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 .
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
multiply(divide(subtract(floor(divide(21, const_2)), const_1), subtract(21, const_1)), divide(floor(divide(21, const_2)), 21))