Problem 39 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(2 n\) boys are randomly divided into two subgroups containing \(n\) boys each. The probability that the two tallest boys are in different groups is (A) \(\frac{n}{2 n-1}\) (B) \(\frac{n-1}{2 n-1}\) (C) \(\frac{2 n-1}{4 n^{2}}\) (D) none of these

Step-by-Step Solution

Answer

Option (A) \(\frac{n}{2n-1}\).

Step 1: Understanding the Problem

We need to find the probability that the two tallest boys among the group of \(2n\) boys end up in different subgroups. We know we are dividing \(2n\) boys into two groups of \(n\) each.

Step 2: Calculate Total Ways

Calculate the total ways to divide \(2n\) boys into two groups of \(n\) each. The total ways are given by a combination formula: \(\binom{2n}{n}\).

Step 3: Possible Ways for Tallest Boys in Different Groups

If the two tallest boys are separated, choose one group for the first tallest and place the other tallest in the second group. Considering the contribution of all other boys: choose \(n-1\) boys to join the first tallest in one subgroup. The number of ways to do this is \(\binom{2n-2}{n-1}\).

Step 4: Calculate the Probability

The probability is the ratio of favorable outcomes to the total possible outcomes: \( \frac{\binom{2n-2}{n-1}}{\binom{2n}{n}} \).

Step 5: Simplify the Expression

Simplify the probability expression: \[ \frac{\binom{2n-2}{n-1}}{\binom{2n}{n}} = \frac{\frac{(2n-2)!}{(n-1)!(n-1)!}}{\frac{(2n)!}{n!n!}} = \frac{(2n-2)! n! n!}{(n-1)!(n-1)!(2n)!} \]. By simplifying, we find \(\frac{n}{2n-1}\).

Step 6: Choose the Correct Answer

Comparing the simplified expression against the options provided, we select (A) \(\frac{n}{2n-1}\).

Key Concepts

Combinatorics

Combinatorics is a branch of mathematics that deals with counting, arranging, and combination of objects. It explores how to count the different possible ways structured groups or sets can be formed. In this problem, we are tasked with forming two subgroups from a larger set of boys.Here, we have a total of \(2n\) boys and we need to divide them into two equal groups. The number of ways to do this is given by the combination formula: \( \binom{2n}{n} \). Combinations tell us how many ways we can choose \(n\) boys from \(2n\) without regard to the order.- **Key Points** - Combination formulas like \( \binom{2n}{n} \) are central in combinatorics for counting selections of items. - In a typical problem, you decide how many items are needed from the total set and apply the formula.This combinatorial approach allows us to systematize counting in structured ways, which is crucial for working with probability where these counts form the basis for probability calculations.

Binomial Theorem

The binomial theorem provides a method to expand expressions that are raised to a power, such as \((x + y)^n\). Although not directly used in this exercise, the concept of choosing a subset of items — fundamental to combinations — is linked to this powerful theorem.- The basic combination definition \( \binom{n}{k} \) within the context of binomial expansion reveals how coefficients in binomial expansions are determined.- It tells us the number of ways to choose \(k\) successes (or objects) from a total of \(n\), analogous to picking \(n\) objects for a subgroup.- **Application** - This exercise involves dividing a group of objects, very much akin to determining coefficients in a binomial expansion where selection order does not matter.Understanding the binomial theorem helps solidify why combination calculations are structured as they are, and how they integrate into probability determinations.

Probability Theory

Probability theory allows us to compute the likelihood of different outcomes in random events. In this example, we are interested in the probability that two specific individuals (the tallest boys) are placed in different groups.- **Basic Concept**: Probability is calculated by dividing the number of favorable outcomes by the total number of possible outcomes.To solve our problem:\- **Total Possible Ways**:\ We use \( \binom{2n}{n} \) to calculate the various ways to arrange all \(2n\) boys into two groups.- **Favorable Ways**: \ For the tallest boys to be placed in different groups, we calculate \( \binom{2n-2}{n-1} \), which counts the various combinations of the remaining boys split with one tallest in each group.The probability expression \( \frac{\binom{2n-2}{n-1}}{\binom{2n}{n}} \) simplifies to \( \frac{n}{2n-1} \), giving us the chances of this particular distribution.- **Simplifying Probability** - Correct simplifications rely on factoring combinations and recognizing patterns. - It's essential to clearly comprehend formula relationships to efficiently simplify probability expressions.