There are $\binom{10}{2}$ possible pairs for the defective cars.
Now, if the 6th car is defective there are $\binom{9}{1}$ options for the other defective car.
$\therefore$ The required probability is $\frac{\binom{9}{1}}{\binom{10}{2}}$=$\frac{9}{\frac{9 \times10}{2}}$=$\frac{2}{10}$=$0.2$