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Mathematics 7 Online
OpenStudy (anonymous):

If you roll two number cubes, what is the probability of their sum being seven or higher?

OpenStudy (anonymous):

Yea

OpenStudy (anonymous):

Right

OpenStudy (anonymous):

ok so if u roll a dice u will get two numbers

OpenStudy (anonymous):

Yea

OpenStudy (anonymous):

Equal to

OpenStudy (anonymous):

or higher

OpenStudy (anonymous):

even 5+6 = 11 and 6+6 = 12

OpenStudy (anonymous):

Yeah,

OpenStudy (anonymous):

What?

OpenStudy (anonymous):

easiest method is to count

OpenStudy (anonymous):

That'd give you the probability?

OpenStudy (anonymous):

fill in a dice table, let me see if i can find one count how many, then divide by 36

OpenStudy (anonymous):

here i feel it would be 22

OpenStudy (anonymous):

22/36?

OpenStudy (anonymous):

The answers to choose from are 5/7, 5/12, 7/5, and 7/12

OpenStudy (anonymous):

Which would reduce to 5/12

OpenStudy (anonymous):

i am still looking for a table, but the idea is to count how many of each there are 6 ways to get a 7 5 ways to get an 8 4 ways to get a 9 3 ways to get a 10 2 want to get an 11 1 way to get a 12 for a total of \(6+5+4+3+2+1=21\) ways to get a number 7 or higher

OpenStudy (anonymous):

so your answer is \(\frac{21}{36}\) which you can reduce

OpenStudy (anonymous):

So the answer would be 7/12, Thank You!!

OpenStudy (anonymous):

still can't find one, but it should look like this 1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12

OpenStudy (anonymous):

now everything is a matter of counting how many times you see the sum in the table

OpenStudy (anonymous):

lol i have to remember dices at least they didn't call them "number cubes"

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