Exercise Solutions Available
Step-by-step answers for all textbook questions in this chapter
Chapter Introduction
Probability is Chapter 14 of CBSE Class 11 Mathematics, falling under the Statistics and Probability unit. This chapter formalizes the intuitive idea of chance you encountered in earlier classes. You learn the language of random experiments, sample spaces, and events, and build probability on an axiomatic foundation. This chapter typically carries 6-8 marks in board exams and is essential for both Class 12 probability and competitive exams.
The key idea: probability quantifies uncertainty. It assigns a number between 0 and 1 to every event, where 0 means impossible and 1 means certain.
Key Definitions
| Term | Definition |
|---|---|
| Random Experiment | An experiment whose outcome cannot be predicted with certainty before it is performed |
| Sample Space (S) | The set of all possible outcomes of a random experiment |
| Sample Point | Each element (outcome) of the sample space |
| Event | A subset of the sample space |
| Impossible Event | The empty set (null set); probability = 0 |
| Sure Event | The entire sample space S; probability = 1 |
| Simple Event | An event consisting of exactly one sample point |
| Compound Event | An event consisting of more than one sample point |
| Complementary Event | For event A, the complement A' = S - A; P(A') = 1 - P(A) |
| Mutually Exclusive Events | Events with no common outcomes: A intersection B = null |
| Exhaustive Events | Events whose union equals the entire sample space S |
| Equally Likely Outcomes | Outcomes that have the same chance of occurring |
Detailed Concepts
1. Random Experiments and Sample Space
A random experiment has three properties: 1. It can be repeated under identical conditions. 2. The set of all possible outcomes is known in advance. 3. The exact outcome of a specific trial is unpredictable.
Examples of sample spaces:
| Experiment | Sample Space | n(S) |
|---|---|---|
| Tossing one coin | {H, T} | 2 |
| Tossing two coins | {HH, HT, TH, TT} | 4 |
| Tossing three coins | {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} | 8 |
| Rolling one die | {1, 2, 3, 4, 5, 6} | 6 |
| Rolling two dice | 36 ordered pairs | 36 |
Key rule: When two experiments are performed together, the total number of outcomes = n(S1) x n(S2).
2. Types of Events
| Event Type | Definition | Example (one die) |
|---|---|---|
| Impossible | Empty set | {7} — rolling a 7 on a standard die |
| Sure | Entire sample space | {1, 2, 3, 4, 5, 6} |
| Simple (Elementary) | Single outcome | {3} — rolling a 3 |
| Compound | Multiple outcomes | {2, 4, 6} — rolling an even number |
| Complementary | S - A | {1, 3, 5} if A = {2, 4, 6} |
3. Algebra of Events
Given events A and B:
| Operation | Meaning | Result |
|---|---|---|
| A union B | A or B or both occur | Outcomes in A or B or both |
| A intersection B | Both A and B occur | Outcomes common to A and B |
| A' (complement) | A does not occur | Outcomes in S but not in A |
| A - B | A occurs but B does not | Outcomes in A but not in B |
Mutually exclusive: A intersection B = null (no common outcomes) Exhaustive: A union B = S (together cover all outcomes)
4. Axiomatic Approach to Probability
The three axioms of probability (Kolmogorov):
Axiom 1: P(A) >= 0 for every event A (non-negativity)
Axiom 2: P(S) = 1 for the sure event S (normalization)
Axiom 3: If A and B are mutually exclusive, then P(A union B) = P(A) + P(B) (additivity)
From these axioms, all probability rules can be derived.
5. Probability of an Event
P(E) = n(E) / n(S) (when all outcomes are equally likely)
Properties directly derived from axioms:
- P(impossible event) = 0
- P(sure event) = 1
- 0 <= P(E) <= 1 for any event E
- P(A') = 1 - P(A)
- P(A union B) = P(A) + P(B) - P(A intersection B)
- If A and B are mutually exclusive: P(A union B) = P(A) + P(B)
6. Mutually Exclusive and Exhaustive Events
Mutually exclusive events: A_1, A_2, ..., A_n are mutually exclusive if A_i intersection A_j = null for all i != j.
Exhaustive events: A_1, A_2, ..., A_n are exhaustive if A_1 union A_2 union ... union A_n = S.
Mutually exclusive and exhaustive: If both conditions hold, then P(A_1) + P(A_2) + ... + P(A_n) = 1.
7. Probability of Complement
P(A') = 1 - P(A)
This is one of the most useful rules. When finding P(A) is difficult, find P(A') and subtract from 1.
Example: Probability of getting at least one head in 3 coin tosses.
P(at least one head) = 1 - P(no heads) = 1 - P(TTT) = 1 - 1/8 = 7/8
8. Odds in Favour and Odds Against
Odds in favour of E = P(E) : P(E') = n(E) : n(E')
Odds against E = P(E') : P(E) = n(E') : n(E)
If odds in favour are a : b, then P(E) = a/(a + b). If odds against are a : b, then P(E) = b/(a + b).
Event Types — Summary Table
| Event Type | Set Notation | Probability |
|---|---|---|
| Impossible | null set | 0 |
| Sure (certain) | S | 1 |
| Simple (elementary) | {omega} (single point) | 1/n(S) if equally likely |
| Compound | subset with >1 point | n(E)/n(S) |
| Complementary | A' = S - A | 1 - P(A) |
| Mutually exclusive (A, B) | A intersection B = null | P(A) + P(B) for union |
| Exhaustive (A, B) | A union B = S | P(A) + P(B) - P(A intersection B) = 1 |
Key Probability Rules
| Rule | Formula | Condition |
|---|---|---|
| Classical probability | P(E) = n(E)/n(S) | Equally likely outcomes |
| Complement rule | P(A') = 1 - P(A) | Always |
| Addition rule | P(A union B) = P(A) + P(B) - P(A intersection B) | Always |
| Addition (mutually exclusive) | P(A union B) = P(A) + P(B) | A intersection B = null |
| Range | 0 <= P(E) <= 1 | Always |
| P(S) | P(S) = 1 | Always |
| P(null) | P(null) = 0 | Always |
| Odds in favour a:b | P(E) = a/(a+b) | a, b > 0 |
| Odds against a:b | P(E) = b/(a+b) | a, b > 0 |
Important Numericals
Example 1: Coin probability
Two coins are tossed. Find the probability of getting at least one head.
S = {HH, HT, TH, TT}, n(S) = 4
E = at least one head = {HH, HT, TH}, n(E) = 3
P(E) = 3/4
Example 2: Dice probability
A die is thrown. Find the probability of getting a prime number.
S = {1, 2, 3, 4, 5, 6}, n(S) = 6
E = prime number = {2, 3, 5}, n(E) = 3
P(E) = 3/6 = 1/2
Example 3: Two dice
Two dice are thrown. Find the probability that the sum is 7.
n(S) = 36
Favourable outcomes: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6
P(sum = 7) = 6/36 = 1/6
Example 4: Card probability
A card is drawn from a well-shuffled deck of 52 cards. Find the probability of drawing a king.
n(S) = 52
n(king) = 4
P(king) = 4/52 = 1/13
Example 5: Mutually exclusive events
If P(A) = 0.3, P(B) = 0.5, and A and B are mutually exclusive, find P(A union B).
Since A and B are mutually exclusive: P(A union B) = P(A) + P(B) = 0.3 + 0.5 = 0.8
Example 6: Using complement rule
Three dice are thrown. Find the probability that at least one shows 5.
P(at least one 5) = 1 - P(no 5)
P(no 5 on one die) = 5/6
P(no 5 on three dice) = (5/6)^3 = 125/216
P(at least one 5) = 1 - 125/216 = 91/216
Example 7: Odds in favour
If odds in favour of event E are 3 : 5, find P(E).
P(E) = 3/(3 + 5) = 3/8
Common Mistakes to Avoid
1. Not listing the complete sample space: For two dice, n(S) = 36 (not 21). (1,2) and (2,1) are different outcomes. Always count ordered pairs unless the problem specifies otherwise.
2. Confusing "at least one" with "exactly one": "At least one head" means 1 or 2 or 3 heads. "Exactly one head" means precisely one head. Use the complement rule for "at least one" problems.
3. Assuming events are mutually exclusive: Do not assume A and B are mutually exclusive unless the problem states it. Use P(A union B) = P(A) + P(B) - P(A intersection B) in general, and the simplified version only when you know A intersection B = null.
4. Counting errors in card problems: A standard deck has 52 cards: 13 each of hearts, diamonds, clubs, spades. There are 4 aces, 4 kings, 4 queens, 4 jacks. 12 face cards. 26 red cards. Know these numbers cold.
5. Probability exceeding 1: If your calculated P(E) > 1, you have made an error. Check your counting.
6. Forgetting equally likely condition: P(E) = n(E)/n(S) works only when outcomes are equally likely. If a coin is biased, you cannot use 1/2 for each outcome.
7. Wrong complement: The complement of "at least one" is "none" (not "at most one"). The complement of "all" is "at least one does not." Be precise.
Exam Tips
- For 1-mark questions, know: definitions of sample space, event types, P(A') = 1 - P(A), and the three axioms.
- For 3-mark questions, practice: coin/dice problems, card problems, mutually exclusive event calculations.
- For 5-mark questions, practice: two-dice problems, complement rule applications, problems combining multiple events, odds calculations.
- Always write the sample space S first, then identify the favourable outcomes.
- For card problems, memorize the deck composition: 52 cards, 4 suits, 13 cards per suit.
- When the problem says "two dice," always count ordered pairs: n(S) = 36.
- For "at least one" problems, use the complement rule: P(at least one) = 1 - P(none). This is much faster than counting directly.
- Draw Venn diagrams for problems involving A union B and A intersection B.
Quick Revision Notes
NCERT Exercise Problem Patterns
| Exercise | Topic | Typical Question Type |
|---|---|---|
| 14.1 | Sample space, events | List sample space, identify event types |
| 14.2 | Probability calculation | P(E) = n(E)/n(S), coins, dice, cards |
| 14.3 | Mutually exclusive, exhaustive, complement | Addition rule, P(A'), odds |
Derivations and Key Proofs
Derivation 1: P(A') = 1 - P(A)
Since A and A' are mutually exclusive and exhaustive:
A union A' = S, and A intersection A' = null
By Axiom 3: P(S) = P(A) + P(A')
By Axiom 2: 1 = P(A) + P(A')
Derivation 2: Addition rule for non-mutually exclusive events
For any two events A and B:
(A union B) = A union (B - A), where B - A = B intersection A'
These are mutually exclusive, so: P(A union B) = P(A) + P(B - A) = P(A) + P(B intersection A')
Now, B = (B intersection A) union (B intersection A'), which are mutually exclusive.
P(B) = P(B intersection A) + P(B intersection A')
Therefore P(B intersection A') = P(B) - P(A intersection B)
Substituting: P(A union B) = P(A) + P(B) - P(A intersection B)
Derivation 3: For mutually exclusive events, P(A union B) = P(A) + P(B)
When A and B are mutually exclusive: A intersection B = null, so P(A intersection B) = 0.
From the addition rule: P(A union B) = P(A) + P(B) - 0 = P(A) + P(B)
Card Deck Composition — Quick Reference
| Category | Count | Cards |
|---|---|---|
| Total cards | 52 | — |
| Suits | 4 | Hearts, Diamonds, Clubs, Spades |
| Cards per suit | 13 | A, 2-10, J, Q, K |
| Red cards | 26 | Hearts + Diamonds |
| Black cards | 26 | Clubs + Spades |
| Face cards | 12 | J, Q, K of each suit (4 x 3) |
| Aces | 4 | One per suit |
| Honours | 16 | A, K, Q, J of each suit (4 x 4) |
| Number cards | 36 | 2-10 of each suit (4 x 9) |
Additional Solved Examples
Example 8: Two dice — specific condition
Two dice are thrown. Find the probability that the sum is at most 5.
n(S) = 36
Favourable outcomes (sum <= 5): (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1) = 10
P(sum at most 5) = 10/36 = 5/18
Example 9: Card problem — multiple conditions
A card is drawn from a deck. Find the probability of getting a face card or a red card.
n(face cards) = 12, n(red cards) = 26, n(red face cards) = 6
P(face or red) = 12/52 + 26/52 - 6/52 = 32/52 = 8/13
Example 10: Probability of complement
A bag contains 5 red and 3 blue balls. Two balls are drawn at random. Find the probability that at least one is red.
P(at least one red) = 1 - P(both blue)
Total ways = C(8,2) = 28. Ways to pick 2 blue = C(3,2) = 3
P(both blue) = 3/28
P(at least one red) = 1 - 3/28 = 25/28
Example 11: Odds problem
If P(E) = 3/7, find the odds in favour of and odds against E.
P(E') = 1 - 3/7 = 4/7
Odds in favour = P(E) : P(E') = 3/7 : 4/7 = 3 : 4
Odds against = P(E') : P(E) = 4 : 3
Practice Checklist
- List complete sample spaces for: 1 coin, 2 coins, 3 coins, 1 die, 2 dice
- Solve 3 coin problems (various conditions)
- Solve 3 dice problems (sum, product, specific outcomes)
- Solve 3 card problems (suit, rank, colour, face)
- Use the complement rule in at least 2 "at least one" problems
- Apply the addition rule for 2 non-mutually exclusive events
- Convert between probability and odds (2 problems each way)
Quick Revision Notes
1. Sample space S: set of all possible outcomes; P(S) = 1 2. Event: subset of S; P(null) = 0, P(S) = 1, 0 <= P(E) <= 1 3. P(E) = n(E)/n(S) (equally likely outcomes) 4. Complement: P(A') = 1 - P(A) — most useful shortcut 5. Addition rule: P(A union B) = P(A) + P(B) - P(A intersection B) 6. Mutually exclusive: A intersection B = null; P(A union B) = P(A) + P(B) 7. Exhaustive: A union B = S 8. Odds in favour a:b -> P(E) = a/(a+b) 9. Odds against a:b -> P(E) = b/(a+b) 10. Two coins: n(S) = 4; Two dice: n(S) = 36 11. Cards: 52 total, 4 of each rank, 13 of each suit, 26 red, 26 black, 12 face cards 12. "At least one": always use 1 - P(none) instead of counting directly