1. The problem statement, all variables and given/known data 23 One fund-raising activity of a particular sports club is a weekly draw. Participants choose two different numbers between 1 and 15 inclusive. When the draw is made, two numbers between 1 and 15 inclusive are selected at random. The prize money is shared by anyone who has chosen these two numbers. If there are no winners, the prize fund rolls over to the following week. When last week's winning numbers were announced as 3 and 10, someone pointed out that it was the fifth successive week that the two numbers consisted of a total of three digits, all different. What is the probability that when two different numbers between 1 and 15 inclusive are selected at random they will consist of three digits all different? 2. Relevant equations 3. The attempt at a solution I had attempted the questions with two different methods First method: There are 10C2 = 105 ways to choose two number for 1 to 15. There are 8+ 7 x 4 = 36 ways to of choosing two numbers that fulfill the stated condition. Therefore, the probability is 36/105= 12/35 Second method: To choose the set of numbers to fulfill the stated condition; One of the number needs to be double digits (10,11, ... ,15) excluding 11 Therefore, if the first number is: 10 it can be paired with 2,3,4,5,6,7,8,9 the probability will then be 1/15 x 8/14 12 it can be paired with 3,4,5,6,7,8,9 the probability will then be 1/15 x 7/14 13 it can be paired with 2,4,5,6,7,8,9 the probability will then be 1/15 x 7/14 14 it can be paired with 2,3,5,6,7,8,9 the probability will then be 1/15 x 7/14 15 it can be paired with 2,3,4,6,7,8,9 the probability will then be 1/15 x 7/14 By summing all the probability: (1/15)(1/14)(8+7x4) = 36/210 = 6/35 I know that my second solution is incorrect, but can someone kindly explain to me what is the error in it?