NIMCET 2015 — Mathematics PYQ

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Question

NIMCET 2015 — Mathematics PYQ

NIMCET | Mathematics | 2015

Out of $2n + 1$ ticket, which are consecutively numbered, three are drawn at random. Then, the probability that, the numbers on them are in arithmetic progression is:

Choose the correct answer:

Explanation

1. Total Cases ( $n(S)$ ):

$(2n + 1)$ tickets mein se $3$ select karne ke tareeke:

n(S) = \binom{2n+1}{3} = \frac{(2n+1)(2n)(2n-1)}{6} = \frac{n(2n+1)(2n-1)}{3}

2. Favorable Cases ( $n(E)$ ):

Teen numbers $x, y, z$ A.P. mein tabhi honge jab $x + z = 2y$ . Iska matlab $x + z$ hamesha even hona chahiye.

$x + z$ even tabhi hoga jab dono numbers ya toh Odd hon ya dono Even hon.

Tickets $\{1, 2, 3, \dots, 2n+1\}$ mein:

Odd numbers: $(n + 1)$
Even numbers: $n$

Ab combinations nikaalte hain:

Dono Odd choose karne ke tareeke: $\binom{n+1}{2} = \frac{n(n+1)}{2}$
Dono Even choose karne ke tareeke: $\binom{n}{2} = \frac{n(n-1)}{2}$

n(E) = \frac{n(n+1)}{2} + \frac{n(n-1)}{2} = \frac{n}{2}(n + 1 + n - 1) = \frac{n(2n)}{2} = n^2

3. Probability ( $P$ ):

P = \frac{n(E)}{n(S)} = \frac{n^2}{\frac{n(2n+1)(2n-1)}{3}}

P = \frac{3n^2}{n(4n^2 - 1)}

P = \frac{3n}{4n^2 - 1}

Correct Option: (d)