Two distinct numbers a and b are chosen randomly from the set of first 20 positive integers. What is the probability that |a–b| would not be divisible by 3?	Options.
A) 49/380 B) 21/95 C) 49/190 D) 42/95 E) 133/190
Any number could be written in the form of one among the following:   (i) 3k   (ii) 3k + 1   (iii) 3k + 2   Only if the 2 numbers are of different types would |a – b| not be divisible by 3:   Numbers of type 3k --&gt; 3, 6, 9, ......... 18 --&gt; Total 6 numbers   Numbers of type 3k + 1 --&gt; 1, 4, 7, .... 19 --&gt; Total 7 numbers   Numbers of type 3k + 2 --&gt; 2, 5, 8, 11 ...20 --&gt;Total 7 numbers   Total possible outcomes = 20 * 19= 
<gadget id="calculator">20 * 19</gadget>
<output>380</output>
380   Favorable Cases:   1) Picking from Type 1 and Type 2: 2 * (6 * 7)= 84   2) Picking from Type 1 and Type 3: 2 * (6 * 7) = 84   3) Picking from Type 2 and Type 3: 2 * (7 * 7) = 98   Hence, the required probability = (84 + 84 + 98) / 380, and 133/190 is the correct answer.
<result>E</result>
Two distinct numbers a and b are chosen randomly from the set of first 20 positive integers. What is the probability that |a–b| would not be divisible by 3?