#probability(self doubt)

Question

An automobile showroom has 10 cars, 2 of which are defective. If you are going to buy the 6th car sold that day at random, then the probability of selecting a defective car is??

Answer 1

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 $\frac{\binom{9}{1}}{\binom{10}{2}}$=$\frac{9}{\frac{9 \times10}{2}}$=$\frac{2}{10}$=$0.2$

Answer 2

Probability of $6th$ car will be defective will be $\frac{^{2}\textrm{C}_{1}}{^{10}\textrm{C}_{1}}=\frac{2}{10}=\frac{1}{5}=0.2$