probability

Question

Out of (2n + 1) tickets consecutively numbered three are drawn at random. Find the probability that the numbers on them are in arithmetic progression?

• 3n/4n²-1

• 2n/4n²-1
• 2n/4n²+1
• 2n/2n²+1

Answer 1

Lets the tickets be numbered 1, 2, 3, 4.........................(2n+1)

Now, sample space = $\binom{2n+1}{3}$

As stated in some other answer, terms are in AP iff the terms are: {a-r, a, a+r}

where, 'r' is some constant, 'a' is one of the terms.

Now, we consider favourable events.

r=1 $\Rightarrow$ 1 2 3 4 5 6..................... (2n-1) (2n) (2n+1)

Only the highlighted terms can be selected. 1 doesnt have left value and (2n+1) doesnt have right value.

So, total terms = (2n+1) - 2

r=2 $\Rightarrow$ 1 2 3 4 5 6..................... (2n-2)(2n-1) (2n) (2n+1)

Total terms = (2n+1) - 4

.

.

.

r=n (max possible value of 'r') $\Rightarrow$ 1 2 3 4 5 6.........(n+1)............ (2n-1) (2n) (2n+1)

Total terms = (2n+1) - 2*n

Thus, total favourable cases

= n(2n+1) - [2+4+6........+2n]

= n(2n+1) - 2*[1+2+3....n]

= n²

Thus, probablity

= $\frac{n^{2}}{\binom{2n+1}{3}}$

So, option A is the answer.

Answer 2

Let 2n + 1  numbers be

 A-n  ,  A-n+1 , A-n+2  .....   A-1 ,  A  , A1 , A2 , A3 , A4 ...... An-1 ,  An (  Consecutive numbers )

Three numbers are in AP

Let say  {  a - r , r  , a + r  }     where a is middle term of AP.

Now total ways to select 3 numbers out of (2n + 1 )   :   C( 2n + 1 ,  3 )      

 where C( n , r )  =   binomial coefficient.

C(n,r)   =  (n/r) * C(n-1, r - 1)   =  (n/r) * ( (n-1)/(r-1) ) * ((n-2)/((r-2))....and so on till r !=  1.

Using above binomial formula :

Total ways  :      ((2n + 1 )  * ( 2n )  * (2n-1))/6  =   (n * (4n^2 -1) )/3

Now number of favourable cases.

Let say middle term  a =  A (  middle tern of (2n+1) numbers.

Then { a-r  , a  + r  }  will be { ( A-1 ,  A1 )  ( A-2 , A2)  (A-3 , A3) ....  (A-n+1 , An-1) , (A-n , An ) }

Total N ways.

Now if middle term r = A1  then  we have  (N-1) ways : {  (A , A2 ) , ( A-1 , A3 ) .... and so on }

In this manner total ways :   n  +   2 * ( (n-1)  + (n-2) + (n-3) ....  3 + 2 + 1 )    

Middle term can be formed n triplets , next and previois term foemed ( n-1 ) triplets and so on...

Total favourable ways :   n + 2 *  (n*(n-1)/2   = n^2

Required Probality :  n^2/ ((n * (4n^2 -1) )/3)  :   3n/(4n^2-1)  

Option (A) is the correct asnwer.

Comments

Plz use subscript and superscript facility. It will look better