What's the probability of rolling a 9 or higher using two fair standard dice?

10/36=5/18=27.bar7%

Explanation:

The rolls you can get with 2 six-sided dice is in this table:

$\left(\begin{matrix}\textcolor{w h i t e}{0} & \underline{1} & \underline{2} & \underline{3} & \underline{4} & \underline{5} & \underline{6} \\ 1 | & 2 & 3 & 4 & 5 & 6 & 7 \\ 2 | & 3 & 4 & 5 & 6 & 7 & 8 \\ 3 | & 4 & 5 & 6 & 7 & 8 & 9 \\ 4 | & 5 & 6 & 7 & 8 & 9 & 10 \\ 5 | & 6 & 7 & 8 & 9 & 10 & 11 \\ 6 | & 7 & 8 & 9 & 10 & 11 & 12\end{matrix}\right)$

See that there are 10 numbers that are 9 or higher out of a total number of 36 possibilities, giving a probability of 10/36=5/18=27.bar7%

Dec 26, 2017

total no. of outcomes = 6×6
favourable outcomes require that sum should be $\ge 9$ .
favourable outcomes are
for $9 \left(4 , 5\right) , \left(5 , 4\right) , \left(6 , 3\right) , \left(3 , 6\right)$
for $10 \left(5 , 5\right) , \left(6 , 4\right) , \left(4 , 6\right)$
for $11 \left(5 , 6\right) , \left(6 , 5\right)$
for $12 \left(6 , 6\right)$;
probability =10/36=5/18~~27.7777777778%