Basic Probability

Question

out of 2n+1 tickets consecutively numbered 3 are drawn at random . Find the probability that number on them are in AP ?

Comments

please explain in detail . i am weak in probability .

Answer 1

lets take simple string n=5 : 1 2 3 4 5 6 7 8 9 10 11
Total combination = 10C3
if i select 1 as first number then total combination = 123 , 135 , 147 , 159 ,1611 = n combination
if i select 2 as first number then total combination = 234 , 246 , 258 ,2610       = n-1 combination

General way   
The total number of combination (2n+1C3)
 
If 1 is slected total number of combination n
If 2 is slected total number of combination n-1
If 3 is slected total number of combination n-1
.
.
.
If (2n-2) is selected total number of combination 1( i.e. 2n-1 2n 2n+1)
If (2n-2) is selected total number of combination 1( i.e. 2n-2 2n-1 2n)
 
So total number of combination = n+ n-1 + n-1 + n-2 + n-2+ ..............+ 1 + 1 = n + $\frac {(n-1)(n-1+1)}{2}*2$
So probability = n^2/(2n+1C3)

Comments

yo Thanks :) Your example was useful for me !

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since element are 2n+1 then how " n" come if diffrenece is 1.

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@pratap what i mentioned in my example i took 1 as first element then how many combination of AP i can make it with that.
@priti i mis-typed some equation incorrectly now its ok!

Answer 2

2n+1C3 is total selection..
number of groups (with common difference d =1) = 2n-1
number of groups (with common difference d =2) = 2n-3
number of groups (with common difference d =3) = 2n-5
....
.... 
number of groups (with common difference d =n) = 1

Total number of group = 1+2+3+4+5..........+(2n-1)
= sum of first n odd numbers = n*n           // sum of first n odd numbers = n*n : it is Identity

required Probability = n*n/ (2n+1C3) = 3n/(4n*n-1)