Overview
This chapter introduces probability as a mathematical way to measure the likelihood of random events. Starting with simple random experiments (coin tosses, dice rolls, drawing cards/balls), students learn formal terms—experiment, sample space, outcome and event—and two complementary approaches to probability: classical (equally likely outcomes) and experimental (frequency-based). The chapter presents basic properties (probabilities range from 0 to 1, sum of probabilities of all outcomes is 1), the complement rule (P(not E) = 1 − P(E)), and simple combination rules for mutually exclusive events. Emphasis is on modelling real situations, performing simple experiments to estimate probabilities, solving concrete problems, and developing intuition for uncertain situations. Understanding these ideas builds a foundation for probability applications in statistics, risk assessment and problem solving in higher classes.
Learning Objectives
- Define probability of an event using the classical (equally likely outcomes) approach.
- Explain the terms experiment, outcome, sample space and event with relevant examples.
- Determine the sample space for single and combined random experiments (coin tosses, dice, cards, draws).
- Classify events as certain, impossible, mutually exclusive and complementary.
- Compute probabilities of simple events by counting favourable and total outcomes.
- Apply the concept of complementary events to calculate probabilities more efficiently.
- Use the addition rule for mutually exclusive events to find probabilities of unions.
- Solve numerical problems involving probabilities for dice, coins, cards and selection of balls from a bag.
Topics in this chapter
9 topics · tap a topic title to jump straight to it.
Random experiment and Outcome
Random experiment: An action or process that can be repeated under identical conditions and has one or more possible results (outcomes), where the result cannot be predicted with certainty beforehand. Examples: tossing a coin, rolling a die, drawing a card.
Outcome (sample point): A single possible result of a random experiment. The set of all possible outcomes is called the sample space (denoted S). Each outcome in S is an elementary event.
Sample space (S): The collection of all possible outcomes. It can be represented as a list or set. Example: For a fair die S = {1,2,3,4,5,6}; for a coin S = {H, T}.
Equally likely outcomes: Outcomes are equally likely if each has the same chance of occurring (common in symmetric experiments like fair coins, fair dice).
Event: Any subset of the sample space (one or more outcomes). A simple (elementary) event contains a single outcome; a compound event contains several.
Relation to probability (brief): When outcomes are equally likely, the probability of event E is the ratio of the number of favourable outcomes to the total number of outcomes: P(E) = n(E) / n(S). (See formulas below.)
Deterministic vs random: Deterministic experiments always give the same outcome under identical conditions (no randomness). Random experiments may give different outcomes when repeated.
How to list S: Use outcome listing, tree diagrams (for sequences), or systematic counting rules (product rule) to build S. For combined experiments with m outcomes in the first and n in the second, total outcomes = m × n.
- Tossing a fair coin once. S = {H, T}. Each outcome is equally likely.
- Rolling a single fair die. S = {1,2,3,4,5,6}. Outcome '4' is one sample point.
- Tossing two coins. Use a tree or product rule: S = {HH, HT, TH, TT} (4 outcomes).
- Drawing one card from a standard deck. S contains 52 outcomes; outcome 'Ace of Spades' is one sample point.
- Selecting a student at random from a class. Outcomes = the list of all students; each student is a sample point (if equally likely selection).
- Sample space S: set of all outcomes; n(S) denotes number of outcomes in S.
- Probability for equally likely outcomes: P(E) = n(E) / n(S).
- Complement rule: P(E') = 1 - P(E), where E' is the event 'E does not occur'.
- Addition rule for mutually exclusive events: If A and B are disjoint, P(A ∪ B) = P(A) + P(B).
- Product (counting) rule for combined experiments: If first experiment has m outcomes and second has n outcomes, total outcomes = m × n.
Sample space
Definition: In a random experiment, the sample space (denoted by S or Ω) is the set of all possible outcomes. Each element of S is called a sample point. For example, when tossing one fair coin, S = {H, T} where H = head and T = tail.
Why it matters: The sample space is the foundation for defining events and computing probabilities. An event is any subset of S. To find the probability of an event, you must first know what outcomes exist (S) and which of them make the event happen.
Types of sample spaces:
- Finite (discrete): number of outcomes is finite (e.g., tossing a coin, rolling a die).
- Countably infinite: outcomes can be listed in sequence (e.g., number of tosses until first head).
- Uncountable/continuous: outcomes form a continuum (e.g., all possible arrival times between 0 and 60 minutes). In Class 10 most problems use finite or countable sample spaces.
How to list a sample space: Use set notation with commas and braces. For compound experiments, outcomes are ordered pairs (or tuples). Example: tossing two coins, S = {(H,H), (H,T), (T,H), (T,T)}. For rolling two dice, S is the set of 36 ordered pairs (1,1),(1,2),...,(6,6).
Important remarks:
- Order matters when outcomes are ordered pairs. If order does not matter, states should be listed appropriately (e.g., drawing two balls without order).
- When outcomes are equally likely, probability of an event E is n(E)/n(S), where n(X) denotes number of elements of set X.
Examples in words: "Choose one card from a pack" → S = {52 playing cards}; "Observe today’s weather" → S = {Sunny, Cloudy, Rainy, ...}; "Measure lifetime of a bulb" → continuous sample space (positive real numbers).
- Tossing one coin: S = {H, T}.
- Rolling one six-faced die: S = {1, 2, 3, 4, 5, 6}.
- Tossing two coins: S = {(H,H), (H,T), (T,H), (T,T)}.
- Rolling two dice: S = {(i,j) | i = 1..6, j = 1..6} (36 outcomes).
- Drawing one card from a standard deck: S = set of 52 distinct cards (13 ranks × 4 suits).
- Real life — choosing a bus route: S = {Route A, Route B, Route C} (finite); measuring waiting time for a bus between 0 and 30 minutes is continuous.
- Notation: sample space S or Ω; event E ⊂ S.
- If outcomes are equally likely: P(E) = n(E) / n(S).
- Complement: P(E') = 1 − P(E), where E' = S \ E.
- Union (general): P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
- For independent sequential experiments: n(S_total) = n(S1) × n(S2) × ... (rule of product).
Event
Definition: In probability, an event is any subset of the sample space (S) of a random experiment. An event E contains one or more outcomes (elements) of S. If S is the set of all possible outcomes, then E ⊆ S.
Types of events:
- Simple (elementary) event: contains exactly one outcome (e.g., rolling a 4).
- Compound event: contains more than one outcome (e.g., rolling an even number).
- Sure (certain) event: E = S; P(E) = 1.
- Impossible event: E = ∅; P(E) = 0.
- Complementary event: Ec (or E') is the set of outcomes in S not in E; P(Ec) = 1 − P(E).
- Mutually exclusive (disjoint) events: A and B are mutually exclusive if A ∩ B = ∅ (they cannot occur together).
- Independent events: occurrence of one does not affect the probability of the other; P(A ∩ B) = P(A)P(B).
How probability of an event is computed (equally likely outcomes): If all outcomes in S are equally likely, P(E) = n(E) / n(S), where n(E) is number of favourable outcomes and n(S) is total number of outcomes.
Alternative (empirical) approach: For experiments repeated many times, P(E) ≈ (number of times E occurred) / (total trials).
Basic properties: P(∅)=0, P(S)=1, 0 ≤ P(E) ≤ 1. For any events A and B: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). If A and B are mutually exclusive: P(A ∪ B) = P(A) + P(B). For complement: P(Ec) = 1 − P(E).
- Tossing a fair die: S = {1,2,3,4,5,6}. Event E = “even number” = {2,4,6}. P(E) = 3/6 = 1/2.
- Tossing a fair coin twice: S = {HH, HT, TH, TT}. Event A = “exactly one head” = {HT, TH}. P(A) = 2/4 = 1/2.
- Drawing one card from a well-shuffled 52-card deck: Event B = “an ace” has n(B)=4, so P(B) = 4/52 = 1/13.
- Bag with 5 red and 3 blue identical balls: S has 8 outcomes (choosing one ball). Event R = “red” so P(R) = 5/8. The complement (not red) is blue with P = 3/8.
- Combined experiment (independence): Toss a coin and roll a die. Event C = “head and an even number”. P(head)=1/2, P(even)=1/2, independent so P(C)=1/2 × 1/2 = 1/4.
- P(E) = n(E) / n(S) (for equally likely outcomes)
- 0 ≤ P(E) ≤ 1, P(S) = 1, P(∅) = 0
- P(E<sup>c</sup>) = 1 − P(E) (complement rule)
- P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
- If A and B are mutually exclusive: P(A ∪ B) = P(A) + P(B)
- If A and B are independent: P(A ∩ B) = P(A) · P(B)
Types of events and relations
Basic terms: A random experiment is an action with uncertain outcomes. The sample space S is the set of all possible outcomes. An event A is any subset of S (one or more outcomes).
Types of events:
- Sure (certain) event: an event that always occurs: A = S. P(A) = 1.
- Impossible event: an event that cannot occur: A = Ø. P(A) = 0.
- Elementary (simple) event: contains exactly one outcome (e.g., getting 5 on a die).
- Compound event: contains two or more outcomes (e.g., getting an even number on a die: {2,4,6}).
- Equally likely events: outcomes with the same probability (e.g., fair coin faces, fair die faces).
- Mutually exclusive (disjoint) events: cannot occur together: A ∩ B = Ø. Example: getting 2 and getting 5 on a single die roll.
- Exhaustive events: a collection of events whose union is the whole sample space (they cover all outcomes).
- Complementary events: A and A' (not A) are complements: A ∪ A' = S and A ∩ A' = Ø. P(A') = 1 − P(A).
- Independent events: occurrence of one does not affect the probability of the other. For A and B independent: P(A ∩ B) = P(A)·P(B).
Relations and important rules:
- Addition (general): P(A ∪ B) = P(A) + P(B) − P(A ∩ B). This avoids double counting the intersection.
- Addition (mutually exclusive): If A and B are mutually exclusive, P(A ∩ B) = 0, so P(A ∪ B) = P(A) + P(B).
- Multiplication (independent): If A and B are independent, P(A ∩ B) = P(A)P(B).
- Conditional probability: P(A|B) = P(A ∩ B) / P(B), provided P(B) > 0. If P(A|B) = P(A) then A and B are independent.
- Complement rule: P(A') = 1 − P(A).
- De Morgan’s laws: (A ∪ B)' = A' ∩ B' and (A ∩ B)' = A' ∪ B'.
- Classical probability (for equally likely outcomes): If n(S) is finite and equally likely, P(A) = n(A) / n(S).
- Coin toss: S = {H, T}. Events: A = {H} (elementary), B = {H, T} (sure). H and T are mutually exclusive and exhaustive; P(H) = 1/2.
- Single die roll: S = {1,2,3,4,5,6}. Event E = {even} = {2,4,6} (compound). P(E) = 3/6 = 1/2.
- Drawing a card from a standard 52-card deck: A = {ace} has n(A)=4, P(A)=4/52=1/13. A and A' (not an ace) are complementary.
- Two coin tosses (with replacement): events A = 'first toss is H', B = 'second toss is H'. A and B are independent, P(A ∩ B) = 1/2 · 1/2 = 1/4.
- Drawing two balls from a bag without replacement: bag has 2 red, 3 blue. Event A = 'first is red', B = 'second is red'. A and B are not independent; P(B|A) changes after the first draw.
- Weather example: A = 'it rains today', A' = 'it does not rain today' are complementary. Two outcomes (rain/no rain) are not exhaustive of all weather details but give a simple complementary pair.
- P(A) = n(A) / n(S) (when all outcomes are equally likely and n(S) is finite)
- P(A') = 1 − P(A)
- P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
- If A and B are mutually exclusive: P(A ∩ B) = 0 and P(A ∪ B) = P(A) + P(B)
- If A and B are independent: P(A ∩ B) = P(A)·P(B)
- Conditional probability: P(A|B) = P(A ∩ B) / P(B), provided P(B) > 0
Classical definition of probability
Classical definition: If an experiment has a finite sample space S in which all outcomes are equally likely, and E is an event (a set of favourable outcomes), then the probability of E is
P(E) = n(E) / n(S)
Here n(E) is the number of favourable outcomes and n(S) is the total number of possible outcomes. The classical definition applies only when all individual outcomes have the same chance of occurring (equally likely).
How to apply (steps):
- Describe the sample space S and count n(S).
- Identify the favourable outcomes that make up event E and count n(E).
- Compute P(E) = n(E)/n(S). Simplify the fraction.
Properties (direct consequences):
- P(E) is between 0 and 1: 0 ≤ P(E) ≤ 1.
- Probability of the certain event (sample space): P(S) = 1.
- Probability of the impossible event (empty set): P(∅) = 0.
- Complement rule: P(E') = 1 − P(E), where E' is the complement of E in S.
- Addition for mutually exclusive events A and B: P(A ∪ B) = P(A) + P(B) (if A ∩ B = ∅).
When classical definition is not valid: If outcomes are not equally likely (for example, a biased coin, or selecting days weighted by activity), the classical formula n(E)/n(S) is not valid and a different approach (relative frequency, axiomatic) is needed.
- Toss a fair coin once. Sample space S = {H, T}, so n(S)=2. Event E = {H} (getting a head), n(E)=1. P(H) = 1/2.
- Roll a fair six-sided die. S = {1,2,3,4,5,6}, n(S)=6. Probability of rolling a 4: n(E)=1, P(4)=1/6.
- Pick one card from a well-shuffled 52-card deck. Probability of drawing an ace: n(E)=4, n(S)=52, P(ace)=4/52=1/13.
- A bag has 5 red and 3 blue identical-looking marbles (total 8). If one marble is drawn at random and all marbles equally likely, probability of red = 5/8.
- Spin a fair spinner divided into 8 equal sectors, 3 of which are green. Probability of landing on green = 3/8.
- P(E) = n(E) / n(S)
- 0 ≤ P(E) ≤ 1
- P(S) = 1
- P(∅) = 0
- P(E') = 1 − P(E)
- For mutually exclusive A, B: P(A ∪ B) = P(A) + P(B)
Basic properties of probability
Definition and setup: For a random experiment, the sample space S is the set of all possible outcomes. An event is any subset E of S. The probability P(E) is a number assigned to E that measures how likely E is to occur. In the classical (equally likely) case, P(E) = (number of favourable outcomes)/(total number of outcomes).
Interpretations: Two common interpretations are the classical (combinatorial) interpretation—useful when outcomes are equally likely—and the relative frequency (empirical) interpretation—P(E) ≈ (frequency of E)/N for large N repeats of the experiment.
Basic properties (with brief reasoning):
- Non-negativity: 0 ≤ P(E) for every event E. (Probabilities cannot be negative.)
- Normalization: P(S) = 1. The probability that some outcome in the sample space occurs is certain.
- Null event: P(∅) = 0. The empty event (no outcome) has probability zero.
- Complement rule: P(E') = 1 − P(E), where E' (or E^c) is the complement of E in S. (Either E occurs or it does not.)
- Addition rule (two events): For any two events A and B, P(A ∪ B) = P(A) + P(B) − P(A ∩ B). This avoids double-counting outcomes that are in both A and B.
- Mutually exclusive events: If A and B are mutually exclusive (A ∩ B = ∅), then P(A ∪ B) = P(A) + P(B).
- Monotonicity: If A ⊆ B then P(A) ≤ P(B). (A smaller event cannot be more likely than a larger one containing it.)
- Difference rule: P(B \ A) = P(B) − P(A ∩ B). (Probability of outcomes in B but not in A.)
Remarks: These properties follow directly from the set relationships among events and from the basic idea that probability distributes over disjoint unions. For Class 10 problems you usually apply the classical formula when outcomes are equally likely and then use the properties above to combine events.
- Coin toss: For a fair coin, S = {H, T}. P(H) = 1/2 and P(T) = 1/2. Complement: P(not H) = 1 − P(H) = 1/2.
- Single die: S = {1,2,3,4,5,6}. Event A = {odd outcomes} = {1,3,5}, so P(A) = 3/6 = 1/2. Non-negativity: 0 ≤ 1/2 ≤ 1.
- Card example (52 cards): Let A = {heart} (13 cards) and B = {ace} (4 cards). P(A) = 13/52, P(B) = 4/52, A ∩ B = {ace of hearts} (1 card). So P(A ∪ B) = 13/52 + 4/52 − 1/52 = 16/52 = 4/13.
- Mutually exclusive event example: Drawing one card, A = {king}, B = {queen}. A ∩ B = ∅, so P(A ∪ B) = P(A) + P(B) = 4/52 + 4/52 = 8/52 = 2/13.
- Empirical example: If in 100 days it rained on 18 days, the empirical probability of rain ≈ 18/100 = 0.18. Complement: probability of no rain ≈ 0.82.
- Classical probability: P(E) = (number of favourable outcomes)/(total number of outcomes)
- 0 ≤ P(E) ≤ 1
- P(S) = 1
- P(∅) = 0
- Complement: P(E') = 1 − P(E)
- Addition rule (two events): P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Methods to compute probability
Overview: Probability measures how likely an event is to occur. There are three common methods to compute probability taught in Class 10: the classical (theoretical) method, the experimental (empirical or relative-frequency) method, and the axiomatic approach. In many problems you also use counting techniques (combinations/permutations), complementary rules, and addition/multiplication rules.
1. Classical (Theoretical) Method
Use when all outcomes of a random experiment are equally likely. If S is the sample space with n(S) equally likely outcomes and E is an event with n(E) favourable outcomes,
P(E) = n(E) / n(S)
Counting n(E) and n(S) often uses combinations/permutations (e.g., number of ways to choose or arrange objects).
2. Experimental (Empirical or Relative-frequency) Method
Use when you perform actual trials and record results. If an experiment is repeated N times and event E occurs f(E) times, the experimental probability is:
Experimental P(E) ≈ f(E)/N
As N becomes large, f(E)/N tends to the true probability (law of large numbers).
3. Axiomatic Approach
Probability is defined on a sample space S with a function P satisfying the axioms: (i) 0 ≤ P(E) ≤ 1, (ii) P(S) = 1, (iii) For disjoint events A and B, P(A ∪ B) = P(A) + P(B). From these axioms several useful rules follow (complement, addition, conditional, multiplication rules).
Useful rules and techniques
- Complement rule: P(E') = 1 − P(E) (useful for 'at least one' problems).
- Addition rule: For any events A and B, P(A ∪ B) = P(A) + P(B) − P(A ∩ B). If A and B are mutually exclusive, P(A ∪ B) = P(A) + P(B).
- Multiplication and conditional: P(A ∩ B) = P(A) · P(B|A). If A and B are independent, P(A ∩ B) = P(A)·P(B).
- Counting: Use combinations C(n, r) and permutations P(n, r) to count outcomes when items are chosen/arranged.
When to use which method: Use the classical method for well-defined, equally likely outcomes (coins, fair dice, cards). Use experimental method when you actually perform trials or when theoretical model is unavailable. Use axiomatic approach for rigorous properties and when combining events.
- Coin toss (classical): Find P(head) when a fair coin is tossed. S = {H, T}, n(S)=2; n(E)=1 ⇒ P(H)=1/2.
- Single die (classical and complementary): Find P(not 4) when a fair die is rolled. n(S)=6, n(E')=5 ⇒ P(not 4)=5/6 = 1 − 1/6.
- Bag with balls (counting): A bag has 3 red and 2 blue balls. One ball drawn at random. P(red) = 3/5. If drawing two without replacement, P(both red) = (3/5)·(2/4)=3/10.
- Card example (addition rule): From a 52-card deck, find P(card is a heart or a king). P(heart)=13/52, P(king)=4/52, P(heart ∩ king)=1/52 ⇒ P = 13/52 + 4/52 − 1/52 = 16/52 = 4/13.
- Experimental (relative-frequency): Toss a coin 1000 times and get 528 heads. Experimental probability of head ≈ 528/1000 = 0.528. With more tosses this ratio should approach 0.5.
- Classical probability: P(E) = n(E) / n(S)
- Experimental (relative frequency): P(E) ≈ f(E) / N (as N → ∞, approaches true probability)
- Complement rule: P(E') = 1 − P(E)
- Addition rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B); if A and B are mutually exclusive, P(A ∪ B) = P(A) + P(B)
- Conditional probability: P(A|B) = P(A ∩ B) / P(B), provided P(B) > 0
- Multiplication rule: P(A ∩ B) = P(A) · P(B|A); if independent, P(A ∩ B) = P(A)·P(B)
Standard examples and applications
What probability is: Probability measures how likely an event is to occur. For equally likely outcomes, probability of an event E is P(E) = n(E) / n(S), where n(E) is number of favourable outcomes and n(S) is total number of outcomes in the sample space S.
Basic terms: sample space (all possible outcomes), event (a subset of outcomes), complement (E' = outcomes where E does not happen), mutually exclusive events (cannot occur together), independent events (one does not affect the other).
Standard examples: many class‑10 problems use a small finite sample space with equally likely outcomes: coin tosses, throws of a die, picking cards from a 52‑card deck, selecting numbered tickets, and drawing balls from an urn. These examples teach counting favourable outcomes, using complements and simple product/addition rules.
Common techniques:
- List the sample space explicitly for small experiments (e.g., for two coin tosses S = {HH, HT, TH, TT}).
- Use counting (combinations/permutations) when sample space is larger (e.g., choosing 2 from n without order).
- Use the complement rule P(E) = 1 − P(E') when it is easier to count E'.
- Use tree diagrams for sequential experiments (with or without replacement).
Applications in real life: predicting weather (chance of rain), quality control (probability of defective item), games of chance (cards, dice), insurance and risk assessment (estimating loss probabilities), genetics (probability of inheriting traits), and reliability of systems (probability of component failure).
How to approach CBSE problems: identify whether outcomes are equally likely, write S and E, count n(S) and n(E), simplify the fraction. For sequential draws check if replacement makes trials independent; without replacement use conditional counting (reduce totals after each draw).
- Coin tossed twice. S = {HH, HT, TH, TT}. Probability of exactly one head = 2/4 = 1/2.
- One fair die rolled. Probability of getting an even number = {2,4,6} → 3/6 = 1/2.
- Two dice rolled. Probability that the sum is 7 = 6/36 = 1/6 (favourable pairs: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1)).
- One card drawn from 52. Probability of getting a heart = 13/52 = 1/4.
- Urn with 5 red and 3 blue balls. Two balls drawn without replacement. Probability both are red = (5/8) * (4/7) = 20/56 = 5/14.
- Number chosen at random from 1–100. Probability it is divisible by 3 = floor(100/3)/100 = 33/100.
- Classical probability: P(E) = n(E) / n(S) for equally likely outcomes.
- Complement rule: P(E') = 1 − P(E).
- Addition (mutually exclusive): If A and B are mutually exclusive, P(A ∪ B) = P(A) + P(B).
- General addition: P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
- Independent events (multiplication): If A and B are independent, P(A ∩ B) = P(A) · P(B).
- Sequential without replacement (example): P(first success and second success) = (m/n) · ((m−1)/(n−1)), where m successes in n total initially.
Problem‑solving strategies
Problem‑solving strategies in probability are systematic ways to translate a real situation into mathematical statements and compute probabilities correctly. The basic classical idea used in Class 10 is: Probability of an event E = number of favourable outcomes / number of equally likely outcomes (P(E) = n(E)/n(S)).
Follow these stepwise strategies for most problems:
- Read and understand: Identify the random experiment, what one trial is, and what the required event E means in plain words.
- Define the sample space S: List all equally likely outcomes (coin tosses, faces of a die, balls in a bag, ordered pairs for two experiments, etc.). Use a table, tree diagram, or grid when helpful.
- Identify favourable outcomes: Write the outcomes that satisfy the event. Count them carefully; watch for order (ordered vs unordered), replacement vs no replacement.
- Use appropriate counting tools: Use the basic counting principle (product rule), simple permutations/combinations if needed, or complementary counting to simplify counting of favourable outcomes.
- Apply probability formulas: Compute P(E)=n(E)/n(S). Use complement rule P(E') = 1 − P(E), addition rule for unions, and multiplication rule for independent events where appropriate.
- Check edge cases and reasonableness: Probability should be between 0 and 1. If result seems odd, try solving with complementary or direct counting again.
Common techniques/shortcuts:
- Complement rule: If counting E is hard, count E' (the complement) and use P(E)=1−P(E').
- Symmetry: If outcomes are symmetric (e.g., fair coin, fair die), exploit symmetry to reduce counting.
- Tree diagrams: Useful for sequential experiments (with/without replacement). They visually encode outcome probabilities or counts.
- Tables and grids: Use a 6x6 grid for two dice, or a matrix for two-stage experiments, to list ordered pairs.
- Venn diagrams: Helpful when dealing with unions, intersections and complements of events.
Keep clear whether order matters and whether trials are independent. In "without replacement" situations the sample space changes after each draw; then count ordered outcomes or use conditional probabilities.
Worked mini example (conceptual): A bag has 3 red and 2 blue balls. One ball is drawn at random. S = {R,R,R,B,B} (5 equally likely outcomes). Probability of red = 3/5. If two balls are drawn without replacement and order matters, S has 5x4 = 20 ordered outcomes; count favourable ordered pairs or use complementary/conditional reasoning.
Use these strategies together: careful modeling, choosing counting method, using complement/symmetry, and then applying the classical formula. This approach will solve most Class 10 probability problems reliably.
- Example 1 — Single die: Problem: Find probability of getting an even number when a fair die is rolled once. Solution: S = {1,2,3,4,5,6}, n(S)=6. Favourable E={2,4,6}, n(E)=3. P(E)=3/6=1/2.
- Example 2 — Two coins: Problem: Two fair coins are tossed. Find probability of exactly one head. Strategy: Use tree or list sample space S={(H,H),(H,T),(T,H),(T,T)}. Favourable = {(H,T),(T,H)}, n(E)=2, n(S)=4. P(E)=2/4=1/2.
- Example 3 — Bag without replacement: Problem: A bag has 3 red and 2 blue balls. Two balls are drawn one after other without replacement. Find probability both are red. Strategy: Use multiplication (conditional) or count ordered outcomes. P(first red)=3/5. Given first red, P(second red)=2/4=1/2. So P(both red)= (3/5)*(1/2)=3/10.
- Classical probability: P(E) = n(E) / n(S)
- Complement rule: P(E') = 1 - P(E)
- Addition rule (mutually exclusive): P(A or B) = P(A) + P(B)
- General addition: P(A or B) = P(A) + P(B) - P(A and B)
- Multiplication rule (independent events): P(A and B) = P(A) * P(B)
- Conditional probability (useful for without replacement): P(A|B) = P(A and B) / P(B)
Key Concepts
- Random experiment
- An action or process that leads to one of several possible outcomes, where the exact outcome cannot be predicted in advance.
- Trial
- A single performance or repetition of a random experiment.
- Outcome (sample point)
- A single possible result of a random experiment.
- Sample space
- The set of all possible outcomes of a random experiment, usually denoted by S.
- Elementary (simple) event
- An event consisting of exactly one outcome (one sample point).
- Compound event
- An event that consists of two or more outcomes (a union of elementary events).
- Favorable outcomes
- Outcomes in the sample space that satisfy the condition of the event being considered.
- Equally likely outcomes
- Outcomes that have the same chance of occurring in a random experiment.
- Theoretical probability
- Probability calculated using reasoning, as (number of favorable outcomes) / (total number of possible outcomes) when outcomes are equally likely.
- Experimental (empirical) probability
- Probability estimated from actual experiments; it equals (number of times event occurred) / (total trials).
- Relative frequency
- The ratio of the number of times an event occurs to the total number of trials; used in experimental probability.
- Complementary event
- For an event A, the complement A' is the event that A does not occur; P(A') = 1 − P(A).
- Mutually exclusive events
- Two events that cannot occur at the same time (their intersection is empty).
- Exhaustive events
- A collection of events that together include all possible outcomes of the experiment (their union is the sample space).
- Certain event
- An event that is sure to occur; its probability is 1.
- Impossible event
- An event that cannot occur; its probability is 0.
- Law of large numbers
- A principle stating that as the number of trials increases, the experimental probability tends to approach the theoretical probability.
- Fair coin/die
- A coin or die with no bias; each outcome is equally likely.
- Probability distribution (discrete)
- A listing of possible values of a discrete random variable with their corresponding probabilities (sum = 1).
- Expected value (mean)
- The weighted average of all possible values of a discrete random variable, using probabilities as weights; gives the long-run average outcome.
End-of-Chapter Trial Paper & Test Questions
Topic-wise questions to test your understanding of every concept in this chapter.
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State the classical definition of probability and the condition under which it is valid. / प्रायिकता की चिरसम्मत परिभाषा तथा वह शर्त बताइए जिसके अंतर्गत यह मान्य है।
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If all outcomes are equally likely, P(E) = n(E)/n(S), where n(E) is favourable outcomes and n(S) is total outcomes; it is valid only when outcomes are equally likely. / यदि सभी परिणाम समसंभाव्य हों तो P(E) = n(E)/n(S), जहाँ n(E) अनुकूल परिणाम तथा n(S) कुल परिणाम हैं; यह केवल तभी मान्य है जब परिणाम समसंभाव्य हों।
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A die is rolled once. Find the probability of getting a number not equal to 4 using the complement rule. / एक पासा एक बार फेंका जाता है। पूरक नियम का उपयोग करके 4 के अतिरिक्त संख्या आने की प्रायिकता ज्ञात कीजिए।
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P(4) = 1/6, so P(not 4) = 1 − 1/6 = 5/6. / P(4) = 1/6, अतः P(4 नहीं) = 1 − 1/6 = 5/6।
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Why must the probability of any event always lie between 0 and 1? / किसी भी घटना की प्रायिकता सदैव 0 और 1 के बीच क्यों होनी चाहिए?
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Favourable outcomes cannot be fewer than zero nor more than the total outcomes, so n(E)/n(S) ranges from 0 (impossible event) to 1 (certain event). / अनुकूल परिणाम न तो शून्य से कम और न ही कुल परिणामों से अधिक हो सकते हैं, अतः n(E)/n(S) का मान 0 (असंभव घटना) से 1 (निश्चित घटना) तक होता है।
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Two coins are tossed together. Find the probability of getting exactly one head. / दो सिक्के एक साथ उछाले जाते हैं। ठीक एक चित आने की प्रायिकता ज्ञात कीजिए।
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Sample space S = {HH, HT, TH, TT}, favourable = {HT, TH}, so P = 2/4 = 1/2. / प्रतिदर्श समष्टि S = {HH, HT, TH, TT}, अनुकूल = {HT, TH}, अतः P = 2/4 = 1/2।
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One card is drawn from a 52-card deck. Find the probability that it is a heart or a king. / 52 पत्तों की गड्डी से एक पत्ता निकाला जाता है। उसके पान या बादशाह होने की प्रायिकता ज्ञात कीजिए।
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Using P(A∪B) = P(A)+P(B)−P(A∩B) = 13/52 + 4/52 − 1/52 = 16/52 = 4/13. / P(A∪B) = P(A)+P(B)−P(A∩B) = 13/52 + 4/52 − 1/52 = 16/52 = 4/13।
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Define mutually exclusive events and state the addition rule for them. / परस्पर अपवर्जी घटनाओं को परिभाषित कीजिए तथा उनके लिए योग नियम लिखिए।
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Mutually exclusive events cannot occur together (A ∩ B = ∅), and for them P(A ∪ B) = P(A) + P(B). / परस्पर अपवर्जी घटनाएँ एक साथ घटित नहीं हो सकतीं (A ∩ B = ∅), तथा इनके लिए P(A ∪ B) = P(A) + P(B)।
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A bag has 5 red and 3 blue balls. Two balls are drawn without replacement. Find the probability that both are red. / एक थैले में 5 लाल और 3 नीली गेंदें हैं। बिना प्रतिस्थापन के दो गेंदें निकाली जाती हैं। दोनों के लाल होने की प्रायिकता ज्ञात कीजिए।
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P(both red) = (5/8) × (4/7) = 20/56 = 5/14. / P(दोनों लाल) = (5/8) × (4/7) = 20/56 = 5/14।
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How does experimental probability relate to theoretical probability as the number of trials increases? / जैसे-जैसे परीक्षणों की संख्या बढ़ती है, प्रायोगिक प्रायिकता का सैद्धांतिक प्रायिकता से क्या संबंध होता है?
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By the law of large numbers, as the number of trials increases, the experimental probability f(E)/N tends to approach the theoretical probability. / बड़ी संख्याओं के नियम के अनुसार, जैसे-जैसे परीक्षणों की संख्या बढ़ती है, प्रायोगिक प्रायिकता f(E)/N सैद्धांतिक प्रायिकता के निकट पहुँचने लगती है।
Related Laws & Principles
Explore allFoundational laws & principles behind this chapter. Each one opens a full page — what it says, why it matters, five practice questions and the mistakes to avoid.