jmc

There are $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$ ways to choose where the red lamps go, and $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$ ways to choose which lamps are on. If the left lamp is blue and off, and the right lamp is red and on, there are $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ ways to choose which of the remaining lamps are red, and $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ ways to choose which of the remaining lamps are on. Therefore, the probability is $\frac{6\cdot 6}{20\cdot 20}=\boxed{\frac{9}{100}}$.