Event in Probability
In Class XI Probability, the concept of an event forms the foundation for all further topics such as probability calculation, conditional probability, and random variables. Understanding events precisely is crucial for CBSE Board Exams as well as competitive exams like JEE, NEET, CUET.
A random experiment is an action whose outcome cannot be predicted with certainty, but all possible outcomes are known. The set of all possible outcomes is called the sample space, denoted by \(S\).
An event is defined as any subset of the sample space.
\[ \boxed{E \subseteq S} \]
This means an event represents a collection of outcomes in which we are interested.
Types of Events (Highly Important for Exams)
- Simple Event: Contains only one outcome.
Example: Getting 3 on a die → \(E = \{3\}\) - Compound Event: Contains more than one outcome.
Example: Even numbers → \(E = \{2,4,6\}\) - Sure Event: Event equal to sample space.
\(E = S\) - Impossible Event: Event with no outcomes.
\(E = \varnothing\)
Visual Understanding (SVG Illustration)
The event \(E\) is always a subset inside the sample space \(S\)
Worked Examples (Board + JEE Level)
Example 1 (Basic):
A die is thrown once.
\[
S = \{1,2,3,4,5,6\}
\]
Event: Getting an even number
\[
E = \{2,4,6\}
\]
Since \(E \subseteq S\), it is a valid event.
Example 2 (Competitive Level):
A card is drawn from a standard deck of 52 cards.
Event \(A\): Getting a red card
\[
n(A) = 26
\]
Event \(B\): Getting a king
\[
B = \{4 \text{ kings}\}
\]
Both are subsets of sample space → hence events.
Example 3 (Conceptual Trap - JEE/NEET):
If \(S = \{1,2,3\}\), then:
• \(E = \{1,4\}\) is not an event because 4 ∉ S.
Key Properties (Exam Ready)
- \( \varnothing \subseteq S \) → Impossible event always exists
- \( S \subseteq S \) → Sure event always exists
- Every event is a set → Set theory is directly applied
- Number of possible events = \(2^n\) (where n = number of outcomes)
Why This Topic is Important
- Foundation of entire probability chapter
- Directly used in conditional probability and Bayes theorem (Class XII)
- Frequently asked in MCQs (Boards + CUET)
- Concept clarity required for JEE Main logical questions
AI Insight (Smart Shortcut)
If any element of a set is outside the sample space → it is NOT an event.
Always check: Is every element inside S?
Practice Booster (Try Yourself)
A coin is tossed twice.
Sample space:
\[
S = \{HH, HT, TH, TT\}
\]
- Find event of getting exactly one head
- Find event of getting at least one tail
Tip: Convert statements into sets → then check subset condition.
Types of Events in Probability
The classification of events is a core scoring area in Class XI Probability and forms the base for solving advanced problems in JEE Main, NEET, CUET and higher probability concepts like conditional probability and Bayes’ theorem.
From an exam perspective, questions are frequently asked to identify event types, check logical validity, and apply set operations. Strong conceptual clarity here gives a direct advantage in MCQs.
-
Simple (Elementary) Event:
An event containing exactly one outcome.\[ n(E)=1 \]
Example: In a die throw, \[ E=\{3\} \]
Exam Insight: These are the building blocks of probability → each outcome contributes equally in classical probability.
-
Compound Event:
An event containing more than one outcome.\[ n(E)>1 \]
Example: \[ E=\{2,4,6\} \]
JEE Focus: Most real probability questions deal with compound events rather than simple ones.
-
Certain (Sure) Event:
An event equal to the entire sample space.\[ E=S \]
Example: Getting a number less than 7 in a die throw.
Key Result: Probability of a certain event = 1
-
Impossible Event:
An event that has no outcomes.\[ E=\varnothing \]
Example: Getting 7 on a standard die.
Key Result: Probability of an impossible event = 0
-
Complementary Event:
For any event \(E\), the event “not \(E\)” is called its complement.\[ E' = S - E \]
Example: If \(E=\{2,4,6\}\), then \[ E'=\{1,3,5\} \]
Important Identity: \[ P(E) + P(E') = 1 \]
Visual Classification (SVG Diagram)
Inside region → Event \(E\), Outside region → Complement \(E'\)
Advanced Types (JEE / NEET Extension)
- Equal Events: Two events having same outcomes → \(A = B\)
- Equivalent Events: Different events but same number of outcomes → \(n(A)=n(B)\)
- Mutually Exclusive Events: Cannot occur together → \(A \cap B = \varnothing\)
- Exhaustive Events: Cover entire sample space → \(A_1 \cup A_2 \cup ... = S\)
Worked Example (Exam Level)
A die is thrown.
Let:
- \(A = \{2,4,6\}\) (even numbers)
- \(B = \{1,3,5\}\) (odd numbers)
Then:
- \(A \cap B = \varnothing\) → Mutually exclusive
- \(A \cup B = S\) → Exhaustive
- \(B = A'\) → Complementary
AI Shortcut (High Accuracy Trick)
If intersection = ∅ → Mutually Exclusive
If both → Complementary events
Practice Zone (IIT / NEET Pattern)
A card is drawn from a deck of 52 cards.
- Event A: Getting a red card
- Event B: Getting a face card
Identify:
- Are A and B mutually exclusive?
- Find \(A \cap B\)
- Find complement of A
Tip: Always convert events into sets → then apply set operations.
Algebra of Events (Core Probability Engine)
The algebra of events connects probability with set theory operations. This is one of the most high-weightage concepts in CBSE Boards and a frequent testing ground in JEE Main, NEET, CUET, especially through logical MCQs and multi-step problems.
Mastery of union, intersection, complement, and difference allows you to convert real-world statements into mathematical probability expressions instantly.
Complementary Events
Let \(S\) be the sample space and \(A\) an event. The complement of \(A\) contains all outcomes not in \(A\).
\[ A' = S - A \]
Interpretation:
- \(A\): Event occurs
- \(A'\): Event does not occur
Exactly one of \(A\) or \(A'\) must occur in every trial → binary certainty principle.
Key Properties
- \(A \cap A' = \varnothing\) (Mutually exclusive)
- \(A \cup A' = S\) (Exhaustive)
Probability Law
\[ P(A') = 1 - P(A) \]
Event “A or B” (Union)
The event “A or B” means at least one occurs.
\[ A \cup B \]
- Only A occurs
- Only B occurs
- Both occur
Probability Formula (General Case)
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Why subtract? Because common outcomes are counted twice.
Special Case (Mutually Exclusive)
\[ P(A \cup B) = P(A) + P(B) \]
Event “A and B” (Intersection)
The event “A and B” means both events occur together.
\[ A \cap B \]
Contains only the common outcomes of both events.
JEE Insight: Intersection represents logical AND → often hidden inside word problems.
Event “A but not B” (Difference)
This represents outcomes where \(A\) occurs but \(B\) does not.
\[ A \cap B' \quad \text{or} \quad A - B \]
Probability Formula
\[ P(A - B) = P(A) - P(A \cap B) \]
Key Insight: Break complex events into disjoint parts.
Mutually Exclusive Events
Two events are mutually exclusive if they cannot occur together.
\[ A \cap B = \varnothing \]
- If A occurs → B cannot occur
- No overlap between events
Probability Rule
\[ P(A \cup B) = P(A) + P(B) \]
Visual Logic Map (SVG)
Overlap → \(A \cap B\), Entire area → \(A \cup B\)
Worked Example (JEE / NEET Level)
A card is drawn from a deck.
- \(A\): Red card → 26 outcomes
- \(B\): King → 4 outcomes
- \(A \cap B\): Red kings → 2 outcomes
\[ P(A \cup B) = \frac{26}{52} + \frac{4}{52} - \frac{2}{52} = \frac{28}{52} \]
AI Exam Strategy
AND → Use intersection
NOT → Use complement
“Only” → Use difference (subtract overlap)
Practice Drill (Exam Pattern)
A die is thrown:
- \(A = \{2,4,6\}\)
- \(B = \{4,5,6\}\)
- Find \(A \cup B\)
- Find \(A \cap B\)
- Find \(A - B\)
- Check if mutually exclusive
Tip: Always draw a quick mental Venn diagram before solving.
Exhaustive Events (Complete Coverage Concept)
The concept of exhaustive events is fundamental in probability theory and plays a crucial role in solving board-level numericals and JEE/NEET MCQs. It ensures that no possible outcome is left unaccounted for.
Let \(S\) be the sample space and \(A, B, C, \ldots\) be events. These events are said to be exhaustive if their union equals the entire sample space.
\[ A \cup B = S \]
\[ A \cup B \cup C = S \]
Interpretation: At least one of the events must occur in every trial.
Conceptual Understanding
- No outcome lies outside the given events
- All possibilities are completely covered
- Useful in breaking complex problems into complete cases
Visual Representation (SVG)
Combined region of \(A\) and \(B\) completely fills sample space \(S\)
Probability Result (Core Formula)
If \(A\) and \(B\) are exhaustive, then:
\[ P(A \cup B) = P(S) = 1 \]
Using the addition rule:
\[ P(A) + P(B) - P(A \cap B) = 1 \]
This identity is frequently used in MCQs where one probability is missing.
Special Case (Exhaustive + Mutually Exclusive)
If events are both exhaustive and mutually exclusive:
\[ A \cup B = S \quad \text{and} \quad A \cap B = \varnothing \]
\[ P(A) + P(B) = 1 \]
Worked Examples (Boards + JEE Level)
Example 1 (Basic):
A die is thrown.
Let:
\(A = \{1,2,3\}\), \(B = \{4,5,6\}\)
Then:
- \(A \cup B = S\) → Exhaustive
- \(A \cap B = \varnothing\) → Mutually exclusive
- \(P(A)+P(B)=1\)
Example 2 (Competitive Level):
A card is drawn.
\(A\): Red card, \(B\): Black card
- Every card is either red or black → exhaustive
- No overlap → mutually exclusive
Therefore: \(P(A) + P(B) = 1\)
Common Mistake (Exam Trap)
Always check: Does their union equal entire sample space?
AI Shortcut Strategy
- If union = S → Exhaustive
- If intersection = ∅ → Mutually Exclusive
- If both → Directly use \(P(A)+P(B)=1\)
Practice Drill (IIT / NEET Pattern)
A number is chosen from 1 to 10.
- \(A\): Even numbers
- \(B\): Odd numbers
- Check if exhaustive
- Check if mutually exclusive
- Verify \(P(A)+P(B)=1\)
Tip: Always convert into sets → then test union and intersection.
Axiomatic Approach to Probability (Mathematical Foundation)
The axiomatic approach to probability, introduced by Kolmogorov, provides a rigorous mathematical framework for probability. Unlike classical probability, this approach does not depend on equally likely outcomes and is therefore widely used in advanced mathematics, statistics, AI, and machine learning.
Exam Importance: Frequently tested in CBSE Boards (theory + reasoning) and forms the conceptual base for Class XII topics like conditional probability and Bayes’ theorem.
Basic Structure
Let \(S\) be the sample space of a random experiment. A probability function \(P\) assigns a real number to each event \(E \subseteq S\).
\[ P : \mathcal{P}(S) \rightarrow [0,1] \]
where \( \mathcal{P}(S) \) is the power set of \(S\).
Kolmogorov’s Three Axioms
- Non-Negativity: \[ P(E) \geq 0 \]
- Normalization: \[ P(S) = 1 \]
- Additivity (for disjoint events): \[ P(E \cup F) = P(E) + P(F), \quad \text{if } E \cap F = \varnothing \]
Derived Results (Very Important)
1. Probability of Impossible Event
\[ P(\varnothing) = 0 \]
Derived using additivity axiom → frequently asked in theory questions.
2. Range of Probability
\[ 0 \leq P(E) \leq 1 \]
3. Complement Rule
\[ P(E') = 1 - P(E) \]
4. Addition Rule (General Case)
\[ P(E \cup F) = P(E) + P(F) - P(E \cap F) \]
Finite Sample Space Model
Let:
\[ S = \{\omega_1, \omega_2, \ldots, \omega_n\} \]
Each outcome \( \omega_i \) has a probability:
\[ 0 \leq P(\omega_i) \leq 1 \]
Total probability:
\[ \sum_{i=1}^{n} P(\omega_i) = 1 \]
For any event \(A\):
\[ P(A) = \sum_{\omega_i \in A} P(\omega_i) \]
Visual Representation (SVG)
Each outcome has an assigned probability → total equals 1
Worked Example (Board + JEE Level)
A biased coin has: \[ P(H) = 0.7, \quad P(T) = 0.3 \]
- Check: \(P(H)+P(T)=1\) → Valid probability distribution
- Event \(A\): Getting head → \(P(A)=0.7\)
- Complement: \(P(A')=0.3\)
Why Axiomatic Approach Matters
- Works even when outcomes are not equally likely
- Foundation of modern probability theory
- Used in data science, AI, statistics
- Essential for higher mathematics (Class XII & beyond)
AI Exam Strategy
1. Probability ≥ 0
2. Total probability = 1
3. Use complement when direct calculation is hard
Practice Drill (IIT / NEET Pattern)
A random experiment has outcomes with probabilities: \[ 0.2,\; 0.3,\; x \]
- Find \(x\)
- Check if valid probability distribution
- Find probability of complement of first outcome
Tip: Sum must be 1 → solve instantly.
Probability of an Event (Classical Approach)
Probability is a numerical measure that quantifies the likelihood of occurrence of an event. It is one of the most important concepts in Class XI Mathematics and serves as a direct gateway to advanced topics in statistics, data science, artificial intelligence, and competitive exams like JEE & NEET.
In real-world scenarios such as tossing a coin, rolling a die, or drawing a card, outcomes are uncertain. Probability helps convert this uncertainty into a precise mathematical value.
Basic Terminology (Quick Recall)
- Random Experiment: An action with uncertain outcome
- Sample Space (\(S\)): Set of all possible outcomes
- Event (\(E\)): Subset of sample space
- Favourable Outcomes: Outcomes satisfying event \(E\)
Classical Definition of Probability
If all outcomes of a random experiment are equally likely, then the probability of an event \(E\) is defined as:
\[ P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \]
\[ P(E) = \frac{n(E)}{n(S)} \]
Visual Understanding (SVG)
Probability = favourable outcomes / total outcomes
Range of Probability
\[ 0 \leq P(E) \leq 1 \]
- \(P(E)=0\) → Impossible event
- \(P(E)=1\) → Certain event
Worked Examples (Boards + JEE Level)
Example 1 (Basic):
A fair die is thrown.
Find probability of getting an even number.
\[ S = \{1,2,3,4,5,6\}, \quad E = \{2,4,6\} \]
\[ P(E) = \frac{3}{6} = \frac{1}{2} \]
Example 2 (Card Problem):
A card is drawn from a deck of 52 cards.
Find probability of getting a heart.
\[ P(E) = \frac{13}{52} = \frac{1}{4} \]
Example 3 (JEE Trick):
Find probability of not getting a 6 on a die.
\[ P(\text{not 6}) = 1 - P(6) = 1 - \frac{1}{6} = \frac{5}{6} \]
Conditions for Applying Formula
- Outcomes must be equally likely
- Sample space must be finite
- Events must be clearly defined
Why This Topic is Important
- Direct formula-based questions in CBSE exams
- High-speed MCQs in JEE Main & NEET
- Foundation for conditional probability (Class XII)
- Used in real-world decision making & AI systems
AI Shortcut Strategy
Step 2: Identify favourable outcomes
Step 3: Apply \(n(E)/n(S)\)
Step 4: Simplify fraction
Practice Drill (IIT / NEET Pattern)
Two dice are thrown. Find probability of:
- Sum = 7
- Sum is even
- At least one die shows 6
Tip: Total outcomes = 36 → use counting smartly.
Example 2: Mutually Exclusive Events (Two Dice)
Two dice are thrown and the sum is noted. Consider the events:
- \(A\): Sum is even
- \(B\): Sum is a multiple of 3
- \(C\): Sum is less than 4
- \(D\): Sum is greater than 11
Objective: Identify which pairs are mutually exclusive.
Step 1: Sample Space Insight
Total outcomes when two dice are thrown:
\[ n(S) = 6 \times 6 = 36 \]
Each outcome is an ordered pair \((i,j)\), where \(i,j \in \{1,2,3,4,5,6\}\).
Step 2: Convert Events Using Sum Logic (Smart Method)
Instead of listing all 36 outcomes, use sum-based classification (JEE shortcut).
- Event A (Even sum): 2, 4, 6, 8, 10, 12
- Event B (Multiple of 3): 3, 6, 9, 12
- Event C (Sum < 4): 2, 3
- Event D (Sum > 11): 12
Visual Understanding (SVG)
Overlapping region → common outcomes → not mutually exclusive
Step 3: Check Intersections
Check \(C \cap D\):
- \(C = \{2,3\}\)
- \(D = \{12\}\)
\[ C \cap D = \varnothing \]
Therefore, \(C\) and \(D\) are mutually exclusive.
Step 4: Check Other Pairs (Quick Elimination)
- \(A \cap B \neq \varnothing\) → common sums: 6, 12
- \(A \cap C \neq \varnothing\) → common sum: 2
- \(B \cap C \neq \varnothing\) → common sum: 3
- \(A \cap D \neq \varnothing\) → common sum: 12
- \(B \cap D \neq \varnothing\) → common sum: 12
Hence, none of these pairs are mutually exclusive.
Final Answer
\(C\) and \(D\)
Why This Question is Important
- Tests understanding of mutually exclusive events
- Common MCQ pattern in JEE & NEET
- Builds speed using sum-based reasoning
AI Shortcut Strategy
Step 2: Check intersection quickly
Step 3: If no common element → mutually exclusive
Practice Extension
Two dice are thrown. Define:
- \(E\): Sum is prime
- \(F\): Sum is odd
Check if \(E\) and \(F\) are mutually exclusive.
Tip: Prime sums = 2, 3, 5, 7, 11 → compare with odd sums.
Example 3: Mutually Exclusive & Exhaustive Events (3 Coin Tosses)
A coin is tossed three times. Consider the events:
- \(A\): No head appears
- \(B\): Exactly one head appears
- \(C\): At least two heads appear
Objective: Check whether these events are mutually exclusive and exhaustive.
Step 1: Construct Sample Space
Total outcomes when a coin is tossed 3 times:
\[ n(S) = 2^3 = 8 \]
\[ S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\} \]
Step 2: Define Events Clearly
Classify outcomes based on number of heads:
- Event A (0 heads): \(\{TTT\}\)
- Event B (exactly 1 head): \(\{HTT, THT, TTH\}\)
- Event C (≥ 2 heads): \(\{HHT, HTH, THH, HHH\}\)
Visual Understanding (SVG)
Disjoint regions covering entire sample space
Step 3: Check Mutual Exclusivity
Two events are mutually exclusive if their intersection is empty.
\[ A \cap B = \varnothing, \quad A \cap C = \varnothing, \quad B \cap C = \varnothing \]
No outcome is common → Events are mutually exclusive.
Step 4: Check Exhaustiveness
Events are exhaustive if their union equals the sample space:
\[ A \cup B \cup C = S \]
Every outcome belongs to exactly one event → Events are exhaustive.
Final Conclusion
✔ Mutually Exclusive
✔ Exhaustive
Why This Example is Important
- Classic example of partition of sample space
- Frequently asked in CBSE theory + MCQs
- Builds base for probability distribution (Class XII)
AI Shortcut Strategy
✔ No overlap → Mutually Exclusive
✔ Full coverage → Exhaustive
Practice Extension (JEE / NEET Level)
A coin is tossed 4 times. Define:
- \(A\): Exactly 2 heads
- \(B\): At most 1 head
- \(C\): At least 3 heads
- Check if mutually exclusive
- Check if exhaustive
Tip: Count heads → classify systematically.
Example 4: Probability with Playing Cards
One card is drawn from a well-shuffled deck of 52 cards. Find the probability that the card is:
- (i) a diamond
- (ii) not an ace
- (iii) a black card
- (iv) not a diamond
- (v) not a black card
Step 1: Core Concept
Since the deck is well-shuffled, all outcomes are equally likely.
\[ P(E) = \frac{\text{Favourable Outcomes}}{52} \]
Visual Understanding (SVG Structure of Deck)
4 suits → Each has 13 cards → Total = 52
(i) Probability of a Diamond
Number of diamond cards = 13
\[ P(\text{diamond}) = \frac{13}{52} = \frac{1}{4} \]
(ii) Probability of Not an Ace
Number of aces = 4
\[ P(\text{ace}) = \frac{4}{52} = \frac{1}{13} \]
\[ P(\text{not ace}) = 1 - \frac{1}{13} = \frac{12}{13} \]
(iii) Probability of a Black Card
Black cards = Clubs + Spades = 26
\[ P(\text{black}) = \frac{26}{52} = \frac{1}{2} \]
(iv) Probability of Not a Diamond
\[ P(\text{not diamond}) = 1 - \frac{1}{4} = \frac{3}{4} \]
(v) Probability of Not a Black Card
\[ P(\text{not black}) = 1 - \frac{1}{2} = \frac{1}{2} \]
Final Answers Summary
- (i) \( \frac{1}{4} \)
- (ii) \( \frac{12}{13} \)
- (iii) \( \frac{1}{2} \)
- (iv) \( \frac{3}{4} \)
- (v) \( \frac{1}{2} \)
Why This Example is Important
- Classic card probability model (very common in exams)
- Tests both direct counting and complement method
- Frequently appears in JEE Main & NEET MCQs
AI Shortcut Strategy
Suit-based → 13 cards each
Color-based → 26 cards each
Ace/King → Always 4 cards
Practice Extension (Exam Level)
A card is drawn. Find probability of:
- Getting a face card
- Getting a red king
- Getting neither red nor ace
Tip: Convert into counts → then apply probability formula.
Example 5: Probability with Colored Discs
A bag contains 9 discs: 4 red, 3 blue, and 2 yellow. A disc is drawn at random. Find the probability that it is:
- (i) red
- (ii) yellow
- (iii) blue
- (iv) not blue
- (v) either red or blue
Step 1: Core Understanding
Since all discs are identical in shape and size, each outcome is equally likely.
\[ \text{Total outcomes} = 9 \]
- Red = 4
- Blue = 3
- Yellow = 2
Visual Representation (SVG)
Distribution of discs in the bag
(i) Probability of Red
\[ P(R) = \frac{4}{9} \]
(ii) Probability of Yellow
\[ P(Y) = \frac{2}{9} \]
(iii) Probability of Blue
\[ P(B) = \frac{3}{9} = \frac{1}{3} \]
(iv) Probability of Not Blue
Using complement:
\[ P(\text{not }B) = 1 - \frac{1}{3} = \frac{2}{3} \]
(v) Probability of Red or Blue
Since a disc cannot be both red and blue:
\[ R \cap B = \varnothing \]
So, using addition rule:
\[ P(R \cup B) = P(R) + P(B) \]
\[ = \frac{4}{9} + \frac{1}{3} = \frac{4}{9} + \frac{3}{9} = \frac{7}{9} \]
Final Answers Summary
- (i) \( \frac{4}{9} \)
- (ii) \( \frac{2}{9} \)
- (iii) \( \frac{1}{3} \)
- (iv) \( \frac{2}{3} \)
- (v) \( \frac{7}{9} \)
Why This Example Matters
- Tests basic probability + complement
- Uses mutually exclusive addition rule
- Common pattern in CBSE & entrance exams
AI Shortcut Strategy
OR → Add probabilities (if no overlap)
Always simplify fractions
Practice Extension (JEE / NEET Level)
A bag has 5 red, 4 green, 3 blue balls. Find:
- Probability of not green
- Probability of red or blue
- Probability of neither red nor blue
Tip: Convert “neither” → complement of (red ∪ blue)
Example 6: Probability of Qualifying (Two Events)
Two students Anil and Ashima appeared in an examination.
- \(P(E)\): Probability Anil qualifies = 0.05
- \(P(F)\): Probability Ashima qualifies = 0.10
- \(P(E \cap F)\): Probability both qualify = 0.02
Find the required probabilities.
Step 1: Key Formula (Union)
\[ P(E \cup F) = P(E) + P(F) - P(E \cap F) \]
\[ = 0.05 + 0.10 - 0.02 = 0.13 \]
Visual Understanding (SVG)
Overlap represents \(P(E \cap F)\)
(a) Both Do Not Qualify
This is:
\[ E' \cap F' = (E \cup F)' \]
\[ P(E' \cap F') = 1 - P(E \cup F) \]
\[ = 1 - 0.13 = 0.87 \]
(b) At Least One Does Not Qualify
Complement of both qualifying:
\[ P(\text{at least one fails}) = 1 - P(E \cap F) \]
\[ = 1 - 0.02 = 0.98 \]
(c) Only One Qualifies
This means:
\[ (E \cap F') \cup (E' \cap F) \]
Using formula:
\[ P = P(E) - P(E \cap F) + P(F) - P(E \cap F) \]
\[ = (0.05 - 0.02) + (0.10 - 0.02) \]
\[ = 0.03 + 0.08 = 0.11 \]
Final Answers
- (a) \(0.87\)
- (b) \(0.98\)
- (c) \(0.11\)
Why This Example is Important
- Combines union, intersection, complement
- Very common in JEE Main & NEET MCQs
- Tests logical interpretation of phrases like “at least” and “only”
AI Shortcut Strategy
At least one fails → complement of both pass
Only one → subtract overlap from both sides
Practice Extension (Exam Level)
Given: \[ P(A)=0.6,\; P(B)=0.5,\; P(A \cap B)=0.3 \]
- Find probability neither occurs
- Find probability exactly one occurs
- Find probability at least one occurs
Tip: Use union and complement smartly.
Example 7: Committee Selection (Combinatorics + Probability)
A committee of 2 persons is selected from 2 men and 2 women. Find the probability that the committee has:
- (a) no man
- (b) one man
- (c) two men
Step 1: Total Possible Committees
Total persons = 4 → selecting 2
\[ n(S) = {}^{4}C_{2} = 6 \]
Visual Understanding (SVG)
Selection from two distinct groups
(a) Probability of No Man (Both Women)
Choose 2 women from 2:
\[ n(E) = {}^{2}C_{2} = 1 \]
\[ P = \frac{1}{6} \]
(b) Probability of Exactly One Man
Choose:
- 1 man from 2 → \( {}^{2}C_{1} = 2 \)
- 1 woman from 2 → \( {}^{2}C_{1} = 2 \)
\[ n(E) = 2 \times 2 = 4 \]
\[ P = \frac{4}{6} = \frac{2}{3} \]
(c) Probability of Two Men
Choose 2 men from 2:
\[ n(E) = {}^{2}C_{2} = 1 \]
\[ P = \frac{1}{6} \]
Final Answers Summary
- (a) \( \frac{1}{6} \)
- (b) \( \frac{2}{3} \)
- (c) \( \frac{1}{6} \)
Concept Insight (Very Important)
- This is a combination-based probability problem
- Uses concept of favourable cases / total cases
- Also represents a probability distribution
AI Shortcut Strategy
Step 2: Count favourable combinations
Step 3: Divide and simplify
Step 4: Check sum = 1 (sanity check)
Check: \( \frac{1}{6} + \frac{2}{3} + \frac{1}{6} = 1 \) ✔
Practice Extension (JEE / NEET Level)
A committee of 3 is selected from 3 men and 2 women. Find probability that:
- All are men
- At least one woman is selected
- Exactly two men are selected
Tip: Use complement for “at least” type questions.
✦ OR fraction like 3/13 to analyse