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Mathematics 9 Online
OpenStudy (anonymous):

two fair dice are rolled what is the probability that the sum of the dice equal to 5. ( next ? sum of dice is at least 10 or more)

OpenStudy (anonymous):

for this we count how many ways to get 5 on two dice? (1,4),(2,3), (3,2), (4,1) four ways out of the total of 36

OpenStudy (anonymous):

so your probability is \(\frac{4}{36}=\frac{1}{9}\)

OpenStudy (anonymous):

where are you gettiing the 36 from

OpenStudy (anonymous):

six possibilities for first die six possibilities for second die counting principle says total ways is \(6\times 6=36\)

OpenStudy (anonymous):

While rolling \(n\) dices the sample space is given by \(6^n \)

OpenStudy (anonymous):

what first die? im confused

OpenStudy (anonymous):

hold on i will send you a table

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

OpenStudy (anonymous):

you see you have 36 possibilities for the two dice to be rolled

OpenStudy (anonymous):

ahh ok got it

OpenStudy (anonymous):

usually they are described as (1,1) , (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) , (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) , (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) , (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) , (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) , (6,2) (6,3) (6,4) (6,5) (6,6)

OpenStudy (anonymous):

so any probability question you have with dice amounts to counting what you want and dividing by 36

OpenStudy (anonymous):

so for ten or more, count how many ways you can get ten or more and divide by 36

OpenStudy (anonymous):

ok i got 1/6

OpenStudy (anonymous):

yeah that'll do it

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