Ever wondered why a tossed coin seems to “choose” heads or tails at random?
💡 In Simple Words: Probability tells us how likely something is to happen. If you have 2 equally possible outcomes, like a fair coin, each one has a 1 in 2 chance, or 0.5 probability.
What is Probability?
In everyday talk, probability is just a fancy word for “chance”. In maths, we define it as the ratio of favourable outcomes (the ways you want the event to happen) to total possible outcomes (all the ways anything can happen). Think of it like a jar of coloured marbles: if you want a red marble, the probability is the number of red marbles divided by the total marbles.
Key Formula Everyone Uses
The basic formula looks like this:
P(E) = Number of favourable outcomes ÷ Total number of outcomes
Where P(E) means “the probability of event E”. The result is always between 0 (impossible) and 1 (certain). If you prefer percentages, just multiply by 100.
Common Types of Events
- Simple (or elementary) event: Only one outcome is considered, e.g., drawing an ace from a pack of cards.
- Compound event: Two or more simple events together, like getting a head **or** a tail when a coin is tossed.
- Mutually exclusive events: Events that cannot happen at the same time, such as rolling a 3 and a 5 on a single die.
- Independent events: One event does not affect the chance of the other, like tossing a coin twice.
Worked Example 1 – Single Die Roll
Find the probability of getting an even number when a fair six‑sided die is rolled.
- List total outcomes: 1, 2, 3, 4, 5, 6 → 6 possibilities.
- Identify favourable outcomes: 2, 4, 6 → 3 possibilities.
- Apply the formula: P(even) = 3 ÷ 6 = 0.5 (or 50%).
Notice how easy it becomes once you write down the two sets.
Worked Example 2 – Drawing Two Cards Without Replacement
From a standard 52‑card deck, what’s the probability of drawing an ace **and then** a king, without putting the first card back?
- First draw – ace: 4 aces out of 52 cards → P₁ = 4/52 = 1/13.
- Second draw – king: after removing an ace, 51 cards remain, still 4 kings → P₂ = 4/51.
- Because the draws are independent only after the first card is removed, multiply the two probabilities: P = (1/13) × (4/51) = 4/663 ≈ 0.0060 (about 0.6%).
This shows how “without replacement” changes the total number of outcomes for the second step.
Quick Revision Table
| Event Type | Symbol | Typical Example |
|---|---|---|
| Simple event | P(A) | Drawing a red marble from a bag of red & blue marbles |
| Compound event (or) | P(A or B) | Getting a head **or** a tail on a coin toss |
| Compound event (and) | P(A and B) | Rolling a 2 **and** a 4 on two dice |
| Mutually exclusive | P(A ∩ B)=0 | Rolling a 3 **and** a 5 on one die |
| Independent | P(A and B)=P(A)·P(B) | Two separate coin tosses |
Tips for Solving ICSE Probability Questions
- Read the question twice – note whether it says “with replacement” or “without replacement”.
- Write down the sample space (the list of all possible outcomes) before you start counting.
- Separate “or” problems (add probabilities) from “and” problems (multiply probabilities) unless the events overlap.
- Convert fractions to decimals or percentages only at the very end, if the exam asks for it.
📝 Likely Exam Questions
- Question: A box contains 3 red, 2 blue and 5 green balls. One ball is drawn at random. What is the probability of getting a blue ball?
Answer: Total balls = 10. Favourable (blue) = 2. P = 2/10 = 1/5 = 0.2 (20%). - Question: Two dice are rolled. Find the probability that the sum of the numbers is 7.
Answer: Possible pairs for sum 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) → 6 ways. Total outcomes = 6×6 = 36. P = 6/36 = 1/6 ≈ 0.1667 (16.67%). - Question: A coin is tossed three times. What is the probability of getting exactly two heads?
Answer: Number of ways to choose 2 heads out of 3 tosses = C(3,2)=3. Each specific sequence has probability (1/2)³ = 1/8. P = 3×1/8 = 3/8 = 0.375 (37.5%). - Question: From a deck of 52 cards, two cards are drawn one after the other without replacement. Find the probability that both are queens.
Answer: First queen: 4/52 = 1/13. Second queen after one removed: 3/51. P = (1/13)×(3/51)=3/663≈0.0045 (0.45%).