The numbers 1 to 13 are written on 13 pieces of paper and dropped into a box. Three of them are drawn at random. What is the probability that the three pieces of paper picked have numbers that are in arithmetic progression?


Answer:

 

18
143
 

Step by Step Explanation:
  1. This is a little complicated, so follow carefully.
  2. For making the explanation and the equations simpler, think of the number on the pieces of paper in the form of (2n+1).
    Here, we can see from the equation 2n + 1 = 13.
    or, n = 6
  3. The probability of getting 3 numbers in an A.P by selecting 3 numbers randomly between 1 and 13 is the ratio of
    - Number of ways we can get an A.P from 3 random numbers between 1 to 13 and
    - The Number of ways to select 3 random numbers between 1 to 13.
  4. Let's look at the second part first. Three tickets can be drawn from (2n+1) numbers is in [(2 × n) + 1]C3 ways.
    i.e. Number of ways 3 tickets can be drawn =  
    (2n+1)(2n)(2n-1)
    3x2x1
     
    Simplifying this, we get the number of ways to draw 3 numbers between 1 and 13 =  
    n(4n2-1)
    3
     .
    Here n = 6, so we can simplify it as 286.
  5. Now, for the ways we can get an A.P from 3 numbers between Arithmetic Progressions of 3 numbers would be a sequence of 3 numbers that are separated by a common interval e.g. 1, 2, 3 or 3, 5, 7 etc.
    They are in the form (a, a + d, a + 2d), where a is an integer from 1 to (13-2), and d is another integer.
    So, it's helpful to think of the solution in terms of this interval.
    So we'll think of all the sequences that have an interval 1, then sequences with interval 2 and so on.
  6. So, what are the possible sequences with interval 1. They are
    (1, 2, 3)
    (2, 3, 4)
    ...
    (2n-1, 2n, 2n+1)
    There are 2n-1 such possible sequences.
  7. Similarly, let's look at A.P with interval 2 between the terms. They are
    (1, 3, 5)
    (2, 4, 6)
    ....(2n-3, 2n-1, 2n+1)
    There are 2n-3 such possible sequences.
  8. We can generalize this to say that the number of such sequences with interval 'd' is (2n-(2d-1))
    Obviously, the largest possible integer is d = n, with just one sequence (1, n+1, 2n+1).
  9. So the total number of such sequences is:
    (2n-1) + (2n-3) + (2n-5) + ... + 5 + 3 + 1
    This is itself an AP with n terms and d = 2.
    The sum of this sequence is  
    n
    2
      [2 + (n-1)2].

    Simplifying, we get n2.
    Here, n = 6, so this is 36.
  10. So, the probability is  
    18
    143
     .

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