Probability of Simple Events
A simple event is an event that consists of exactly one outcome from the sample space. It is the most basic building block in probability.
When a die is rolled, getting a '4' is a simple event. Getting an even number ({2, 4, 6}) is a compound event because it contains more than one outcome.
Calculating probability of simple events is the first step before tackling compound events. Every compound event can be broken down into simple events.
What is Probability of Simple Events?
Definition: A simple event (or elementary event) is an event that contains exactly one outcome of the experiment.
Properties:
- A simple event has only one element from the sample space.
- Simple events are mutually exclusive — no two can occur simultaneously.
- The sum of probabilities of all simple events = 1.
- Every event (including compound events) is a union of simple events.
Examples of simple events:
- Tossing a coin: {H} and {T} are each simple events.
- Rolling a die: {1}, {2}, {3}, {4}, {5}, {6} are the 6 simple events.
- Drawing a card: Each of the 52 cards is a simple event.
Examples of compound events (NOT simple):
- Getting an even number on a die: {2, 4, 6} — contains 3 outcomes.
- Getting a face card: {J, Q, K of all suits} — contains 12 outcomes.
Probability of Simple Events Formula
Probability of a Simple Event:
P(simple event) = 1 / n(S)
Where:
- n(S) = total number of equally likely outcomes in the sample space
Since a simple event has exactly 1 favourable outcome: n(E) = 1.
For any event E (simple or compound):
P(E) = n(E) / n(S)
Sum rule:
- If S = {e₁, e₂, ..., eₙ}, then P(e₁) + P(e₂) + ... + P(eₙ) = 1.
- For equally likely outcomes: each P(eᵢ) = 1/n.
Derivation and Proof
Why each simple event has probability 1/n(S):
- The sample space S has n(S) outcomes, all equally likely.
- Every outcome must have the same probability — call it p.
- The total probability must equal 1 (something must happen).
- So: n(S) × p = 1.
- Therefore: p = 1/n(S).
- Each simple event {eᵢ} has probability p = 1/n(S).
Building compound events from simple events:
- If E = {e₁, e₃, e₅}, then P(E) = P(e₁) + P(e₃) + P(e₅) = 3 × (1/n(S)) = 3/n(S).
- This is exactly the formula P(E) = n(E)/n(S) with n(E) = 3.
Important: Simple events are mutually exclusive — the occurrence of one rules out every other. When rolling a die, if the outcome is 3, it cannot simultaneously be 5. This is why we can ADD probabilities of simple events.
Types and Properties
Classification of events by number of outcomes:
| Type | Definition | Example (die roll) |
|---|---|---|
| Simple event | Exactly 1 outcome | {4} — getting a 4 |
| Compound event | More than 1 outcome | {2, 4, 6} — getting even |
| Impossible event | 0 outcomes (empty set) | {} — getting 7 |
| Sure event | All outcomes | {1,2,3,4,5,6} — getting < 7 |
Common experiments and their simple events:
- One coin: 2 simple events — {H}, {T}. Each has P = 1/2.
- One die: 6 simple events — {1}, {2}, {3}, {4}, {5}, {6}. Each has P = 1/6.
- Two coins: 4 simple events — {HH}, {HT}, {TH}, {TT}. Each has P = 1/4.
- One card from deck: 52 simple events. Each has P = 1/52.
- Bag with n objects: n simple events. Each has P = 1/n.
Methods
Steps to solve probability of simple events:
- Define the experiment: What is being done? (toss, roll, draw, pick)
- Write the sample space: List ALL possible outcomes.
- Verify equally likely: Confirm each outcome has the same chance.
- Identify the simple event: Which single outcome is asked about?
- Apply the formula: P = 1/n(S).
For compound events built from simple events:
- Identify which simple events form the compound event.
- Count them: n(E).
- Apply: P(E) = n(E)/n(S).
Common pitfalls:
- Treating "at least one head in two tosses" as a simple event — it is compound: {HH, HT, TH}.
- Listing unequal outcomes as if they were equally likely.
- Forgetting to count repeated elements. In MISSISSIPPI, S appears 4 times — each letter position is a separate simple event.
Solved Examples
Example 1: Simple Event on a Die
Problem: A die is rolled. Find the probability of the simple event {3}.
Solution:
Given:
- S = {1, 2, 3, 4, 5, 6}, n(S) = 6
- E = {3} — a simple event with 1 outcome
Using P = 1/n(S):
- P({3}) = 1/6
Answer: P(getting 3) = 1/6
Example 2: Simple Event on a Coin
Problem: A fair coin is tossed. Find the probability of getting a tail.
Solution:
Given:
- S = {H, T}, n(S) = 2
- E = {T} — simple event
Using the formula:
- P(T) = 1/2
Answer: P(Tail) = 1/2
Example 3: Drawing a Specific Card
Problem: A card is drawn from a deck. Find the probability of drawing the Ace of Spades.
Solution:
Given:
- n(S) = 52 (total cards)
- E = {Ace of Spades} — exactly 1 card
Using the formula:
- P(Ace of Spades) = 1/52
Answer: P(Ace of Spades) = 1/52
Example 4: Picking a Specific Ball
Problem: A box contains 15 balls numbered 1 to 15. One is drawn at random. Find the probability of getting ball number 9.
Solution:
Given:
- n(S) = 15
- E = {9} — simple event
Using P = 1/n(S):
- P({9}) = 1/15
Answer: P(ball number 9) = 1/15
Example 5: Simple Events in Two Coin Tosses
Problem: Two coins are tossed simultaneously. Find the probability of the simple event {HT} (first coin Head, second coin Tail).
Solution:
Given:
- S = {HH, HT, TH, TT}, n(S) = 4
- E = {HT} — a specific ordered outcome
Using the formula:
- P({HT}) = 1/4
Note: {HT} and {TH} are different simple events. "Exactly one head" is the compound event {HT, TH} with P = 2/4 = 1/2.
Answer: P(HT) = 1/4
Example 6: Sum of Probabilities of All Simple Events
Problem: Verify that the sum of probabilities of all simple events for a die roll equals 1.
Solution:
Simple events: {1}, {2}, {3}, {4}, {5}, {6}
- P({1}) = 1/6
- P({2}) = 1/6
- P({3}) = 1/6
- P({4}) = 1/6
- P({5}) = 1/6
- P({6}) = 1/6
Sum:
- 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 6/6 = 1
Answer: Sum = 1. Verified.
Example 7: Compound Event from Simple Events
Problem: A die is rolled. Find the probability of getting a number divisible by 3.
Solution:
Given:
- S = {1, 2, 3, 4, 5, 6}, n(S) = 6
- Numbers divisible by 3: {3, 6}
- This is a compound event = union of simple events {3} and {6}
Using P(E) = n(E)/n(S):
- P(divisible by 3) = 2/6 = 1/3
Alternatively: P({3}) + P({6}) = 1/6 + 1/6 = 2/6 = 1/3.
Answer: P(divisible by 3) = 1/3
Example 8: Letter from a Word
Problem: A letter is chosen at random from the word 'SCHOOL'. Find the probability of choosing the letter 'O'.
Solution:
Given:
- Letters: S, C, H, O, O, L → 6 letters total
- Each letter position is a simple event → n(S) = 6
- Positions with 'O': 4th and 5th → n(E) = 2
Using the formula:
- P(O) = 2/6 = 1/3
Answer: P(letter O) = 1/3
Example 9: Numbered Cards
Problem: Cards numbered 11 to 30 are in a bag. One card is drawn at random. Find the probability of getting card number 25.
Solution:
Given:
- Cards: 11, 12, 13, ..., 30 → total = 30 − 11 + 1 = 20 cards
- E = {25} — simple event
Using P = 1/n(S):
- P(card 25) = 1/20
Answer: P(card number 25) = 1/20
Example 10: Spinner Problem
Problem: A spinner has 8 equal sections numbered 1 to 8. Find the probability of landing on 5.
Solution:
Given:
- n(S) = 8 (equal sections, so equally likely)
- E = {5} — simple event
Using the formula:
- P(landing on 5) = 1/8
Answer: P(5) = 1/8
Real-World Applications
Lottery and Raffle Draws:
- Each ticket is a simple event. Probability of winning with 1 ticket out of n = 1/n.
Random Selection:
- Selecting a student at random from a class of 40: each student is a simple event with P = 1/40.
Quality Testing:
- Picking one item from a batch — each item is a simple event. Used to estimate defect rates.
Computer Science:
- Random number generators produce each number as a simple event with equal probability.
Genetics:
- Each allele combination in a Punnett square is a simple event.
Key Points to Remember
- A simple event has exactly one outcome from the sample space.
- For equally likely outcomes: P(simple event) = 1/n(S).
- Simple events are mutually exclusive — only one can occur at a time.
- The sum of probabilities of all simple events in a sample space = 1.
- A compound event is a collection of two or more simple events.
- P(compound event) = sum of probabilities of its constituent simple events.
- Each card in a 52-card deck, each face of a die, and each ball in a bag is a simple event.
- "At least one" and "at most one" describe compound events, not simple events.
- Always count the total outcomes carefully — errors in n(S) propagate to every calculation.
- Simple events form the foundation for all probability calculations in CBSE Class 10.
Practice Problems
- A die is rolled. Find the probability of the simple event {5}.
- A bag has 12 marbles of different colours. One marble is drawn at random. What is the probability of drawing any specific marble?
- Two coins are tossed. List all 4 simple events and find the probability of each.
- Cards numbered 1 to 25 are in a box. Find the probability of drawing card number 17.
- A spinner has 10 equal sections labelled A to J. Find the probability of landing on section D.
- A class has 45 students. One is chosen at random to answer a question. Find the probability that a specific student (say, Ravi) is chosen.
- A die is thrown. Express the event 'getting a number less than 4' as a union of simple events and find its probability.
- From the letters of the word 'EXAMPLE', one letter is chosen at random. Find the probability of choosing the letter 'E'.
Frequently Asked Questions
Q1. What is a simple event in probability?
A simple event is an event with exactly one outcome. Rolling a die and getting 4 is a simple event {4}. Getting an even number {2, 4, 6} is NOT a simple event — it is a compound event.
Q2. What is the difference between simple and compound events?
A simple event has exactly one outcome. A compound event has two or more outcomes. Example: {3} is simple; {3, 6} is compound. Every compound event can be expressed as a union of simple events.
Q3. Can two simple events happen at the same time?
No. Simple events are mutually exclusive. When a die is rolled, the outcome is exactly one number. Getting 3 and getting 5 cannot happen simultaneously in a single roll.
Q4. What is the probability of each simple event for a fair die?
For a fair die with 6 faces, each simple event has probability 1/6. The six simple events are {1}, {2}, {3}, {4}, {5}, {6}.
Q5. How do you find the probability of a compound event using simple events?
Add the probabilities of the simple events that make up the compound event. P(even number on die) = P({2}) + P({4}) + P({6}) = 1/6 + 1/6 + 1/6 = 3/6 = 1/2.
Q6. Why must the sum of all simple event probabilities equal 1?
Because one outcome MUST occur. The sample space covers every possibility. Since simple events are all the individual outcomes and they are mutually exclusive, their probabilities must add up to 1.
Q7. Is 'getting a head' on a coin toss a simple event?
Yes. The sample space is {H, T}. The event {H} contains exactly one outcome, making it a simple event with P(H) = 1/2.
Q8. How many simple events are there when two dice are rolled?
36 simple events. Each simple event is an ordered pair (a, b) where a and b are each from {1, 2, 3, 4, 5, 6}. Examples: (1,1), (1,2), ..., (6,6). Each has probability 1/36.
Q9. Are simple events always equally likely?
Not necessarily. Simple events are equally likely only when the experiment is fair (unbiased). For a loaded die, {1}, {2}, ..., {6} are still simple events, but they may not have equal probabilities.
Q10. Is this topic important for CBSE board exams?
Yes. Understanding simple events is essential for solving all probability problems in NCERT Class 10. Board questions require identifying outcomes, counting favourable cases, and applying P(E) = n(E)/n(S).
