Probability of Compound Events
A compound event consists of two or more simple events occurring together or in sequence. Examples include rolling two dice, drawing two cards, or tossing a coin and rolling a die simultaneously.
The sample space of a compound event is the set of all possible combined outcomes. For two dice, the sample space has 6 × 6 = 36 outcomes. For a coin and a die, it has 2 × 6 = 12 outcomes.
In Class 10, compound event problems focus on listing the combined sample space and computing probabilities by counting favourable outcomes.
Compound events are fundamental to understanding real-world probability, where multiple things happen simultaneously or in sequence. The key challenge is correctly enumerating the combined sample space and identifying favourable outcomes.
In Class 10, compound event problems primarily involve dice, coins, and cards. The multiplication principle (fundamental counting principle) determines the total number of outcomes.
The concept of compound events bridges the gap between simple probability (one experiment) and the multiplication rule of probability studied in higher classes. In Class 10, the focus is on systematically enumerating compound outcomes and counting favourable ones.
The most important skill is systematic enumeration — using grids, tables, or tree diagrams to list all possible outcomes without missing any or counting any twice. Problems involving two dice are by far the most common type in CBSE board exams.
Common mistakes include confusing ordered with unordered outcomes (treating (3,5) and (5,3) as one outcome), and forgetting to use the complement method for "at least" problems.
The key difference from simple events is that the sample space grows multiplicatively. While a single die has 6 outcomes, two dice have 36, and three dice have 216. Each additional experiment multiplies the sample space size by the number of outcomes of that experiment. This rapid growth means systematic enumeration methods (grids, trees) become essential for accurate counting.
What is Probability of Compound Events?
Definition: A compound event is an event whose sample space is formed by combining the outcomes of two or more individual experiments.
Key facts:
- If experiment 1 has m outcomes and experiment 2 has n outcomes, the combined sample space has m × n outcomes.
- Each combined outcome is an ordered pair (or tuple) of individual outcomes.
- The probability of a compound event = (Number of favourable combined outcomes) / (Total combined outcomes).
- All combined outcomes are assumed equally likely in Class 10 problems.
Tree Diagram Approach:
- A tree diagram represents outcomes as branches from left to right.
- The first level shows outcomes of experiment 1.
- From each outcome of experiment 1, branches show all outcomes of experiment 2.
- Each complete path from root to leaf represents one compound outcome.
- The total number of leaves = total compound outcomes.
Example — Tree for Coin + Die:
- Level 1: H and T (2 branches).
- From each: 1, 2, 3, 4, 5, 6 (6 branches each).
- Total leaves = 2 × 6 = 12 compound outcomes.
Independent Events:
- Two events are independent if the outcome of one does not affect the outcome of the other.
- Rolling two dice: the result on die 1 does not change the probabilities for die 2. They are independent.
- For independent events A and B: P(A and B) = P(A) × P(B).
- This multiplication rule is introduced in Class 10 through compound event problems, though the formal definition comes later.
Mutually Exclusive Events:
- Events that cannot happen simultaneously are mutually exclusive.
- On a single die: "getting 3" and "getting 5" are mutually exclusive (you cannot get both on one roll).
- But on two dice: "die 1 shows 3" and "die 2 shows 5" are NOT mutually exclusive — both can happen together.
- For mutually exclusive events: P(A or B) = P(A) + P(B).
Probability of Compound Events Formula
Probability of a Compound Event:
P(E) = Number of favourable outcomes / Total outcomes in sample space
Total outcomes:
- Two dice: 6 × 6 = 36
- Three coins: 2 × 2 × 2 = 8
- One coin + one die: 2 × 6 = 12
- Two cards drawn (with replacement): 52 × 52 = 2704
Multiplication Principle:
If one experiment has m outcomes and another has n outcomes, the total number of combined outcomes = m × n. This extends to three or more experiments: m × n × p × ...
Complement Strategy for Compound Events:
- For "at least one" problems, it is often easier to calculate the complement.
- P(at least one 6) = 1 − P(no 6 at all).
- P(no 6 on die 1) = 5/6. P(no 6 on die 2) = 5/6.
- P(no 6 on either) = (5/6) × (5/6) = 25/36.
- P(at least one 6) = 1 − 25/36 = 11/36.
For three coins:
- P(at least one head) = 1 − P(no heads) = 1 − P(TTT) = 1 − 1/8 = 7/8.
Derivation and Proof
Constructing the Sample Space for Two Dice:
Die 1 can show: {1, 2, 3, 4, 5, 6}. Die 2 can show: {1, 2, 3, 4, 5, 6}.
- Each outcome is an ordered pair (a, b) where a = die 1 result, b = die 2 result.
- Total outcomes = 6 × 6 = 36.
- The sample space can be written as a 6 × 6 grid:
S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),
(2,1), (2,2), (2,3), ..., (2,6),
...
(6,1), (6,2), ..., (6,6)}
Key Subsets (for common problems):
- Sum = 7: {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)} → 6 outcomes
- Doublets: {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)} → 6 outcomes
- Sum = 2: {(1,1)} → 1 outcome
- Sum = 12: {(6,6)} → 1 outcome
- Sum ≥ 10: {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)} → 6 outcomes
Sample Space for Three Coins:
- Coin 1: {H, T}, Coin 2: {H, T}, Coin 3: {H, T}.
- Total = 2 × 2 × 2 = 8 outcomes.
- S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
- Number of heads: 3(×1), 2(×3), 1(×3), 0(×1) — this is the binomial distribution pattern (1, 3, 3, 1).
Sample Space for Coin + Die:
- Coin: {H, T}, Die: {1, 2, 3, 4, 5, 6}.
- Total = 2 × 6 = 12 outcomes.
- S = {(H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}
Important observation: The number of outcomes that give a particular sum follows a triangular pattern — increasing from 1 to 6 and then decreasing back to 1. This symmetric distribution is characteristic of the sum of two uniform discrete random variables.
Types and Properties
Common Types of Compound Event Problems:
Type 1: Two Dice
- Sample space: 36 outcomes
- Problems ask for P(sum = k), P(doublet), P(product is even), etc.
Type 2: Two or More Coins
- Two coins: 4 outcomes {HH, HT, TH, TT}
- Three coins: 8 outcomes {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Type 3: Coin + Die
- 12 outcomes: {(H,1), (H,2), ..., (H,6), (T,1), ..., (T,6)}
Type 4: Drawing from a Bag (with replacement)
- Two draws with replacement: total outcomes = n², where n = number of items.
| Experiment | Total Outcomes |
|---|---|
| 1 die | 6 |
| 2 dice | 36 |
| 2 coins | 4 |
| 3 coins | 8 |
| 1 coin + 1 die | 12 |
| 1 die + 1 card | 312 |
Summary of Probabilities for Two Dice:
| Sum | Outcomes | Count | P(sum) |
|---|---|---|---|
| 2 | (1,1) | 1 | 1/36 |
| 3 | (1,2),(2,1) | 2 | 2/36 |
| 4 | (1,3),(2,2),(3,1) | 3 | 3/36 |
| 5 | (1,4),(2,3),(3,2),(4,1) | 4 | 4/36 |
| 6 | (1,5),(2,4),(3,3),(4,2),(5,1) | 5 | 5/36 |
| 7 | (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) | 6 | 6/36 |
| 8 | (2,6),(3,5),(4,4),(5,3),(6,2) | 5 | 5/36 |
| 9 | (3,6),(4,5),(5,4),(6,3) | 4 | 4/36 |
| 10 | (4,6),(5,5),(6,4) | 3 | 3/36 |
| 11 | (5,6),(6,5) | 2 | 2/36 |
| 12 | (6,6) | 1 | 1/36 |
The distribution is symmetric around 7, which has the highest probability.
Methods
Steps to solve compound event problems:
- Identify the individual experiments and their outcomes.
- Calculate the total number of combined outcomes using the multiplication principle.
- List or describe the sample space systematically (grid, tree diagram, or ordered pairs).
- Define the event E clearly — what condition must the combined outcome satisfy?
- Count the favourable outcomes (those satisfying the condition).
- Compute P(E) = favourable / total.
Tools for counting:
- Grid/Table: Best for two dice — list all 36 pairs in a 6×6 table.
- Tree diagram: Best for sequential events (coin then die).
- Systematic listing: Best for small sample spaces (2–3 coins).
Tips for Two-Dice Problems:
- Always write outcomes as ordered pairs (a, b) where a = first die, b = second die.
- (3, 5) and (5, 3) are different outcomes — do not combine them.
- For sum = k, the number of favourable outcomes follows a pattern: 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 for sums 2 through 12.
- Sum = 7 has the maximum probability (6/36 = 1/6).
Common Mistakes:
- Confusing ordered and unordered outcomes: When rolling two dice, (3, 5) ≠ (5, 3). Total is 36, not 21.
- Forgetting outcomes: Systematically list using a grid or tree diagram to avoid missing cases.
- Double counting: When finding P(A or B), ensure you don't count outcomes satisfying both A and B twice.
Systematic Grid Method for Two Dice:
Create a 6 × 6 grid with die 1 values (1-6) as rows and die 2 values (1-6) as columns. Each cell represents one outcome. To find P(event), highlight all cells satisfying the condition and count them. This method guarantees no outcomes are missed or double-counted.
Example — Grid for Sums:
- Shade all cells where row + column = 9: cells (3,6), (4,5), (5,4), (6,3) → 4 shaded cells.
- P(sum = 9) = 4/36 = 1/9.
Solved Examples
Example 1: Sum of Two Dice Equals 9
Problem: Two dice are thrown simultaneously. Find the probability that the sum is 9.
Solution:
Given:
- Total outcomes = 36
Favourable outcomes (sum = 9):
- (3, 6), (4, 5), (5, 4), (6, 3)
- Count = 4
P(sum = 9) = 4/36 = 1/9
Example 2: Doublet of Even Numbers
Problem: Two dice are rolled. Find the probability of getting a doublet of even numbers.
Solution:
Given:
- Total outcomes = 36
Favourable outcomes (doublet AND even):
- (2, 2), (4, 4), (6, 6)
- Count = 3
P(even doublet) = 3/36 = 1/12
Example 3: Product Greater Than 20
Problem: Two dice are thrown. Find P(product of numbers > 20).
Solution:
Total outcomes = 36
Listing products > 20:
- (4, 6): 24 ✓
- (5, 5): 25 ✓
- (5, 6): 30 ✓
- (6, 4): 24 ✓
- (6, 5): 30 ✓
- (6, 6): 36 ✓
- Count = 6
P(product > 20) = 6/36 = 1/6
Example 4: Three Coins — At Least Two Heads
Problem: Three coins are tossed simultaneously. Find P(at least 2 heads).
Solution:
Sample space (8 outcomes):
- {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Favourable (≥ 2 heads):
- Exactly 2 heads: HHT, HTH, THH → 3
- Exactly 3 heads: HHH → 1
- Total favourable = 4
P(at least 2 heads) = 4/8 = 1/2
Example 5: Coin and Die — Head and Even Number
Problem: A coin is tossed and a die is rolled. Find P(head and even number).
Solution:
Total outcomes = 2 × 6 = 12:
- {(H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}
Favourable (H and even):
- (H, 2), (H, 4), (H, 6)
- Count = 3
P(H and even) = 3/12 = 1/4
Example 6: Sum of Two Dice is Prime
Problem: Two dice are thrown. Find P(sum is a prime number).
Solution:
Possible sums: 2 to 12. Primes in this range: 2, 3, 5, 7, 11.
Counting favourable outcomes:
- Sum = 2: (1,1) → 1
- Sum = 3: (1,2), (2,1) → 2
- Sum = 5: (1,4), (2,3), (3,2), (4,1) → 4
- Sum = 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6
- Sum = 11: (5,6), (6,5) → 2
- Total = 1 + 2 + 4 + 6 + 2 = 15
P(prime sum) = 15/36 = 5/12
Example 7: Neither Sum is 7 Nor 11
Problem: Two dice are rolled. Find P(sum is neither 7 nor 11).
Solution:
Total = 36
P(sum = 7):
- (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes
P(sum = 11):
- (5,6), (6,5) → 2 outcomes
P(sum = 7 or 11) = (6 + 2)/36 = 8/36
P(neither 7 nor 11) = 1 − 8/36 = 28/36 = 7/9
Example 8: Two Coins — Exactly One Head
Problem: Two coins are tossed together. Find P(exactly one head).
Solution:
Sample space = {HH, HT, TH, TT}
Total outcomes = 4
Favourable (exactly one H):
- HT, TH → 2 outcomes
P(exactly one head) = 2/4 = 1/2
Example 9: Difference of Numbers on Two Dice
Problem: Two dice are thrown. Find P(difference between numbers is 2).
Solution:
Total = 36
Favourable (|a − b| = 2):
- (1,3), (2,4), (3,5), (4,6): die 1 is smaller → 4
- (3,1), (4,2), (5,3), (6,4): die 1 is larger → 4
- Total = 8
P(difference = 2) = 8/36 = 2/9
Example 10: At Most One Six on Two Dice
Problem: Two dice are rolled. Find P(at most one 6 appears).
Solution:
Total = 36
Complement: P(both show 6):
- Only (6, 6) → 1 outcome
- P(both 6) = 1/36
P(at most one 6) = 1 − 1/36 = 35/36
Real-World Applications
Games and Gambling:
- Board games like Ludo, Monopoly, and Snakes & Ladders use dice. The probability of specific sums determines game strategy.
Quality Control:
- Testing two components together — the combined probability of both passing determines batch acceptance rates.
Weather Forecasting:
- The probability of rain on two consecutive days is a compound event.
Genetics:
- Punnett squares model compound events — combining alleles from two parents to predict offspring traits.
Sports:
- The probability of winning a best-of-3 series requires computing compound probabilities across multiple games.
Insurance:
- Insurers assess the compound probability of multiple independent risk events occurring together to price policies.
Medicine:
- The probability of two independent medical tests both giving false negatives is a compound event calculation crucial for diagnostic reliability.
Cryptography:
- Password strength depends on compound probabilities — the chance of guessing each character correctly in sequence.
Key Points to Remember
- A compound event combines two or more simple experiments.
- Total outcomes = product of individual outcomes: m × n.
- Two dice: 36 outcomes. Three coins: 8 outcomes. Coin + die: 12 outcomes.
- P(E) = favourable compound outcomes / total compound outcomes.
- Use a grid for two dice, a tree diagram for sequential events.
- Sum of 7 on two dice has the highest probability (6/36 = 1/6).
- Sum of 2 and sum of 12 each have the lowest probability (1/36).
- For "at least" problems, use the complement: P(at least one) = 1 − P(none).
- Order matters: (3, 5) and (5, 3) are different outcomes when rolling two dice.
- All compound outcomes are assumed equally likely.
- The probability distribution of two-dice sums is symmetric around 7.
- The number of favourable outcomes for sum = k follows the pattern: 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 for k = 2 through 12.
- For "or" problems: P(A or B) = P(A) + P(B) if A and B are mutually exclusive.
- For "and" problems with independent events: P(A and B) = P(A) × P(B).
- A tree diagram is the best visual tool for listing all compound outcomes systematically.
- A grid/table is ideal for two-dice problems — plot die 1 values on rows and die 2 on columns.
- When events are independent, the probability of both occurring = product of individual probabilities.
- The complement approach (P(at least one) = 1 − P(none)) saves significant counting effort.
Practice Problems
- Two dice are thrown. Find the probability that the sum is less than 5.
- Two dice are rolled. Find P(both numbers are odd).
- Three coins are tossed. Find P(at most one tail).
- A coin and a die are used. Find P(tail and a number greater than 4).
- Two dice are thrown. Find P(sum is a multiple of 3).
- Two dice are thrown. Find P(at least one die shows 5).
- Three coins are tossed. Find P(no heads).
- Two dice are thrown. Find the probability that the product is 12.
Frequently Asked Questions
Q1. What is the difference between a simple event and a compound event?
A simple event involves a single experiment with a single outcome (rolling one die, tossing one coin). A compound event involves two or more experiments, and the outcome is a combination of individual outcomes.
Q2. Why does rolling two dice give 36 outcomes, not 12?
Each die has 6 faces. The outcome of die 1 is independent of die 2. By the multiplication principle, total outcomes = 6 × 6 = 36. Each outcome is an ordered pair (a, b).
Q3. Are (3, 5) and (5, 3) different outcomes?
Yes. (3, 5) means die 1 shows 3 and die 2 shows 5. (5, 3) means die 1 shows 5 and die 2 shows 3. They are distinct outcomes in the sample space, even though the sum is the same.
Q4. How do I handle 'at least' type problems efficiently?
Use the complement: P(at least one) = 1 − P(none). For example, P(at least one 6 on two dice) = 1 − P(no 6 on either) = 1 − (5/6)(5/6) = 1 − 25/36 = 11/36.
Q5. What is the most common sum when rolling two dice?
The sum 7 has the highest probability: 6/36 = 1/6. The six favourable outcomes are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).
Q6. Can compound events involve more than two experiments?
Yes. For example, tossing 3 coins gives 2³ = 8 outcomes. Rolling 3 dice gives 6³ = 216 outcomes. The multiplication principle extends to any number of independent experiments.
Q7. What is a tree diagram?
A tree diagram shows all possible outcomes of sequential events as branches. Each branch represents one possible outcome. The end of each complete path (from root to leaf) represents one compound outcome.
Q8. How is this topic different from probability with playing cards?
Playing card problems typically involve drawing from a single deck (52 cards). Compound events involve combining outcomes from two or more independent experiments. Card problems become compound events when two cards are drawn.
Q9. Is compound probability asked in CBSE board exams?
Yes. Two-dice problems and multi-coin problems appear frequently. Typical questions ask for P(sum = k), P(doublet), P(at least one head), etc. They carry 2–4 marks.
Q10. What does 'simultaneously' mean in probability questions?
When two dice are thrown 'simultaneously' or two coins are tossed 'together', it means the outcomes are independent. The mathematical treatment is the same whether the events happen at the same time or one after the other.










