# Probability Qn

1. Oct 3, 2015

### icystrike

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?

Last edited: Oct 3, 2015
2. Oct 3, 2015

### Ray Vickson

(36)/(15*14) = 6/35.

3. Oct 3, 2015

### icystrike

My apologies, my answer is still incorrect and different from that of method one.

4. Oct 3, 2015

### haruspex

The discrepancy is a factor of 2. This should suggest you are double counting or half counting something. And you are.
What is the probability that a particular pair of numbers is chosen? In your second method, what have you taken the probability of the pair (10, 2) to be?

5. Oct 3, 2015

### Ray Vickson

In such cases I often like to fall back on construction of an appropriate sample space. Typically, in such problems, the number-choosing mechanism is sequential. Perhaps a local celebrity reaches into a barrel and extracts a numbered ticket, then reaches in for another ticket; perhaps some numbered balls are spun around in a cage and then drop one-by-one into a bin; or perhaps you are writing a Monte-Carlo simulation model of the process to run on a computer. In all these cases it makes sense to regard the sample space as the set of pairs like (i,j), 1 <= i,j <= 15, i ≠ j. So, the ordered pairs (10,7) and (7,10) both occur in the sample space, but the order is ignored when tallying up the outcome. However, having both possibilities affects the probability of the pair {10,7}.