Correct. The probability of choosing a red ball first is $\displaystyle\frac{{10}}{{30}}=\frac{{1}}{{3}}\usepackage{cmbright}\Huge\$ and the probability that the second ball is green is $\displaystyle\frac{{15}}{{29}}\usepackage{cmbright}\Huge\$ , so the answer is $\displaystyle\frac{{1}}{{3}}\cdot\frac{{15}}{{29}}=\frac{{15}}{{87}}\usepackage{cmbright}\Huge\$ .