NIMCET 2015 Question Paper with solutions

Question : 48 of 120

Marks: +1, -0

Out of 2n + 1 tickets, which are consecutivelynumbered, three are drawn at random. Then theprobability that the numbers on them are in arithmeticprogression is;

$n^2/{4n^2-1}$
$n/{4n^2-1}$
${3n^2}/{4n^2-1}$
${3n}/{4n^2-1}$

Validate

Solution:

We have,
Total number of ticket

=(2n+1)

Total number of outcomes of 3 numbers are drawn from

(2n+1)

^{2n+1}C_{3}

=\;\;{(2n+1)}/{3!(2n-2)!}=\;\;{(2n+1)(2n)(2n-1)}/{3!}

=\;\;{n(4n^{2}-1)}/{3}

Favourable outcomes are if three numbers are in AP

(2n-1+2n-3+2n-5...+5+3+1)=n^2

∴ Required probability=

n^2/{n(4n^2-1)}/3={3n}/{4n^2-1}

Go to Question:

Prev Question

Next Question