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Mathematics 18 Online
OpenStudy (anonymous):

You roll 2 dice. What is the probability that the sum of the dice is odd or 1 die shows a 5?

OpenStudy (anonymous):

count.

OpenStudy (anonymous):

you get lots.

OpenStudy (anonymous):

for the odd it would b 3/6 1/6

OpenStudy (anonymous):

(5,1) (5,2)(5,3) 5,4) (5,5) (5,6) (1,5)(2,5)(3,5)(4,5)(6,5) (1,2) (1,4) (1,6) (2,1) (2,3) (3,2) (2,4) (3,6) (4,1) (4,3) (61) (6,3) first row is all of the first condition next rows are all that give a total of odd not recounting what is in first row

OpenStudy (anonymous):

check that i didn't miss any but i get \[\frac{23}{36}\]

OpenStudy (anonymous):

you can use some sort of formula but with dice easier to count

OpenStudy (anonymous):

Isn't it asking about rolling two dice together one time?

OpenStudy (anonymous):

i think so

OpenStudy (anonymous):

OK, I missred your earlier post.

OpenStudy (anonymous):

yes i take it to mean that. i listed all the ordered pairs that fit your description and i could 23 of them.

OpenStudy (anonymous):

@skittles the probability that you get an odd number is 1/2 but the probability that ONE die shows at 5 is not 1/6. that is the probability that the first die (or second) shows a 5.

OpenStudy (anonymous):

so how do i get the finmal answer?

OpenStudy (anonymous):

amistre, think u can help? long time no c :P

OpenStudy (amistre64):

You roll 2 dice. What is the probability that the sum of the dice is odd or 1 die shows a 5 1 2 3 4 5 6 1 x x x x 2 x x x 3 x x x x 4 x x x 5 x x x x x x 6 x x x might have gotten them all

OpenStudy (amistre64):

count the xs and put it over 36

OpenStudy (anonymous):

23/36

OpenStudy (amistre64):

thats what I get also if i did it right ;)

OpenStudy (anonymous):

That's what I got, too.

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