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.  Answers.
A) 5 / 21
B) 9 / 42
C) 6 / 22
D) 4 / 21
E) 3 / 21

Rationale: since there are 10 even numbers in the given range , the probability that the first toy shows an even number = 10 / 21 since the toy is not replaced , there are now 9 even numbered toys and total 20 toys left . hence , probability that the second toy shows an even number = 9 / 20 . required probability = 10 / 21 x 9 / 20 = 9 / 42 . answer : b
Answer: B