Problem 13 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

Out of \((2 n+1)\) consecutively numbered tickets, three are drawn at random. The chance that the numbers on them are in A.P. is (a) \(4 n^{2}-1 / 3 n\) (b) \(4 n^{2}+1 / 3 n\) (c) \(3 n / 4 n^{2}-1\) (d) \(3 n / 4 n^{2}+1\) (b) If the smallest number is 1 , the groups of three numbers in A.P. are as \(1,2,3 ; 1,3\), \(5 ; 1,4,7 ; \ldots ; 1, n+1,2 n+1 ;\) and they are \(n\) in number. If the smallest number selected is 2, the possible groupings are \(2,3,4 ; 2,4,6 ; 2,5,8 ; \ldots ; 2, n+1,2 n\) and their number is \(n-1\). If the lowest number is 3, the groupings are \(3,4,5 ; 3,5,7 ; 3,6,9 ; \ldots ; 3, n+2,2 n+1\) their number being \(n-1\). Similarly, it can be seen that if the lowest numbers selected are \(4,5,6,2 n-2,2 n-1\), the numbers of selections, respectively, are \(n-2, n-2, n-3, n-3, \ldots, 2,2,1,1\). Thus, the favourable ways for 2,3 are the same and similarly they are the same for 4,5 and so on. Hence, number of favourably ways $$ \begin{aligned} M &=2(1+2+3+\ldots n-1)+n \\ &=2 \times \frac{(n-1) n}{2}+n=n^{2}-n+n=n^{2} \end{aligned} $$ Also the total number of ways $$ \begin{aligned} N &={ }^{2 n+1} C_{3}=\frac{(2 n+1) \times 2 n \times(2 n-1)}{1 \times 2 \times 3} \\ &=\frac{n\left(4 n^{2}-1\right)}{3} \end{aligned} $$Hence, the required probability $$ =\frac{M}{N}=\frac{3 n^{2}}{n\left(4 n^{2}-1\right)}=\frac{3 n}{4 n^{2}-1} $$

Step-by-Step Solution

Answer

The probability is \(\frac{3n}{4n^2-1}\), so option (c) is correct.

Step 1: Understanding the Problem

We need to find the probability that three randomly drawn tickets have numbers that are in arithmetic progression (A.P.). The total number of tickets is \(2n + 1\).

Step 2: Calculate Favorable Outcomes

For each possible smallest number, we count how many sets of three numbers can form an A.P. Starting with the smallest number being 1, there are \(n\) possibilities: \(1, 2, 3; 1, 3, 5; \ldots; 1, n+1, 2n+1\). This pattern continues with 2 having \(n-1\) and so on.

Step 3: Sum of Favorable Ways

The sum of the number of possible A.P. sequences for each starting number is \(M = 2(1+2+3+\ldots+(n-1))+n\). Simplify this to get \(M = n^2\).

Step 4: Total Possible Combinations

Calculate the total ways to choose any 3 tickets from \(2n+1\) tickets. Thus, \(N = \binom{2n+1}{3} = \frac{(2n+1) \cdot 2n \cdot (2n-1)}{1 \cdot 2 \cdot 3} = \frac{n(4n^2-1)}{3}\).

Step 5: Calculate Probability

The probability is the ratio of the favorable outcomes to the total outcomes: \(P = \frac{M}{N} = \frac{n^2}{\frac{n(4n^2-1)}{3}} = \frac{3n}{4n^2-1}\).

Step 6: Select the Correct Option

Comparing the calculated probability \(\frac{3n}{4n^2-1}\) with the given options, the correct answer is option (c).

Key Concepts

Arithmetic Progression

An Arithmetic Progression (A.P.) is a sequence of numbers in which the difference between any two consecutive terms is always the same. This fixed difference is known as the "common difference." For example, in the sequence 2, 4, 6, 8, the common difference is 2.
The formula for finding the general term in an arithmetic progression is:

• First term: \( a \)
• Common difference: \( d \)
• n-th term: \( a_n = a + (n-1) \cdot d \)

Understanding Arithmetic Progression is crucial because it helps in identifying patterns within sequences, which is key for solving probability problems like the one given in the exercise. We make use of A.P. to determine the number of sequences that fit a particular pattern, such as combinations of tickets drawn.

Combinatorics

Combinatorics involves the study of counting, arranging, and grouping objects. It's a fundamental part of probability and is used to determine the number of possible outcomes in various scenarios. In this exercise, we're trying to find out the number of ways we can pick three numbers from a group of tickets and have them in an arithmetic progression.
Several principles used in combinatorics include:

• Permutations: Arrangements where the order matters.
• Combinations: Groupings where the order does not matter.
• Factorials: A product of all positive integers up to a certain number \( n \), expressed as \( n! \).

By understanding and applying these principles, one can efficiently solve problems where the task is to count specific groupings, such as the ticket scenario provided.

Binomial Coefficient

The binomial coefficient is a key concept in combinatorics, represented as \( \binom{n}{k} \), and reads as "n choose k." It calculates the number of ways to pick \( k \) items from \( n \) items, regardless of the order of selection. The formula for calculating the binomial coefficient is:\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]In the context of our problem, we use the binomial coefficient to determine the total number of ways to choose any three tickets from the given set of \( 2n+1 \) tickets. The calculation is carried out as follows:\[ \binom{2n+1}{3} = \frac{(2n+1)\cdot(2n)\cdot(2n-1)}{1\cdot2\cdot3} \]Understanding the use of binomial coefficients allows us to precisely calculate possibilities and is especially useful in probability, where determining total and favorable outcomes accurately defines the solution.