JEE Main 2021 (24 February Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Probability: The probability that two randomly selected subsets of the set ${1, 2, 3, 4, 5}$ have exactly two elements in their intersection, is:

Select Answer:

Visualized Solution

Visualizing the Set and Subsets

Given set $S = {1, 2, 3, 4, 5}$ with $n = 5$ elements.
Two subsets $A$ and $B$ are selected independently from $S$ .
We need to find $P (∣ A \cap B ∣ = 2)$ .

Mapping Elements to Regions

For any element $x \in S$ , there are 4 mutually exclusive possibilities:
1. $x \in A$ and $x \in / B$ (Region $R_{1}$ )
2. $x \in A \cap B$ (Region $R_{2}$ )
3. $x \in / A$ and $x \in B$ (Region $R_{3}$ )
4. $x \in / A$ and $x \in / B$ (Region $R_{4}$ )

Calculating Total Outcomes

Total number of pairs of subsets $(A, B)$ :
Each of the 5 elements has exactly 4 choices ( $R_{1}, R_{2}, R_{3}, R_{4}$ ).
Total outcomes $= 4 \times 4 \times 4 \times 4 \times 4 = 4^{5}$ .

Simplifying Total Outcomes

Total outcomes $= 4^{5}$ .
Since $4 = 2^{2}$ , we can write this as $(2^{2})^{5}$ .
Total outcomes $= 2^{10} = 1024$ .

Defining Favorable Condition

Favorable condition: $∣ A \cap B ∣ = 2$ .
This means exactly 2 elements must fall into Region $R_{2}$ (the intersection).
The remaining $5 - 2 = 3$ elements must fall into any of the other 3 regions ( $R_{1}, R_{3},$ or $R_{4}$ ).

Choosing Elements for Intersection

Number of ways to choose 2 elements for $A \cap B$ :
We select 2 elements out of 5.
Ways $= (2 5)$ .

Evaluating Intersection Choices

Ways $= (2 5) = \frac{5 \times 4}{2 \times 1}$ .
Ways $= 10$ .

Arranging Remaining Elements

For each of the remaining 3 elements, there are 3 choices (Regions $R_{1}, R_{3}, R_{4}$ ).
They cannot go into $R_{2}$ , otherwise the intersection would have more than 2 elements.
Number of ways for remaining elements $= 3 \times 3 \times 3 = 3^{3}$ .

Calculating Favorable Outcomes

Total favorable outcomes $= (2 5) \times 3^{3}$ .
Total favorable outcomes $= 10 \times 27 = 270$ .

Final Probability Calculation

Probability $P = \frac{Favorable Outcomes}{Total Outcomes}$ .
$P = \frac{270}{4 ^{5}} = \frac{270}{1024}$ .

Simplifying the Fraction

$P = \frac{270}{1024}$ .
Dividing numerator and denominator by 2:
$P = \frac{135}{512} = \frac{135}{2 ^{9}}$ .

Conclusion and Key Takeaway

Key Takeaway: For any 2 subsets of a set with $n$ elements, there are $4^{n}$ total pairs of subsets.
The number of pairs $(A, B)$ such that $∣ A \cap B ∣ = r$ is $(r n) 3^{n - r}$ .
Final Answer: $\frac{135}{2 ^{9}}$ .

00:00 / 00:00

The Sigma Insight: Classical Definition of Probability

Solution Diagram

Analyzing the Setup

The 'element-wise' perspective is a powerful tool in your JEE arsenal. Instead of viewing subsets

A

and

B

as monolithic entities, we focus on the individual elements of the set

S = {1, 2, 3, 4, 5}

Each element acts as a "free agent" that can occupy one of four distinct regions in a Venn diagram: 1. In

A

but not

B

. 2. In

B

but not

A

. 3. In both

A

and

B

(the intersection). 4. In neither

A

nor

B

Since each of the

5

elements has

4

independent choices, the total number of ways to form the ordered pair

(A, B)

is:

4^{5} = 1024

This value represents the foundation of our sample space.

The Combinatorial Dance

To satisfy the condition

∣ A \cap B ∣ = 2

, we must perform a two-step selection process.

First, we choose exactly

2

elements out of

5

to occupy the intersection. The number of ways to do this is given by the combination formula:

(2 5) = \frac{5 !}{2 ! 3 !} = 10

Next, we consider the remaining

3

elements. These elements are constrained such that they cannot be in the intersection. Therefore, each of these

3

elements has only

3

remaining choices (in

A

only, in

B

only, or in neither).

The number of ways to distribute these

3

elements is:

3^{3} = 27

Final Calculation

By the fundamental principle of counting, the total number of favorable outcomes is the product of our two steps:

10 \times 27 = 270

The probability

P

is the ratio of favorable outcomes to the total sample space:

P = \frac{270}{1024}

Simplifying this fraction by dividing both the numerator and the denominator by

2

, we arrive at the final result:

$P = \frac{135}{512}$

This method generalizes beautifully. For a set of size

n

and an intersection size

r

, the probability is given by the formula:

\frac{( r n ) 3 ^{n - r}}{4 ^{n}}

Common Traps & FAQs

Why is the total number of outcomes

4^{5}

and not something like

2^{5} \times 2^{5}

Great question! Think about it this way: each element has 4 choices (only in A, only in B, in both, or in neither). Since there are 5 elements, it is

4 \times 4 \times 4 \times 4 \times 4 = 4^{5}

. This is equivalent to

2^{5} \times 2^{5}

because

4^{5} = (2^{2})^{5} = 2^{10} = 2^{5} \times 2^{5}

. It is just two different ways of looking at the same sample space! Both are correct, but the 4-region approach is often more intuitive for intersection problems.

Why do we use

(2 5)

Because the problem requires *exactly* two elements in the intersection. We aren't just placing elements; we are selecting which specific elements belong to that intersection.

(2 5)

counts the number of ways to pick those 2 lucky elements out of the 5 available.

What happens to the other 3 elements?

They are the 'non-intersection' elements. They have 3 choices each: they can be in

A

only,

B

only, or outside both. They just cannot be in the intersection, because that would make the intersection size greater than 2! So, for each of these 3 elements, there are 3 valid regions, leading to

3^{3}

total arrangements for them.

Animated Solution for Mathematics - Probability: The probability that two randomly selected subsets of the set ${1, 2, 3, 4, 5}$ have exactly two elements in their intersection, is:

Select Answer:

Visualized Solution

Visualizing the Set and Subsets

Mapping Elements to Regions

Calculating Total Outcomes

Simplifying Total Outcomes

Defining Favorable Condition

Choosing Elements for Intersection

Evaluating Intersection Choices

Arranging Remaining Elements

Calculating Favorable Outcomes

Final Probability Calculation

Simplifying the Fraction

Conclusion and Key Takeaway

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

The Combinatorial Dance

Final Calculation

Similar Questions

$A$ is a set containing $n$ elements. A subset $P$ of $A$ is chosen at random. The set $A$ is reconstructed by replacing the elements of $P$ . A subset $Q$ of $A$ is again chosen at random. Find the probability that $P$ and $Q$ have no common elements.

If 10 different balls has to be placed in 4 distinct boxes at random, then the probability that two of these boxes contain exactly 2 and 3 balls is :

If 10 different balls are to be placed in 4 distinct boxes at random, then the probability that two of these boxes contain exactly 2 and 3 balls is :

Two numbers are selected randomly from the set $S = {1, 2, 3, 4, 5, 6}$ without replacement one by one. The probability that minimum of the two numbers is less than 4 is

One die has two faces marked 1, two faces marked 2, one face marked 3 and one face marked 4. Another die has one face marked 1, two faces marked 2, two faces marked 3 and one face marked 4. The probability of getting the sum of numbers to be 4 or 5, when both the dice are thrown together, is

The probability that a randomly selected 2-digit number belongs to the set ${n \in N : (2^{n} - 2) is a multiple of 3}$ is equal to

If two different numbers are taken from the set ${0, 1, 2, 3, \dots, 10}$ ; then the probability that their sum as well as absolute difference are both multiple of 4, is:

Let $S = {1, 2, \dots, 20}$ . A subset $B$ of $S$ is said to be "nice", if the sum of the elements of $B$ is 203. Then the probability that a randomly chosen subset of $S$ is "nice" is:

Two squares are chosen at random on a chessboard (see figure). The probability that they have a side in common is :

The probability that a randomly chosen 5-digit number is made from exactly two digits is :

Select Answer:

Visualized Solution

Visualizing the Set and Subsets

Mapping Elements to Regions

Calculating Total Outcomes

Simplifying Total Outcomes

Defining Favorable Condition

Choosing Elements for Intersection

Evaluating Intersection Choices

Arranging Remaining Elements

Calculating Favorable Outcomes

Final Probability Calculation

Simplifying the Fraction

Conclusion and Key Takeaway

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

The Combinatorial Dance

Final Calculation

Similar Questions

If 10 different balls has to be placed in 4 distinct boxes at random, then the probability that two of these boxes contain exactly 2 and 3 balls is :

If 10 different balls are to be placed in 4 distinct boxes at random, then the probability that two of these boxes contain exactly 2 and 3 balls is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Two numbers are selected randomly from the set S={1,2,3,4,5,6} without replacement one by one. The probability that minimum of the two numbers is less than 4 is

One die has two faces marked 1, two faces marked 2, one face marked 3 and one face marked 4. Another die has one face marked 1, two faces marked 2, two faces marked 3 and one face marked 4. The probability of getting the sum of numbers to be 4 or 5, when both the dice are thrown together, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The probability that a randomly selected 2-digit number belongs to the set {n∈N:(2n−2) is a multiple of 3} is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If two different numbers are taken from the set {0,1,2,3,…,10}; then the probability that their sum as well as absolute difference are both multiple of 4, is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S={1,2,…,20}. A subset B of S is said to be "nice", if the sum of the elements of B is 203. Then the probability that a randomly chosen subset of S is "nice" is:

Two squares are chosen at random on a chessboard (see figure). The probability that they have a side in common is :

The probability that a randomly chosen 5-digit number is made from exactly two digits is :

Two numbers are selected randomly from the set $S = {1, 2, 3, 4, 5, 6}$ without replacement one by one. The probability that minimum of the two numbers is less than 4 is

The probability that a randomly selected 2-digit number belongs to the set ${n \in N : (2^{n} - 2) is a multiple of 3}$ is equal to

If two different numbers are taken from the set ${0, 1, 2, 3, \dots, 10}$ ; then the probability that their sum as well as absolute difference are both multiple of 4, is:

Let $S = {1, 2, \dots, 20}$ . A subset $B$ of $S$ is said to be "nice", if the sum of the elements of $B$ is 203. Then the probability that a randomly chosen subset of $S$ is "nice" is: