Overview
This chapter introduces Probability as a mathematical way to quantify uncertainty of outcomes of random experiments. Starting with experiments, sample space and events, it builds the classical definition of probability for equally likely outcomes and develops basic rules (complements, addition rule) used to compute probabilities. The chapter also shows how simple counting methods (permutations and combinations) and Venn diagrams help evaluate probabilities in common situations (coins, dice, cards, and selection problems). Understanding these fundamentals prepares students for applications in statistics, decision making and later topics in probability theory.
Learning Objectives
- Define random experiment, sample space, event, outcome, and elementary event with suitable examples.
- Describe and represent sample spaces for finite experiments using lists, Venn diagrams and tree diagrams.
- Distinguish between mutually exclusive (disjoint), exhaustive and independent events with illustrative examples.
- State the axioms of probability and apply them to verify or compute probabilities in given situations.
- Apply the classical definition of probability for equally likely outcomes using counting techniques (permutations and combinations).
- Use the complement rule (P(A') = 1 − P(A)) to simplify and compute probabilities in exam problems.
- Compute probabilities of union and intersection of two events using the addition rule and inclusion–exclusion principle.
- Define conditional probability and apply the multiplication theorem to compute probabilities of successive events.
Topics in this chapter
9 topics · tap a topic title to jump straight to it.
Random experiment, Outcome and Sample space
Random experiment: A random (or stochastic) experiment is a process or action which can be repeated under the same conditions and whose exact result cannot be predicted with certainty in advance. Examples: tossing a coin, rolling a die, measuring the lifetime of a bulb.
Outcome (Elementary outcome): An outcome is a single possible result of a random experiment. Each repetition produces exactly one outcome. Notation: use a lowercase letter (e.g., ω) or list form (e.g., 3 when a die shows 3).
Sample space (S): The sample space of an experiment is the set of all possible outcomes. It is usually denoted by S or Ω. Example notations: S = {H, T} for a coin toss; S = {1,2,3,4,5,6} for a fair die; S = [0, 60] seconds for measuring waiting time up to one minute.
Types of sample spaces
- Finite: S has a finite number of outcomes (e.g., die has 6 outcomes).
- Countably infinite (discrete): S has infinitely many countable outcomes (e.g., number of tosses until first head).
- Uncountable (continuous): S is an interval of real numbers (e.g., all real values between 0 and 1).
Events: An event is any subset of the sample space (one or more outcomes). An elementary event contains exactly one outcome. The impossible event is ∅ and the sure event is S.
Set notation and basic relationships: If E and F are events (subsets of S), then E ∪ F (union) and E ∩ F (intersection) are also events. The complement of E is Ec = S \ E.
Understanding sample spaces and outcomes is the first step to assigning probabilities and solving probability problems: list S clearly, identify the favourable outcomes for the event of interest, then compute probabilities (using appropriate rules) depending on whether outcomes are equally likely or not.
- Tossing a fair coin once: Outcome = H or T; Sample space S = {H, T}.
- Rolling one fair die: Outcomes = 1,2,3,4,5,6; S = {1,2,3,4,5,6}.
- Tossing two coins: Outcomes = {HH, HT, TH, TT}; S has 4 elements.
- Drawing one card from a standard deck: Sample space has 52 outcomes (each individual card).
- Measuring the time (in seconds) until a bus arrives within 0–30 minutes: S = [0, 1800] — a continuous sample space.
- Sample space notation: S (or Ω) = set of all possible outcomes.
- Event: E ⊆ S. Elementary event = {ω} where ω ∈ S.
- If outcomes are equally likely: P(E) = n(E) / n(S), where n(·) denotes number of outcomes.
- Complement: P(E^c) = 1 − P(E).
- Additivity: P(∅) = 0, P(S) = 1, and 0 ≤ P(E) ≤ 1 for any event E.
- Union formula: P(E ∪ F) = P(E) + P(F) − P(E ∩ F).
Events
Definition: In probability, an event is any collection (subset) of possible outcomes of a random experiment. The set of all possible outcomes is the sample space S. An event E is therefore E \u2286 S.
Types of events:
- Simple (Elementary) event: Contains exactly one outcome (e.g., getting '4' on a die).
- Compound event: Contains two or more outcomes (e.g., getting an even number on a die).
- Sure (Certain) event: The event S that always occurs (P(S)=1).
- Impossible event: The empty set Ø that never occurs (P(Ø)=0).
- Complementary event: For event A, A' (or A^c) is the set of outcomes in S not in A; P(A')=1−P(A).
- Mutually exclusive (disjoint) events: A and B are mutually exclusive if A∩B=Ø (they cannot both occur).
- Exhaustive events: A collection of events whose union is the whole sample space S.
- Independent events: Two events A and B are independent if the occurrence of one does not affect the probability of the other: P(B|A)=P(B).
Set notation and relations: Events are handled with set operations: union (A \u222a B), intersection (A \u2229 B), complement (A'), difference (A \u2216 B). Venn diagrams are commonly used to visualize these relations.
Probability of an event (equally likely outcomes): If outcomes are equally likely and S is finite, P(E) = n(E)/n(S), where n(E) is the number of favourable outcomes and n(S) the total outcomes.
Logic and laws: De Morgan's laws for events: (A \u2229 B)' = A' \u222a B' and (A \u222a B)' = A' \u2229 B'. These help simplify complements of unions/intersections.
Connection to conditional probability: For general events, intersection and conditional probability relate by P(A \u2229 B) = P(A)P(B|A). For independent events this reduces to P(A \u2229 B) = P(A)P(B).
Understanding events and their relations is the foundation for computing probabilities of complex situations by breaking them into unions, intersections, complements and sequences.
- Toss a fair coin once. Sample space S = {H, T}. Event E = {H} (getting a head). P(E) = 1/2.
- Roll a fair six-sided die. Event A = {2,4,6} (getting an even number). n(A)=3, n(S)=6 so P(A)=3/6=1/2.
- Draw one card from a standard deck. Event B = {all hearts}. n(B)=13, n(S)=52, so P(B)=13/52=1/4.
- Two students chosen independently; event C = {both pass}. If P(pass)=0.8 and independent, P(C)=0.8*0.8=0.64.
- From a box with 2 defective and 8 good items, pick one at random. Event D = {defective}. P(D)=2/10=0.2. If two are picked without replacement, events are dependent and P(both defective)= (2/10)*(1/9)=1/45.
- Mutually exclusive example: In one die roll, A={1} and B={2} are mutually exclusive. P(A∪B)=P(A)+P(B)=1/6+1/6=1/3.
- Event relation: E \u2286 S ; Ø is impossible event ; S is sure event
- If outcomes equally likely: P(E)=n(E)/n(S)
- Complement: P(A') = 1 - P(A)
- Addition rule (general): P(A \u222a B) = P(A) + P(B) - P(A \u2229 B)
- Addition rule (mutually exclusive): If A \u2229 B = Ø, then P(A \u222a B) = P(A) + P(B)
- Multiplication rule (general): P(A \u2229 B) = P(A) * P(B|A)
Classical (theoretical) definition of probability
Definition. If a random experiment has a finite sample space S with n(S) equally likely outcomes and E is an event with n(E) favourable outcomes, then the probability of E is
P(E) = n(E) / n(S)
Here "equally likely" means every elementary outcome has the same chance of occurring. The classical definition reduces probability to counting favourable outcomes divided by total outcomes.
How to apply (step-by-step):
- Identify the sample space S and verify outcomes are equally likely.
- Count total outcomes n(S).
- Describe event E and count favourable outcomes n(E) (use combinations/permutations if needed).
- Compute P(E) = n(E)/n(S) and simplify.
Important properties (consequences):
- 0 ≤ P(E) ≤ 1 for any event E.
- P(S) = 1 and P(Ø) = 0.
- If A and B are mutually exclusive (disjoint), P(A ∪ B) = P(A) + P(B).
- For any event A, P(Ac) = 1 − P(A), where Ac is the complement.
- General addition rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
Counting tools. For many problems n(E) and n(S) are found using permutations (P) and combinations (C): e.g. if choosing k objects from n without order, use C(n, k) = n!/(k!(n−k)!).
Limitations. The classical definition applies only when (a) the sample space is finite and (b) elementary outcomes are equally likely. It does not apply directly when outcomes have different probabilities or when the sample space is infinite.
- Tossing one fair coin: S = {H, T}, n(S)=2. Probability of head P(H) = 1/2.
- Rolling one fair six-sided die: S = {1,2,3,4,5,6}, n(S)=6. Probability of getting a 4 is 1/6.
- Drawing one card from a well-shuffled 52-card deck: probability of an Ace = n(E)/n(S) = 4/52 = 1/13.
- Drawing 2 cards (without replacement) and getting 2 Aces: n(S)=C(52,2), n(E)=C(4,2) so P = C(4,2)/C(52,2) = 6/1326 = 1/221.
- Choosing one ball from an urn with 3 red and 2 blue balls (all equally likely): P(red) = 3/(3+2) = 3/5.
- P(E) = n(E) / n(S) (classical definition)
- 0 ≤ P(E) ≤ 1
- P(S) = 1, P(Ø) = 0
- P(A^c) = 1 − P(A) (complement rule)
- 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)
Experimental (empirical) probability and relative frequency
Definition: Experimental (or empirical) probability of an event E is the ratio of the number of times the event occurs to the total number of trials performed. It is also called the relative frequency of the event. If an experiment is repeated n times and the event E occurs f(E) times, then the experimental probability of E is P(E) ≈ f(E)/n.
Procedure:
- Specify the experiment and the event E.
- Perform the experiment n times under the same conditions.
- Count f(E), the number of trials in which E occurs.
- Compute the relative frequency P(E) = f(E)/n. This gives an empirical estimate of the probability of E.
Key idea — Law of Large Numbers: As the number of trials n increases, the experimental probability f(E)/n tends to approach the theoretical probability P(E) (if the theoretical probability exists and the trials are independent and identically conducted). In practice, larger n usually gives a more reliable estimate.
Comparison with theoretical probability: Theoretical probability is computed from known equally likely outcomes (for example, 1/6 for a fair die face) without performing experiments. Experimental probability estimates the probability by actual repeated trials; it may differ from the theoretical value for small n but should get closer as n grows.
Sources of error / considerations: Small sample size, biased or non-random trials, dependence between trials (e.g., drawing without replacement) and measurement errors can make the experimental probability deviate from the theoretical value.
Uses: Estimating probabilities when theoretical models are hard or impossible (e.g., reliability testing, medicine, weather, quality control), validating theoretical models, and teaching probability concepts.
- Coin toss: Toss a (fair) coin 100 times. Suppose heads occurs 56 times. Experimental probability of heads = f(heads)/n = 56/100 = 0.56. With more tosses this value should move closer to the theoretical 0.5.
- Die roll: Roll a die 120 times. If the face '4' appears 18 times, experimental probability of rolling a 4 = 18/120 = 3/20 = 0.15. Theoretical probability = 1/6 ≈ 0.1667; the experimental value may approach 0.1667 as rolls increase.
- Drawing balls: A bag contains many balls but unknown proportions of red and blue. Draw a ball, record its color, replace it, and repeat 200 times. If red appears 130 times, experimental probability of red ≈ 130/200 = 0.65. This estimates the fraction of red balls in the bag (replacement keeps trials approximately independent).
- Quality control: From a production line, examine 500 items and find 12 defective. Experimental probability of a defect = 12/500 = 0.024. This helps estimate defect rate for the process.
- Weather example: Over 30 years, suppose it rained on 90 of 180 summer days sampled. Experimental probability of rain on a summer day ≈ 90/180 = 0.5 — an empirical estimate used in climatology.
- Experimental probability (relative frequency): P(E) ≈ f(E) / n, where f(E) = number of times event E occurs, n = total trials.
- As n → ∞ (Law of Large Numbers): f(E)/n → P(E) (theoretical probability), provided trials are independent and identically conducted.
- Complement (empirical): P(E') ≈ 1 - P(E) = 1 - f(E)/n, where E' is the complement of E.
Basic properties of probability
Definition (for equally likely outcomes): If a random experiment has a finite sample space S with equally likely outcomes and an event A has n(A) favourable outcomes, then the probability of A is P(A) = n(A)/n(S).
Fundamental properties:
- Non-negativity: For any event A, P(A) ≥ 0. Probabilities cannot be negative.
- Normalization (total probability): P(S) = 1, where S is the sample space (something in S must happen).
- Probability of impossible event: P(∅) = 0, where ∅ is the null (impossible) event.
- Complement rule: For an event A, its complement A' (event 'A does not occur') satisfies P(A') = 1 − P(A).
- Additivity for disjoint events: If A and B are mutually exclusive (A ∩ B = ∅), then P(A ∪ B) = P(A) + P(B).
- General addition rule: For any two events A and B, P(A ∪ B) = P(A) + P(B) − P(A ∩ B). This removes double counting of outcomes in A ∩ B.
- Monotonicity (inclusion): If A ⊆ B then P(A) ≤ P(B).
- Bounds: From the above, 0 ≤ P(A) ≤ 1 for every event A.
Short derivations / remarks:
- From P(S)=1 and A ∪ A' = S with A and A' disjoint, P(A) + P(A') = 1 ⇒ P(A') = 1 − P(A).
- For general A,B, write A ∪ B = A + (B \ A) where (B \ A) is disjoint from A; hence P(A ∪ B) = P(A) + P(B \ A) = P(A) + P(B) − P(A ∩ B).
- Inequality P(A ∪ B) ≤ P(A) + P(B) follows because P(A ∩ B) ≥ 0.
Why these matter: These properties form the algebra of probability: they let you combine events, compute complements, and ensure consistency (probabilities stay between 0 and 1 and total 1 over the sample space).
- Toss a fair coin once. Sample space S = {H, T}. P(H) = 1/2, P(T) = 1/2, P(∅) = 0, P(S) = 1. Complement: P(H') = P(T) = 1 − P(H) = 1/2.
- Roll a fair six-sided die. Let A = 'roll is even' = {2,4,6} so P(A) = 3/6 = 1/2. Let B = 'roll ≥ 4' = {4,5,6} so P(B) = 3/6 = 1/2. A ∩ B = {4,6} so P(A ∩ B) = 2/6 = 1/3. Use addition rule: P(A ∪ B) = P(A)+P(B)−P(A ∩ B) = 1/2 + 1/2 − 1/3 = 2/3.
- Drawing a card from a 52-card deck. Let A = 'card is a heart' (13/52 = 1/4). Complement: P(not a heart) = 1 − 1/4 = 3/4. Disjoint union: P(heart or spade) = P(heart)+P(spade) = 1/4 + 1/4 = 1/2.
- Real-life: Weather forecast says '30% chance of rain today' means P(rain) = 0.30 and P(no rain) = 0.70. If two different independent sources say 30% and 40% for same day, use rules to reconcile overlapping models (more advanced).
- P(A) = n(A) / n(S) (for equally likely finite outcomes)
- 0 ≤ P(A) ≤ 1
- P(S) = 1
- P(∅) = 0
- P(A') = 1 − P(A)
- If A ∩ B = ∅, then P(A ∪ B) = P(A) + P(B)
Addition rule (probability of union)
What is the addition rule?
The addition rule gives the probability that at least one of two (or more) events occurs — in other words, the probability of the union of events.
Two events (general case):
For any two events A and B in the same sample space,
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Why we subtract P(A ∩ B): When we add P(A) and P(B) we count outcomes in A ∩ B twice, so we subtract P(A ∩ B) once to correct the double counting.
Special case — mutually exclusive events:
If A and B are mutually exclusive (disjoint), then A ∩ B = ∅ and P(A ∩ B) = 0, so
P(A ∪ B) = P(A) + P(B) (if A and B are mutually exclusive)
Complement form:
It is often useful to write the union probability using complements:
P(A ∪ B) = 1 − P(A' ∩ B')
(the probability that not both A and B fail)
Extension to three events (inclusion–exclusion):
For events A, B, C,
P(A ∪ B ∪ C) = P(A)+P(B)+P(C) − P(A∩B) − P(A∩C) − P(B∩C) + P(A∩B∩C)
This pattern generalises to more events (the inclusion–exclusion principle).
Short proof idea for two events:
Write B as (B \ A) ∪ (A ∩ B). Then A ∪ B = A ∪ (B \ A), where A and (B \ A) are disjoint. So P(A ∪ B) = P(A) + P(B \ A) = P(A) + [P(B) − P(A ∩ B)].
Use in practice:
To find P(A ∪ B) you need P(A), P(B) and the overlap probability P(A ∩ B). If the overlap is unknown, find it from data or by counting the outcomes in the intersection.
- Dice example: Roll a fair die. Let A = {even} = {2,4,6}, so P(A)=3/6=1/2. Let B = {multiple of 3} = {3,6}, so P(B)=2/6=1/3. A ∩ B = {6}, P(A ∩ B)=1/6. Using addition rule: P(A ∪ B)=1/2+1/3−1/6=2/3.
- Card example: Draw one card from a 52-card deck. Let A = 'heart' (13/52 = 1/4). Let B = 'face card' (J,Q,K of any suit: 12/52 = 3/13). Intersection A ∩ B = face hearts (J♥,Q♥,K♥): 3/52. Then P(A ∪ B)=1/4 + 3/13 − 3/52 = 11/26.
- Students and subjects: In a class, P(student passes Math)=0.5, P(passes Physics)=0.4, and P(passes both)=0.2. Probability a student passes at least one subject: P(Math ∪ Physics)=0.5+0.4−0.2=0.7.
- Three-event example: Suppose P(A)=0.5, P(B)=0.4, P(C)=0.3, P(A∩B)=0.2, P(A∩C)=0.15, P(B∩C)=0.10, P(A∩B∩C)=0.05. Then P(A∪B∪C)=0.5+0.4+0.3−0.2−0.15−0.10+0.05=0.80.
- General (two events): P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
- Mutually exclusive: if A ∩ B = ∅ then P(A ∪ B) = P(A) + P(B)
- Complement form: P(A ∪ B) = 1 − P(A' ∩ B')
- Three events (inclusion–exclusion): P(A ∪ B ∪ C) = P(A)+P(B)+P(C) − P(A∩B) − P(A∩C) − P(B∩C) + P(A∩B∩C)
- General inclusion–exclusion (n events): alternate sum of probabilities of intersections of 1,2,3,...,n events
Use of counting techniques in probability
What it is: In probability, counting techniques are used to find the number of possible outcomes in the sample space and the number of favourable outcomes for an event. When all outcomes are equally likely, probability = (number of favourable outcomes) / (number of total outcomes). Counting techniques help compute these numbers efficiently.
Key ideas and how to choose a technique:
- Fundamental principle of counting (multiplication rule): If a task can be done in a sequence of steps with m choices for step 1 and n choices for step 2, total choices = m × n (extendable to more steps).
- Order matters → Permutations: Use permutations when different orders are distinct. nPr = n! / (n − r)! gives the number of ordered arrangements of r objects chosen from n.
- Order doesn’t matter → Combinations: Use combinations when only the selection matters (not order). nCr = n! / (r!(n − r)!) gives the number of ways to choose r objects from n.
- Repetition allowed: If repetition is allowed and order matters, use n^r. If repetition allowed but order doesn’t matter, use combinations with repetition: C(n + r − 1, r).
- Identical objects: For permutations of n objects with groups of identical items, total = n!/(n1! n2! …).
- Complementary counting: Sometimes it is easier to count the complement and subtract from the total.
Applying to probability: 1) Identify the sample space and whether outcomes are equally likely. 2) Use the appropriate counting method to find |S| and |E|. 3) Compute P(E) = |E| / |S|. If outcomes are not equally likely, count or compute probabilities for each outcome (often using tree diagrams or binomial formula).
Common pitfalls: confusing when order matters, forgetting to account for replacement (with/without), and not checking whether outcomes are equally likely.
- Example 1 — Selecting a committee: From 10 students, a committee of 3 is chosen. How many possible committees and what is the probability that a particular student A is on the committee? Total ways = C(10,3)=120. Favourable ways where A is included = C(9,2)=36. Probability = 36/120 = 3/10.
- Example 2 — Drawing cards: Two cards are drawn at random (without replacement) from a standard deck of 52. Probability both are aces? Total pairs = C(52,2)=1326. Favourable = C(4,2)=6. Probability = 6/1326 = 1/221.
- Example 3 — Arrangements of letters: How many distinct 4-letter arrangements (order matters) can be made from letters A,B,C,D if repetition is not allowed? Total = 4P4 = 4! = 24. If we ask probability the arrangement starts with A, favourable = 3! = 6, so probability = 6/24 = 1/4.
- Example 4 — Passwords with repetition: A 3-digit PIN using digits 0–9 with repetition allowed. Total possible PINs = 10^3 = 1000. Probability a randomly chosen PIN has all digits equal (e.g., 111, 222)? Favourable = 10, so probability = 10/1000 = 1/100.
- Probability (equally likely outcomes): P(E) = |E| / |S|
- Fundamental counting principle (multiplication rule): If step1 has m choices, step2 has n choices, total = m × n (extendable to more steps)
- Permutation (order matters): nPr = n! / (n − r)!
- Permutation of n distinct objects: n!
- Permutation with identical objects: n! / (n1! n2! ... nk!)
- Combination (order does not matter): nCr = n! / (r!(n − r)!)
Common probability models and examples
This topic summarizes frequently used probability models (both discrete and continuous), when to use each model, their main properties and how they relate. Each model is presented with its defining formula (PMF or PDF), expectation and variance, conditions/assumptions, and a short intuitive description.
Discrete models
- Discrete uniform: all outcomes in a finite set of size N are equally likely. Good for fair dice, shuffled cards, lottery numbers.
- Bernoulli: a single trial with two outcomes: success (probability p) or failure (probability 1-p). Use for a single yes/no experiment (coin toss, pass/fail).
- Binomial: number of successes in n independent Bernoulli trials with same p. Use when you have fixed number of repeated independent trials (e.g., 10 coin tosses, number of defective items in a sample of n).
- Geometric: number of trials until the first success (support 1,2,...). Use when you wait for the first success and trials are independent with same p.
- Hypergeometric: number of successes in n draws without replacement from a finite population containing K successes and N−K failures. Use when sampling without replacement (e.g., drawing cards without replacement).
- Poisson: models count of rare events in a fixed interval if events occur independently and average rate λ is constant. Use for arrivals, rare defects, phone calls per hour when events are rare and many potential opportunities.
Continuous models
- Continuous uniform: every point in interval [a,b] is equally likely (density 1/(b−a)). Use for a random point on a length, random time in an interval when all instants are equally likely.
- Exponential: models waiting time between independent Poisson events with rate λ (memoryless property). Use for lifetimes without aging (constant hazard), e.g., time between phone calls.
- Normal (Gaussian): bell-shaped symmetric distribution determined by mean μ and variance σ2. Many sums of small independent effects are approx. normal (central limit theorem). Use for measurement errors, heights, scores (when approximately symmetric and unimodal).
Key relations and approximations:
- Binomial(n,p) ≈ Poisson(λ = np) when n is large and p is small (np moderate).
- Binomial(n,p) ≈ Normal(mean=np, variance=np(1−p)) when n is large and np and n(1−p) are both ≥ about 5 (use continuity correction for discrete-to-continuous approximation).
- Exponential is the continuous analogue of the geometric distribution (both are memoryless).
Understanding which model fits a situation comes from checking assumptions: discrete vs continuous, independence, identical trials, replacement vs no replacement, constant rate, rarity of events, etc.
- Discrete uniform: Roll a fair six-sided die. Probability of any face k (1≤k≤6) is 1/6.
- Bernoulli: A single biased coin toss with P(success) = p (e.g., probability of 'heads' = 0.3).
- Binomial: In 10 independent tests where each has probability 0.2 of success, probability of exactly 3 successes is C(10,3)(0.2)^3(0.8)^7.
- Geometric: Flipping a fair coin until first head; probability that first head appears on the k-th flip is (1/2) * (1/2)^{k-1} = (1/2)^k.
- Hypergeometric: Drawing 5 cards from a standard 52-card deck without replacement; probability of exactly 2 aces = [C(4,2) * C(48,3)] / C(52,5).
- Poisson: Number of calls arriving at a call center in an hour when average rate is 6 calls/hour: P(X=k) = e^{-6} 6^k / k!.
- Discrete uniform (N outcomes): P(X = x) = 1/N. E(X) = (sum of outcomes)/N, Var(X) = E(X^2) - [E(X)]^2.
- Bernoulli(p): P(X=1)=p, P(X=0)=1-p. E(X)=p, Var(X)=p(1-p).
- \[Binomial(n,p): P(X=k) = C(n,k) p^k (1-p)^{n-k}\]\[k=0,1,...,n\]\[E(X)=np\]\[Var(X)=np(1-p).\]
- \[Geometric(p) (counting trials until first success): P(X=k) = (1-p)^{k-1} p\]\[k=1,2,...\]\[E(X)=1/p\]\[Var(X)=(1-p)/p^2.\]
- Hypergeometric(N,K,n): P(X=k) = [C(K,k) C(N-K, n-k)] / C(N,n), support max(0, n-(N-K)) ≤ k ≤ min(n,K). E(X)=n(K/N), Var(X)=n(K/N)(1-K/N)((N-n)/(N-1)).
- \[Poisson(λ): P(X=k) = e^{-λ} λ^k / k!\]\[k=0,1,2,...\]\[E(X)=λ\]\[Var(X)=λ.\]
Problem-solving strategies and applications
Overview
Problem-solving in probability combines logical reasoning, counting methods and choice of the right model. A reliable approach follows Polya's four steps: (1) understand the problem, (2) devise a plan (choose counting/conditioning/diagrams), (3) carry out the plan (compute probabilities), and (4) review and interpret the result.
Key strategies
- Define the sample space clearly (equally likely outcomes when appropriate). Write S and count n(S).
- Use counting techniques: permutations and combinations for arrangements and selections; multiplication principle for sequential choices.
- Use complements: when "at least one" or "none" type events appear, compute P(A)=1−P(A complement) if easier.
- Condition and partition: split complicated events into simpler conditional cases (use law of total probability).
- Apply Bayes' theorem for reversed conditional questions (find cause given effect).
- Test independence: if P(A∩B)=P(A)P(B), events are independent; otherwise use conditional probabilities.
- Use symmetry to reduce counting (identical objects, identical roles).
- Diagrams and representations: tree diagrams for sequential/conditional problems, Venn diagrams for unions/intersections, sample-space grids for dice/cards.
- Check edge cases and sanity: probabilities must lie in [0,1]; compare with simple bounds (e.g., complement or symmetry).
Applications
Probability techniques apply to games of chance (dice, cards), quality control (defective rates), medical testing (sensitivity/specificity and Bayesian updating), reliability of systems, risk assessment, and simple genetics (Punnett-square probabilities). In modelling, translate the real situation into an appropriate experiment, identify events of interest, and use counting/conditioning to compute probabilities.
How to choose a method
- If outcomes are equally likely and finite, counting (n(A)/n(S)) is often simplest.
- If outcomes occur in sequence or depend on previous draws, consider conditional probability and trees.
- For "at least one" problems use complements.
- For questions asking "given that B occurred, what's the chance of A?" use P(A|B) and Bayes if reversing conditions.
Following these structured steps reduces errors and makes reasoning clear and checkable.
- Example 1 — Dice (counting and sample space): Problem: Two fair dice are rolled. Find probability that the sum is 7. Solution outline: Sample space has 36 equally likely outcomes. Count pairs (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) = 6. So P(sum=7)=6/36=1/6.
- Example 2 — Drawing without replacement (combinations): Problem: An urn has 5 white and 3 black balls. Three balls are drawn without replacement. Find probability of exactly two white. Solution outline: Number of ways to choose 3 balls = C(8,3)=56. Favorable ways = choose 2 white from 5 and 1 black from 3: C(5,2)*C(3,1)=10*3=30. So probability = 30/56 = 15/28.
- Example 3 — Complement trick: Problem: If a fair coin is tossed 4 times, find probability of at least one head. Solution outline: Complement is no heads (all tails) with probability (1/2)^4 = 1/16. So required probability = 1 - 1/16 = 15/16.
- Example 4 — Conditional probability and Bayes (medical test): Problem: A disease affects 1% of a population. Test sensitivity (P(+|disease))=0.99, specificity (P(-|no disease))=0.95. If a randomly chosen person tests positive, what is probability they actually have the disease? Solution outline: Let D = disease, + = positive. P(D)=0.01, P(+|D)=0.99, P(+|D^c)=1-0.95=0.05. Use Bayes: P(D|+)=P(+|D)P(D)/[P(+|D)P(D)+P(+|D^c)P(D^c)] = 0.99*0.01 / (0.99*0.01 + 0.05*0.99) = 0.0099 / (0.0099 + 0.0495) ≈ 0.1667 (about 16.7%).
- P(A) = n(A) / n(S) for equally likely finite outcomes
- Complement rule: P(A^c) = 1 - P(A)
- Addition rule (two events): P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
- Inclusion–exclusion (three events): P(A ∪ B ∪ C) = P(A)+P(B)+P(C) - P(A∩B)-P(B∩C)-P(C∩A) + P(A∩B∩C)
- Multiplication rule (sequential independent events): P(A and B) = P(A)P(B) (if independent)
- General multiplication (dependent): P(A ∩ B) = P(A)P(B|A) = P(B)P(A|B)
Key Concepts
- Random experiment
- A process or action that leads to one of several possible outcomes, where the outcome cannot be predicted with certainty.
- Sample space
- The set of all possible outcomes of a random experiment, usually denoted by S.
- Outcome
- A single possible result of a random experiment, an element of the sample space.
- Event
- A subset of the sample space; one or more outcomes grouped together.
- Simple (elementary) event
- An event consisting of exactly one outcome from the sample space.
- Compound event
- An event containing two or more outcomes from the sample space.
- Mutually exclusive (disjoint) events
- Two events that cannot occur at the same time; their intersection is empty.
- Exhaustive events
- A collection of events whose union equals the entire sample space.
- Complement of an event
- All outcomes in the sample space that are not in the event; denoted E^c.
- Equally likely outcomes
- When every outcome in the sample space has the same probability.
- Classical (theoretical) probability
- Probability defined as |E|/|S| when all outcomes are equally likely.
- Empirical (relative frequency) probability
- Probability estimated from observed frequencies: P(E) ≈ (frequency of E)/(total trials).
- Axiomatic probability
- Probability defined by axioms: non-negativity, P(S)=1, and countable additivity for disjoint events.
- Conditional probability
- Probability of event A given event B has occurred: P(A|B) = P(A ∩ B)/P(B), P(B)>0.
- Independent events
- Two events A and B are independent if P(A ∩ B) = P(A)P(B).
- Addition rule (for union)
- P(A ∪ B) = P(A) + P(B) − P(A ∩ B); simplifies to sum when A and B are disjoint.
- Multiplication rule
- General: P(A ∩ B) = P(A)P(B|A). If A and B are independent, P(A ∩ B) = P(A)P(B).
- Bayes' theorem
- Relates posterior and prior probabilities: P(Ai|B) = P(Ai)P(B|Ai) / Σj P(Aj)P(B|Aj).
- Random variable (discrete)
- A function that assigns a real number to each outcome of a random experiment, taking countable values.
- Expected value (mean)
- The long-run average value of a discrete random variable: E(X) = Σ x·P(X=x).
End-of-Chapter Trial Paper & Test Questions
Topic-wise questions to test your understanding of every concept in this chapter.
-
Define sample space and write the sample space when two coins are tossed together. / प्रतिदर्श समष्टि को परिभाषित कीजिए और जब दो सिक्के एक साथ उछाले जाते हैं तो प्रतिदर्श समष्टि लिखिए।
Show answer
The sample space is the set of all possible outcomes of a random experiment, denoted S; for two coins S = {HH, HT, TH, TT}. / प्रतिदर्श समष्टि किसी यादृच्छिक प्रयोग के सभी संभव परिणामों का समुच्चय है, जिसे S से दर्शाते हैं; दो सिक्कों के लिए S = {HH, HT, TH, TT}।
-
A die is rolled. Let A = {even number} and B = {number ≥ 4}. Find P(A ∪ B) using the addition rule. / एक पासा फेंका जाता है। मान लीजिए A = {सम संख्या} और B = {संख्या ≥ 4}। योग नियम का उपयोग करके P(A ∪ B) ज्ञात कीजिए।
Show answer
A = {2,4,6} so P(A)=1/2, B = {4,5,6} so P(B)=1/2, A∩B = {4,6} so P(A∩B)=1/3; P(A∪B) = 1/2 + 1/2 − 1/3 = 2/3. / A = {2,4,6} अतः P(A)=1/2, B = {4,5,6} अतः P(B)=1/2, A∩B = {4,6} अतः P(A∩B)=1/3; P(A∪B) = 1/2 + 1/2 − 1/3 = 2/3।
-
Distinguish between mutually exclusive events and independent events with an example each. / परस्पर अपवर्जी घटनाओं और स्वतंत्र घटनाओं में एक-एक उदाहरण सहित अंतर बताइए।
Show answer
Mutually exclusive events cannot occur together (A∩B=∅), e.g. getting {1} and {2} in one die roll; independent events do not affect each other's probability (P(A∩B)=P(A)P(B)), e.g. two separate coin tosses. / परस्पर अपवर्जी घटनाएँ एक साथ घटित नहीं हो सकतीं (A∩B=∅), जैसे एक पासा फेंकने पर {1} और {2} प्राप्त करना; स्वतंत्र घटनाएँ एक-दूसरे की प्रायिकता को प्रभावित नहीं करतीं (P(A∩B)=P(A)P(B)), जैसे दो अलग सिक्कों का उछालना।
-
Two cards are drawn without replacement from a 52-card deck. Find the probability that both are aces. / 52 पत्तों की गड्डी से बिना प्रतिस्थापन के दो पत्ते निकाले जाते हैं। दोनों के इक्के होने की प्रायिकता ज्ञात कीजिए।
Show answer
P = C(4,2)/C(52,2) = 6/1326 = 1/221. / P = C(4,2)/C(52,2) = 6/1326 = 1/221।
-
Using the complement rule, find the probability of getting at least one head when a fair coin is tossed 4 times. / पूरक नियम का उपयोग करते हुए, एक न्याय्य सिक्के को 4 बार उछालने पर कम से कम एक चित आने की प्रायिकता ज्ञात कीजिए।
Show answer
The complement (no heads, all tails) has probability (1/2)^4 = 1/16, so P(at least one head) = 1 − 1/16 = 15/16. / पूरक (कोई चित नहीं, सभी पट) की प्रायिकता (1/2)^4 = 1/16 है, अतः P(कम से कम एक चित) = 1 − 1/16 = 15/16।
-
State the three axioms (basic properties) of probability. / प्रायिकता के तीन अभिगृहीत (मूल गुणधर्म) लिखिए।
Show answer
For any event A, P(A) ≥ 0 (non-negativity); P(S) = 1 for the sample space S (normalization); and for mutually exclusive events A and B, P(A ∪ B) = P(A) + P(B) (additivity). / किसी भी घटना A के लिए, P(A) ≥ 0 (अऋणात्मकता); प्रतिदर्श समष्टि S के लिए P(S) = 1 (सामान्यीकरण); और परस्पर अपवर्जी घटनाओं A और B के लिए, P(A ∪ B) = P(A) + P(B) (योज्यता)।
-
From 10 students a committee of 3 is chosen at random. Find the probability that a particular student A is included. / 10 विद्यार्थियों में से यादृच्छिक रूप से 3 की एक समिति चुनी जाती है। एक विशेष विद्यार्थी A के शामिल होने की प्रायिकता ज्ञात कीजिए।
Show answer
Total ways = C(10,3) = 120; favourable ways with A included = C(9,2) = 36; probability = 36/120 = 3/10. / कुल तरीके = C(10,3) = 120; A सहित अनुकूल तरीके = C(9,2) = 36; प्रायिकता = 36/120 = 3/10।
-
What is the difference between classical (theoretical) and experimental (empirical) probability, and how are they related? / चिरसम्मत (सैद्धांतिक) और प्रायोगिक (आनुभविक) प्रायिकता में क्या अंतर है, और वे किस प्रकार संबंधित हैं?
Show answer
Classical probability is computed as n(E)/n(S) for equally likely outcomes without experimentation, while experimental probability is f(E)/n found from actual trials; by the Law of Large Numbers, as n increases the experimental value approaches the theoretical value. / चिरसम्मत प्रायिकता समान रूप से संभावित परिणामों के लिए बिना प्रयोग किए n(E)/n(S) के रूप में निकाली जाती है, जबकि प्रायोगिक प्रायिकता वास्तविक परीक्षणों से प्राप्त f(E)/n होती है; बृहत् संख्या के नियम के अनुसार, जैसे-जैसे n बढ़ता है प्रायोगिक मान सैद्धांतिक मान के निकट पहुँचता 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.