IIT JEE , AIEEE Forum - Mathematics , Algebra

raman_shadow

out of (2n + 1) tickets, consecutively numbered 3 are drawn at random. find the probability that the numbers are in AP

vikrant_sharma

out of (2n+1) tickets 3 tickets can be chosen in = (2n+1)C3

n(S)= (2n+1)C 3

let 3 tickets be drawn a ,b ,c a
a,b,c in A.P means 2b=a+c

2b is an even number , sum of 2 odds or 2 evens gives us a even number
therefore ,a and c must be both even or both odd

out of 2n+1 tickets ( n+1) is odd and n is even

n(E)= (n+1)C 2 + nC 2

probability = n(E) / n(S)

= 3n / [4 (n^2) -1]

iam_quantized

U CAN ALSO SOLVE IT BY CONSIDERING VARIOUS COMMOND DIFFERENCES ....BUT ABOVE METHOD IS SIMPLER I THINK

Asmita

Both of u deserve 2 be saluted.

puneet

hii

well i think the above told answer is not correct .....

let me tell u y ... when u say that the numbers have to be even or odd .. it is true ... but then we never have used the fact that they are in A.P. I mean to say that we can have numbers like this ... 1,4,9 .. now these are not in A.P. even if 1 and 9 both are odd ...

tell me if some of u feel wrong bout this ..

well i think that u can choose the first two at random but the third has only two possiblities .... either bigger than the biggest by the difference of previously chosen two numbers. ... or smaller than the smallest by the difference of previously chosen two numbers ..

so i think the probability should be 2/(2n-1) ..

I invite comments on this .. tell me if someone finds a mistake in this ..

cheers

vikrant_sharma

Puneet sir , I have checked the answer from tata-mcgrawhill . the answer given in the book is 3n / [4 (n^2) -1]

titun

Puneet sir, I think the method provided by Vikrant is correct. Take the following e.g

There are 2n+1 consecutive positive integers. We select 3 nos a,b,c. If a,b,c are in A.P.

2b=a+c. It is obvious that a+c = even integer. Now this implies that,

a & c are both odd integers or even integers.

(since, odd+odd = even & even + even = odd)

Now, that b is the A.M of the two nos a & c, it is obvious that b lies beteween a & c & b being an integer falls in that set of 2n+1 consecutive integers. In this way, all the criteria are satisfies.

e.g. From first 30 consecutive integers, if we select 2 odd nos like 1 & 9, then 5 being the AM of 1 & 9 lies within the first 30 integers.

iam_quantized

THE C.D CAN BE 1 ,2,3....
NO.S WITH C,D 1 (1,2,3),(2,3,4).........(2N-1,2N,2N+1) = 2N-1
NO.S WITH CD 2 (1,3,5),(2,4,6).........(2N-3,2N-1,2N+1)= 2N-3
NO. WITH CD 3 (1,4,7),(2,5,8).........(2N-5,2N-2,2N+1)=2N-5
.
.
.
.
3
NO.S WITH CD N (1,N+1.2N+1) = 1
ADDIND NO. OF CASES POSSIBLE i.e 1+3+5+.....2N-1,2N+1
= (N+1)/2(1+2N+1)=Q(SAY).....THEN ANS IS Q/(2N+1)C3

iam_quantized

oops i made a mistake.....the sum is n/2(2n)=n^2
n prob is
3N/(4N^2-1)

amey

The answer is
3N/(4N^2-1)

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