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

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

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

Step 1: Understand the Problem

We need to divide \(2n\) boys into two groups of \(n\) each, and find the probability that the two tallest boys end up in different groups. This is a problem of combinatorial probability.

Step 2: Total Ways to Form Groups

There are \(\binom{2n}{n}\) ways to choose a group of \(n\) boys from \(2n\) boys. This is the total number of ways to form the two groups when the order within each group does not matter.

Step 3: Specific Outcome Probability

We want the two tallest boys to be in different groups. Choose one of the two tallest boys to be in the first group of \(n\), and the other will automatically be in the second group. Now, choose the remaining \(n-1\) boys for the first group from the \(2n-2\) others available. This can be done in \(\binom{2n-2}{n-1}\) ways.

Step 4: Calculate the Probability

The probability is the number of favorable outcomes divided by the total number of outcomes, calculated as:\[ \frac{\binom{2n-2}{n-1}}{\binom{2n}{n}} \]

Step 5: Simplify the Expression

Using the formula for combinations \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\), we simplify \(\frac{\binom{2n-2}{n-1}}{\binom{2n}{n}}\):1. \(\binom{2n-2}{n-1} = \frac{(2n-2)!}{(n-1)!(n-1)!}\)2. \(\binom{2n}{n} = \frac{(2n)!}{n!n!}\)3. Dividing the two, \[ \frac{\binom{2n-2}{n-1}}{\binom{2n}{n}} = \frac{\frac{(2n-2)!}{(n-1)!(n-1)!}}{\frac{(2n)!}{n!n!}} = \frac{(2n-2)! \times n! \times n!}{(n-1)! \times (n-1)! \times (2n)!} \]4. Simplifying, \[ = \frac{n \times (n!)}{(2n-1)} = \frac{n}{2n-1} \]

Step 6: Verify against Options

The simplified answer \(\frac{n}{2n-1}\) corresponds with option (A). Therefore, option (A) is the correct answer.

Key Concepts

Combinations

Combinations are a key concept in combinatorial probability. They refer to selecting items from a larger pool where order does not matter. For instance, if you have a group of ten boys and want to choose three to form a team, you use combinations. The number of ways to select these items can be calculated using the combinations formula:

• \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \) \( n \) is the total number of items.
• \( r \) represents the number of items to be chosen.
• \( ! \) (factorial) means multiplying all whole numbers from itself down to one.

To understand combinations, consider selecting a committee from club members. The order of selection does not matter; whether Alice is chosen first or last, she is part of the committee. These principles help solve problems where grouping without regard to order is needed. This was used in the exercise to understand that choosing two groups from \( 2n \) boys without order is essential.

Binomial Coefficients

Binomial coefficients are the numerical factors or 'weights' for each term in the expansion of a binomial expression. These coefficients are exactly what combinations produce; in fact, they share the same mathematical formula:

• \( \binom{n}{r} \) represents the binomial coefficient.
• They enumerate the number of ways to choose \( r \) items from \( n \) without considering the order.

The coefficients are often seen in Pascal's Triangle, where each number is the sum of the two directly above it. When it comes to the probability exercises, like the one in question, using binomial coefficients helps to calculate the precise number of ways to split groups, arrange teams, or draw elements from a set. In the given problem, binomial coefficients helped determine both the total ways to arrange the boys and the ways in which the specific arrangement of interest could be achieved.

Probability Calculation

Probability calculation determines the likelihood of a given outcome. This is a central component of solving combinatorial probability problems, and it comprises the following steps:

• Define total possible outcomes. In the exercise, this was the total ways to divide the \( 2n \) boys into two groups, calculated using \( \binom{2n}{n} \).
• Identify favorable outcomes. This relates to the specific condition: ensuring the tallest boys were in separate groups, calculated by \( \binom{2n-2}{n-1} \).

The fundamental formula for determining probability is:

• \( P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \)

By applying this method, we find the likelihood of the tallest boys being in separate groups among the possible configurations. Calculating probabilities in this manner enables us to quantify the expected versus random occurrences as demonstrated when concluding the accurate probability or confirming answers for similar problems.