Problem 115 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

A set \(A\) contains \(n\) elements. \(A\) subset \(P\) of \(A\) is chosen at random and the set \(A\) is reconstructed by replacing the elements of \(P\). Another subset \(\mathrm{Q}\) of \(A\) is now chosen at random. The probability that \(P U Q\) contains exactly \(r\) elements, \(1 \leq r \leq n\) is (A) \(\frac{{ }^{n} C_{r} 3^{r}}{4^{n}}\) (B) \(\frac{{ }^{n} C_{r} 4^{r}}{3^{n}}\)(C) \(\frac{3^{n}}{4^{n}}\) (D) none of these

Step-by-Step Solution

Answer

(A) \(\frac{\binom{n}{r} \times 3^r}{4^n}\).

Step 1: Understand the Problem

You need to determine the probability that the union of two randomly chosen subsets, P and Q, from a set A with n elements, has exactly r elements. The process involves choosing two subsets independently and then finding their union.

Step 2: Counting Total Possibilities

Calculate the total number of ways to choose the subsets P and Q from the set A. Since each element can either be in the subset or not, there are a total of \(2^{n}\) subsets for P and \(2^{n}\) subsets for Q. Therefore, the total number of ways to choose both P and Q is \((2^n) imes (2^n) = 4^n\).

Step 3: Determine Favorable Outcomes

To find the number of favorable outcomes (where \(P \cup Q\) has exactly r elements), note that this union includes exactly r distinct elements from A. First, choose r elements from n to be in the union, which can be done in \(\binom{n}{r}\) ways.

Step 4: Assign Elements between Subsets P and Q

Once r elements are fixed as part of \(P \cup Q\), each can be included in either P, Q, or both, except for not being included at all which is not possible here, giving 3 possibilities per element. Hence, there are \(3^r\) ways to assign r elements to P and Q.

Step 5: Calculate the Probability

Given the favorable outcomes from Step 4 and total possible outcomes from Step 2, the probability that \(P \cup Q\) has exactly r elements is \(\frac{\binom{n}{r} \times 3^r}{4^n}\).

Step 6: Choose Correct Option

Match the calculated probability with the given options. The expression \(\frac{\binom{n}{r} \times 3^r}{4^n}\) corresponds to option (A).

Key Concepts

Set Theory

Set theory is a branch of mathematical logic that studies collections of objects, known as sets. In this exercise, we are dealing with a set \(A\) that contains \(n\) unique elements. Sets are fundamental because they are used to define nearly all mathematical objects. A set is simply a collection of distinct objects, without regard to the order in which they are placed. Objects in a set are called elements, and these can be anything: numbers, letters, symbols, etc.
Set theory also includes operations like union, intersection, and difference, which are used to form new sets from existing sets.

• Union: Combines all elements from two sets, eliminating duplicates.
• Intersection: Includes only elements that are present in both sets.
• Difference: Consists of elements present in one set but not in another.

Understanding these operations is crucial to solving problems involving sets, as they allow us to manipulate and combine sets in numerous ways, like forming new sets \(P\) and \(Q\) from the original set \(A\).

Combinatorics

Combinatorics is the branch of mathematics concerning the counting, arrangement, and combination of objects. It's all about understanding how elements of a set can be selected or arranged under certain conditions. In the given problem, combinatorics helps us to calculate the number of ways to choose subsets \(P\) and \(Q\) from set \(A\).
One main technique used in combinatorics is the binomial coefficient, noted as \(\binom{n}{r}\). This represents the number of ways to choose \(r\) elements from a set of \(n\) elements, without considering the order. This is also known as "n choose r." For instance, to form the union \(P \cup Q\) with exactly \(r\) elements, we first select those \(r\) elements from \(n\), which can be achieved in \(\binom{n}{r}\) ways.
The essence of combinatorics in probability problems is to enumerate the possible outcomes. In our problem, after choosing \(r\) elements, they can be distributed among \(P\) and \(Q\) in \(3^r\) different ways. This comes from the fact that each of the \(r\) elements can exist in \(P\), \(Q\), or both—hence three choices per element.

Union of Sets

The union of sets is a crucial concept in set theory that involves combining all the elements from two sets. The notation for a union is \(P \cup Q\), which represents the set containing all elements that are either in \(P\) or in \(Q\) or in both.
The problem specifically asks for the union of subsets \(P\) and \(Q\) to contain exactly \(r\) elements. To achieve this, we first select \(r\) elements that will be part of this union, regardless of which subset they originate from. Once selected, these elements can be distributed to \(P\) and \(Q\) in a number of ways, except the scenario where an element doesn't appear in the union at all.

• Each chosen element must be included in \(P\), \(Q\), or both, giving three combinations per element.
• The union operation is essential here because it helps determine the presence of elements in any or both of the subsets.

The resulting set from the union operation helps define the overall solution to the problem by ensuring that exactly \(r\) elements are achieved from the prospective combinations of the original set \(A\).