Problem 32 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(X\) be a set containing 10 elements and \(P(X)\) be its power set. If A and B are picked up at random from \(\mathrm{P}(\mathrm{X})\), with replacement, then the probability that \(\mathrm{A}\) and \(\mathrm{B}\) have equal number elements, is: \(\quad\) [Online April 10, 2015] (a) \(\frac{\left(2^{10}-1\right)}{2^{10}}\) (b) \(\frac{20^{0} C_{10}}{2^{10}}\) (c) \(\frac{\left(2^{10}-1\right)}{2^{20}}\) (d) \(\frac{20^{0} C_{10}}{2^{20}}\)

Step-by-Step Solution

Answer

(d) \(\frac{\binom{20}{10}}{2^{20}}\)

Step 1: Calculate Size of Power Set

The power set of a set with \( n \) elements has \( 2^n \) elements. Given \( X \) has 10 elements, its power set \( P(X) \) contains \( 2^{10} = 1024 \) subsets.

Step 2: Determine Possibilities for Pair (A, B)

Since \( A \) and \( B \) are chosen from \( P(X) \) with replacement, the total number of ordered pairs \( (A, B) \) possible is \( 2^{10} \times 2^{10} = 2^{20} \).

Step 3: Calculate Subsets with Equal Number of Elements

A subset of \( X \) can have between 0 and 10 elements. The number of subsets with \( k \) elements is given by \( \binom{10}{k} \). For each \( k \), pairs \( (A, B) \) where both have \( k \) elements can be \( \binom{10}{k}^2 \).

Step 4: Sum Over All Possible k Values

Total pairs where \( A \) and \( B \) have the same number of elements are \( \sum_{k=0}^{10} \binom{10}{k}^2 \). By the identity \[ \sum_{k=0}^{n} \binom{n}{k}^2 = \binom{2n}{n} \], this equals \( \binom{20}{10} \).

Step 5: Compute the Probability

The probability that \( A \) and \( B \) have the same number of elements is \( \frac{\binom{20}{10}}{2^{20}} \), which corresponds to option (d) in the problem statement.

Key Concepts

Power Set

The power set is a fundamental concept in set theory. It represents the set of all possible subsets of a given set, including the empty set and the set itself. If you picture a set as a collection of unique items, the power set encompasses all the different ways you can group these items, no matter the size of the group.
When dealing with a set with \( n \) elements, the power set will have \( 2^n \) elements. This is because each element can either be included or excluded from a subset. For example, given a set \( X \) with 10 elements, the power set \( P(X) \) will contain \( 2^{10} = 1024 \) subsets.

• Includes all subsets: from an empty set to the set itself.
• Total elements: \( 2^n \).

Understanding power sets is crucial when studying different combinations of elements, which is a key part of probability in combinatorics.

Binomial Coefficient

The binomial coefficient, often denoted as \( \binom{n}{k} \), is a key tool in combinatorics. It provides the number of ways to choose \( k \) elements from a set of \( n \) distinct elements, regardless of order. This concept is critical when calculating combinations and is central to binomial expansions.
Mathematically, it is defined as:\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]This formula calculates how many different ways you can pick \( k \) items from \( n \) without concerning the arrangement. For our exercise, each element of a subset of size \( k \) from a set of 10 elements can form the binomial coefficients like \( \binom{10}{k} \) to count subsets.

• Shows different combinations possible.
• Used for permutation and combination calculations.

Using this coefficient helps in figuring out combinatorial probabilities, especially when subsets have identical elements.

Combinatorial Probability

Combinatorial probability involves determining the likelihood of a particular combination occurring, based on the total number of possible combinations. This concept is applied when the outcomes are equally likely, such as when picking subsets from a power set.
Our problem focuses on pairs \( (A, B) \) from a power set. Here the probability we aim to find is for the event where both \( A \) and \( B \) have the same number of elements. To solve this, we'd count all corresponding pairs and divide by the total number of combinations, \( 2^{20} \), which represent all possibilities.

• Focuses on equal likelihood scenarios.
• Calculates probabilities from combinatorial counts.

Understanding combinatorial probabilities lets you predict outcomes in scenarios with a large number of possible combinations.

Set Theory

Set theory forms the foundation of modern mathematics, dealing with the study of sets, which are collections of objects. This concept explores how sets can interact, be combined, and queried. Basic operations include union, intersection, and complement.
In the context of our problem, set theory underpins the entire idea of counting subsets and calculating combinations from them. It provides the language and framework necessary for interpreting the structure and function of sets and their relationships.

• Defines procedures and relationships between sets.
• Helps in understanding the structure of mathematical concepts.

Grasping set theory allows you to handle large datasets and complex mathematical models efficiently, making it essential for combinatorial problems and beyond.