Problem 103 - A Complete Resource Book in Mathematics for JEE Main

Question

A bag contains \(n\) white and \(n\) red balls. Pairs of balls are drawn without replcement until the bag is empty. The probability that each pair consists of one white and one red ball is (A) \(\frac{2^{n-1}}{{ }^{2 n} C_{n}}\) (B) \(\frac{2^{n-1}}{{ }^{2 n} C_{n-1}}\) (C) \(\frac{2^{n}}{{ }^{2 n} C_{n}}\) (D) none of these

Step-by-Step Solution

Answer

The correct probability is \( \frac{2^n}{{}^{2n}C_n}\), which corresponds to option (C).

Step 1: Understanding the Problem

We have a bag with \(n\) white balls and \(n\) red balls, totaling \(2n\) balls. We want to find the probability that every pair drawn (without replacement) contains one white and one red ball.

Step 2: Total Number of Ways to Draw Pairs

The total number of ways to draw these \(2n\) balls into pairs of 2 is given by \(\frac{(2n)!}{2^n n!}\). This formula considers that we're dividing the \(2n\) balls into pairs and accounts for the order within each pair.

Step 3: Drawing Perfect Pairs (One White, One Red)

Each time, you need to choose one white and one red ball to form a pair. For the first pair, you have \(n\) choices for the white ball and \(n\) choices for the red ball, continuing this until no balls are left. Therefore, the number of favorable ways is \(n!\).

Step 4: Calculate Probability

The probability is given by the ratio of favorable outcomes to total outcomes. Therefore, the probability is \(\frac{n!}{\frac{(2n)!}{2^n n!}}\).

Step 5: Simplify the Expression

\[ \frac{n!}{\frac{(2n)!}{2^n n!}} = \frac{n! \cdot 2^n n!}{(2n)!} = \frac{2^n (n!)^2}{(2n)!} \] This can be simplified to \(\frac{2^n}{{}^{2n}C_n} = \frac{2^n}{\binom{2n}{n}}\) using the definition of combinations. Since \( \binom{2n}{n} = \frac{(2n)!}{(n!)^2}\), we arrive at the simplified probability expression.

Key Concepts

Combinatorics

Combinatorics is a branch of mathematics focused on counting, arrangement, and combination of objects. In the given exercise, combinatorial principles help us determine how many ways pairs of balls can be drawn. Combinatorics provides the tools to evaluate possible arrangements in a situation, essential for solving problems involving probability.

When dealing with combinatorics, several fundamental concepts come into play, such as combinations and permutations. These concepts help in understanding how objects can be selected or arranged under various conditions.

• **Combinations** refer to selections of items where order doesn't matter.
• **Permutations** consider the arrangement or order of the selected items.

Understanding these foundational ideas is crucial in calculating probabilities, like in our task of drawing pairs of colored balls with specific constraints.

Binomial Theorem

The Binomial Theorem is a key concept in combinatorics that describes the expansion of powers of a binomial expression. This theorem states that:

\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]

Where \( \binom{n}{k} \) is a binomial coefficient, representing the number of ways to choose \(k\) elements from \(n\) elements.

The binomial theorem is significant in probability because the binomial coefficients (number of ways to choose items) show up in probability calculations. They help to streamline solutions where combinations are involved, just like in our problem of drawing red and white pairs.

Applying the binomial theorem assists in simplifying the expression and calculating the probability, ensuring we account for every possible combination needed in the exercise for drawing pairs.

Permutations and Combinations

Permutations and combinations are foundational concepts in combinatorics, essential for solving probability problems like the one presented in our exercise.

Permutations deal with arrangements of objects, exactly relevant when order matters. However, in our exercise, we focus on combinations because we're interested in selecting combinations of one white ball and one red ball without regard to order.

• **Permutations** calculate the total possible arrangements of a set of items.
• **Combinations** focus on selecting items where the order is not important.

The problem boils down to calculating the probability using combinations, specifically \( \binom{2n}{n} \), which represents the number of ways to select \(n\) items from a pool of \(2n\) without considering order.

By using permutations and combinations efficiently, particularly the formula \( \frac{2^n}{\binom{2n}{n}} \), such problems become manageable and clear, illustrating the power of these concepts in probability and combinatorial calculations.