jmc

The different possible (and equally likely) values Tamika gets are: $8+9=17$ $8+10=18$ $9+10=19$ The different possible (and equally likely) values Carlos gets are: $3\times 5=15$ $3\times 6=18$ $5\times 6=30$ The probability that if Tamika had the sum $17$ her sum would be greater than Carlos's set is $\frac{1}{3}$, because $17$ is only greater than $15$ The probability that if Tamika had the sum $18$ her sum would be greater than Carlos's set is $\frac{1}{3}$, because $18$ is only greater than $15$ The probability that if Tamika had the sum $19$ her sum would be greater than Carlos's set is $\frac{2}{3}$, because $19$ is greater than both $15$ and $18$ Each sum has a $\frac{1}{3}$ possibility of being chosen, so we have $\frac{1}{3}\left(\frac{1}{3}+\frac{1}{3}+\frac{2}{3}\right)=\frac{1}{3}\left(\frac{4}{3}\right)=\frac{4}{9}=\boxed{\frac{4}{9}}$