A bag contains 21 toys numbered 1 to 21. A toy is drawn and then another toy is drawn without replacement.
Find the probability that both toys will show even numbers.

The probability that first toy shows the even number $$ = \frac{{10}}{{21}}$$
Since, the toy is not replaced there are now 9 even numbered toys and total 20 toys left.
Hence, probability that second toy shows the even number $$ = \frac{{9}}{{20}}$$
Required probability,
$$\eqalign{ & = \left( {\frac{{10}}{{21}}} \right) \times \left( {\frac{9}{{20}}} \right) \cr & = \frac{9}{{42}} \cr} $$