set is the wrong size :-p

Answer 35

anyways \(\mathbb{Z}_p\) I have not seen refer to p-adic integers unfortunately

Answer 36

DEWD YOU HAVE PMS FOR A REASON SHTAHP BLOWING UP MY NOTIFICATIONS AND USE YOUR PMS BEFORE I SLAP A HURRR

Answer 37

Reason I say that is because there is no set conclusion as to 0 is in the natural numbers set or not hence including the 1 is a better idea.

Answer 38

@genius12 precisely