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Andhra Pradesh (BIEAP)Class 10 Mathematics★ AP Board (BSEAP) SSC Mathematics · ~4 marks · Board: AP Board

Probability

Classical probability — coins, dice, playing cards. P(E) = n(E)/n(S). Mutually exclusive events. AP Board Class 10 Mathematics.

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By the end of this chapter you'll be able to…

1Apply the classical definition: P(E) = n(E)/n(S)
2List sample spaces for coins (single and multiple) and dice (single and two)
3Solve playing card probability problems knowing the deck structure
4Apply P(not E) = 1 − P(E) for complementary events
5Identify mutually exclusive events and apply P(A∪B) = P(A) + P(B)

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Why this chapter matters

Probability is one of the EASIEST scoring chapters in AP SSC — the questions are predictable (coins, dice, playing cards) and the formula is simple. Once you memorise the deck of cards and understand what counts as a 'favourable outcome', probability questions take under 3 minutes each. AP SSC typically asks 2–4 marks from this chapter and students who are prepared here never lose these marks.

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Probability

Classical Definition: P(E) = n(E)/n(S). 0 ≤ P(E) ≤ 1.

Key Concepts

Sample Space (S) : All possible outcomes. Two dice: 36. Deck of cards: 52.
Complement: P(not E) = 1 − P(E).
Mutually Exclusive: P(A∩B) = 0. Cannot happen together. P(A∪B) = P(A) + P(B).

Playing Cards — Know the Deck COLD

52 cards. 4 suits (♠♥♦♣). 26 RED (♥♦). 26 BLACK (♠♣). Face cards: J, Q, K (12). Aces (4).

Common Mistake: Forgetting that dice outcomes are ORDERED pairs. (1,2) ≠ (2,1).

Theoretical vs Experimental Probability

Theoretical Probability: P(E) = number of favourable outcomes / total number of possible outcomes. Based on the ASSUMED equally likely outcomes.

Experimental Probability: Based on ACTUAL trials or experiments. P(E) = number of times E occurs / total number of trials. As the number of trials increases, experimental probability APPROACHES theoretical probability (Law of Large Numbers).

'AP exams test THEORETICAL probability. Experimental probability is conceptual — understand how it differs from theoretical.'

Worked Examples — Coins

Single Coin

Example 1: A fair coin is tossed once. Find P(H), P(T). Sample space S = {H, T}. n(S) = 2. P(H) = 1/2. P(T) = 1/2.

Two Coins

Example 2: Two coins are tossed. Find the probability of (i) two heads (ii) at least one head (iii) exactly one tail. S = {HH, HT, TH, TT}. n(S) = 4. (i) P(HH) = 1/4. (ii) At least one head: {HH, HT, TH} → 3/4. (iii) Exactly one tail: {HT, TH} → 2/4 = 1/2.

Three Coins

Example 3: Three coins are tossed. Find P(at least two heads). S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. n(S) = 8. At least two heads: {HHH, HHT, HTH, THH} → 4/8 = 1/2.

Worked Examples — Dice

Two Dice

Example 4: Two dice are rolled. Find the probability that the sum is (i) 7 (ii) at most 4 (iii) a prime number. n(S) = 36.

(i) Sum = 7: {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} → 6/36 = 1/6.

(ii) Sum at most 4: sum=2{(1,1)}, sum=3{(1,2),(2,1)}, sum=4{(1,3),(2,2),(3,1)} → 6/36 = 1/6.

(iii) Prime sums: 2,3,5,7,11. Sum=2: (1,1). Sum=3: (1,2),(2,1). Sum=5: (1,4),(2,3),(3,2),(4,1). Sum=7: 6 outcomes. Sum=11: (5,6),(6,5). Total = 1+2+4+6+2 = 15 → 15/36 = 5/12.

Example 5: Two dice are rolled. Find P(doublet i.e. same number on both). Doublets: (1,1),(2,2),(3,3),(4,4),(5,5),(6,6) → 6/36 = 1/6.

Worked Examples — Playing Cards

Example 6: A card is drawn from a well-shuffled deck. Find: (i) P(king) = 4/52 = 1/13. (ii) P(red card) = 26/52 = 1/2. (iii) P(face card) = 12/52 = 3/13. (iv) P(spade) = 13/52 = 1/4. (v) P(queen of hearts) = 1/52. (vi) P(an ace or a king) = 4+4 = 8/52 = 2/13. (vii) P(neither king nor queen) = 1 − P(king or queen) = 1 − 8/52 = 44/52 = 11/13.

'Memorise the card deck: 52 cards, 4 suits, each with 13 cards (A,2-10,J,Q,K). 12 face cards. 26 red. 26 black.'

Conditional Drawing (Without Replacement)

Example 7: Two cards are drawn successively without replacement. Find P(both aces). P(first ace) = 4/52 = 1/13. P(second ace | first ace) = 3/51 = 1/17. P(both aces) = (1/13)×(1/17) = 1/221.

Mutually Exclusive and Complementary Events

Mutually Exclusive: Events that CANNOT happen simultaneously. If A and B are mutually exclusive: P(A∩B) = 0 and P(A∪B) = P(A) + P(B).

Complementary Events: A and not A are always mutually exclusive and exhaustive. P(A) + P(not A) = 1. P(not A) = 1 − P(A).

Example 8 — Not Mutually Exclusive: From a deck, P(king or heart). These are NOT mutually exclusive (king of hearts is both). P(king) = 4/52. P(heart) = 13/52. P(king∩heart) = 1/52. P(king∪heart) = 4/52 + 13/52 − 1/52 = 16/52 = 4/13.

Example 9 — Using Complement: From digits 1 to 20, a number is chosen. Find P(not a multiple of 3). Multiples of 3: {3,6,9,12,15,18} → 6/20 = 3/10. P(not multiple of 3) = 1 − 3/10 = 7/10.

Common Mistakes — Extended

Dice outcomes are ordered: (1,2) and (2,1) are DIFFERENT outcomes. Total outcomes = 36, NOT 21.
'At least one' ≠ complement of 'none': The complement approach: P(at least one) = 1 − P(none). This is OFTEN easier.
Forgetting face cards: 12 face cards (J, Q, K in each of 4 suits). Aces are NOT face cards.
Drawing without replacement: The total reduces by 1 for the second draw.
P(A or B) when not mutually exclusive: P(A∪B) = P(A)+P(B)−P(A∩B). The subtraction prevents double-counting.

AP Exam Focus

Topic	Marks	Frequency
Coin toss problems	2-3	Very Common
Dice problems	3-4	Very Common
Playing cards problems	4	Common
Mutually exclusive events	2-3	Moderate
Complementary events	2-3	Common
Without replacement	4	Moderate

Self-Test Questions

A coin is tossed 3 times. Find P(exactly 2 heads). (Answer: {HHT,HTH,THH} → 3/8)
Two dice are rolled. Find P(sum = 8). (Answer: {(2,6),(3,5),(4,4),(5,3),(6,2)} → 5/36)
From a deck, find P(red face card). (Answer: 6/52 = 3/26)
Two dice are rolled. Find P(product is even). (Answer: P(at least one even) = 1 − P(both odd) = 1 − 9/36 = 27/36 = 3/4)
A bag has 5 red and 3 white balls. One ball is drawn at random. Find P(red). (Answer: 5/8)
From 1 to 30, a number is selected. Find P(multiple of 4 or 5). (Answer: Multiples of 4: 7, multiples of 5: 6, common: 1 (20). P = (7+6−1)/30 = 12/30 = 2/5)
Two dice are rolled. Find P(doublet or sum of 8). (Answer: P(doublet)=6/36, P(sum=8)=5/36, P(both)=1/36 (4,4). P = (6+5−1)/36 = 10/36 = 5/18)

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Probability

CLASSICAL PROBABILITY: P(E) = Number of favourable outcomes / Total possible outcomes = n(E)/n(S). 0 ≤ P(E) ≤ 1. P(E) + P(not E) = 1. SAMPLE SPACES: Single coin: {H, T} → n(S) = 2. Two coins: {HH, HT, TH, TT} → n(S) = 4. Single die: {1,2,3,4,5,6} → n(S) = 6. Two dice: 36 ordered pairs → n(S) = 36. Playing cards: n(S) = 52. 4 suits (♠♥♦♣). Each suit: 13 cards (A,2,3,4,5,6,7,8,9,10,J,Q,K). Red cards: 26 (♥♦). Black: 26 (♠♣). Face cards: 12 (J, Q, K × 4 suits). ACES: 4.

CARD PROBLEMS — KEY FACTS: Spades(♠)=13, Hearts(♥)=13, Diamonds(♦)=13, Clubs(♣)=13. Numbered cards (2–10): 9 per suit × 4 = 36. Face cards (J,Q,K): 3 per suit × 4 = 12. Aces: 1 per suit × 4 = 4. Total = 36+12+4 = 52 ✓. 'Picture cards' = Face cards (J, Q, K) = 12. Note: KING, QUEEN, JACK are face cards; ACE is NOT a face card.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Treating two-dice outcomes as unordered (e.g., counting (1,2) and (2,1) as one outcome)

✓ For TWO DICE, the outcomes are ORDERED PAIRS — (1,2) and (2,1) are DIFFERENT outcomes. Total = 6 × 6 = 36. NOT 21. This is because the dice are distinct (e.g., one red die, one blue die). When you want the sum = 3, favourable outcomes are (1,2) and (2,1) — both count separately. n(S) = 36, always.

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Probability?

1 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

1 questions~2 min

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•CLASSICAL PROBABILITY: P(E) = n(E)/n(S) — number of favourable outcomes ÷ total equally likely outcomes. Conditions: (1) Outcomes must be equally likely. (2) Sample space must be finite and countable. 0 ≤ P(E) ≤ 1. P(certain event) = 1. P(impossible event) = 0.
•COMPLEMENTARY EVENT: P(not E) = 1 − P(E). This is one of the most useful tools. When computing P(at least one) is hard, compute P(none) and subtract from 1. Example: P(at least one head in 3 tosses) = 1 − P(no heads) = 1 − 1/8 = 7/8.
•SAMPLE SPACES: Single coin: S = {H, T}, n(S) = 2. Two coins: S = {HH, HT, TH, TT}, n(S) = 4. Three coins: n(S) = 8. Single die: n(S) = 6. Two dice: n(S) = 36 (ordered pairs). Single card: n(S) = 52. Two cards (without replacement): n(S) = 52 × 51.
•52-CARD DECK STRUCTURE (MEMORISE): 4 suits: Spades♠(black), Clubs♣(black), Hearts♥(red), Diamonds♦(red). Each suit has 13 cards: A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K. Face cards (J, Q, K): 3 per suit × 4 = 12 face cards. Aces: 4 (one per suit). Number cards (2–10): 9 per suit × 4 = 36. Total = 12 + 4 + 36 = 52. Red cards = 26. Black cards = 26.
•TWO DICE — KEY SUMS: n(S) = 36 always (ordered pairs). Sum = 2: only (1,1) → n=1. Sum = 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) → n=6 (highest probability). Sum = 12: only (6,6) → n=1. Sum = 6: (1,5),(2,4),(3,3),(4,2),(5,1) → n=5. P(sum=7) = 6/36 = 1/6 — highest probability sum with two dice.
•MUTUALLY EXCLUSIVE EVENTS: Cannot happen simultaneously. P(A∩B) = 0. P(A∪B) = P(A) + P(B). Example: Drawing a red card AND drawing a black card from one draw — mutually exclusive. P(red or black) = P(red) + P(black) = 1/2 + 1/2 = 1.
•NON-MUTUALLY EXCLUSIVE EVENTS: Can happen simultaneously. P(A∪B) = P(A) + P(B) − P(A∩B). Example: Drawing a king or a red card. King and red: king of hearts and king of diamonds = 2 cards. P(king or red) = 4/52 + 26/52 − 2/52 = 28/52 = 7/13.
•BALLS IN A BAG PROBLEMS: 'A bag contains 3 red and 5 blue balls. One ball drawn at random.' Total = 8. P(red) = 3/8. P(blue) = 5/8. P(not red) = 1 − 3/8 = 5/8. Always compute the total first. 'At random' means equally likely.
•EMPIRICAL PROBABILITY: Based on actual experiments. P(E) = number of times E occurred / total trials. Differs from theoretical probability — empirical probability APPROACHES theoretical as number of trials → infinity. A coin tossed 1000 times gave 510 heads: empirical P(head) = 510/1000 = 0.51 (close to theoretical 0.5).
•IMPOSSIBLE AND CERTAIN EVENTS: Impossible event (∅): P = 0. Example: rolling a 7 on a die. Certain event (whole sample space): P = 1. Example: rolling a number ≤ 6 on a die. Sum of all probabilities in a sample space = 1 always.

Andhra Pradesh (BIEAP) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Insurance and risk assessment

Insurance companies calculate premiums using probability. If the probability of a house in coastal AP flooding in a year is 0.05 (5%), and the average damage is ₹5 lakh, the expected annual loss per house is 0.05 × ₹5 lakh = ₹25,000. The premium is set above this to ensure profitability. Car insurance rates in Visakhapatnam are higher than in interior Anantapur because accident probability is higher in the coastal city. All of this is classical probability applied to historical frequency data.

Quality control in manufacturing

AP's pharmaceutical and manufacturing industries use probability for quality control. If a production line has a defect rate of 2% (P(defective) = 0.02), and a batch of 500 tablets is produced, the expected number of defective tablets = 500 × 0.02 = 10. Statistical sampling — testing 50 random tablets from a batch — uses probability to decide whether to accept or reject the entire batch. This is empirical probability applied to industrial quality management.

Weather forecasting — probability of rain

'70% chance of rain tomorrow' is an empirical probability statement: out of all historically similar atmospheric conditions, 70% resulted in rain. The IMD (India Meteorological Department) computes these probabilities for every district in AP — affecting farmer decisions on when to plant, irrigate, or harvest. Cyclone probability forecasts for AP's coast (Probability of Cyclonic Disturbance within 200 km = X%) are life-safety decisions directly based on probability calculations.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Always list n(S) first: write 'n(S) = [value]' before identifying favourable outcomes. This shows method and prevents counting errors. For cards: 52. For two dice: 36. For two coins: 4. For one die: 6.
Card problems: write the deck structure at the top — '52 cards: 4 suits of 13 each. Face cards: 12. Aces: 4. Red: 26. Black: 26.' Having this reference prevents mid-problem memory errors.
Non-mutually exclusive (2 marks): state whether events can occur simultaneously. If yes: P(A∪B) = P(A)+P(B)−P(A∩B). If no: P(A∪B) = P(A)+P(B). Writing which rule you are applying earns the method mark.
Complementary events: when direct counting is hard, use P(E) = 1−P(not E). Write: 'P(at least one ___) = 1 − P(no ___).' This approach is usually simpler than counting favourable outcomes directly.
Simplify fractions: always reduce P(E) to lowest terms. P = 12/52 → P = 3/13. AP Board mark schemes show simplified fractions — unsimplified fractions may lose a mark.

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Research the Monty Hall Problem — a famous probability puzzle: you are on a game show. Three doors, behind one is a car, behind the others are goats. You choose door 1. The host (who knows what's behind the doors) opens door 3 to reveal a goat. Should you switch to door 2? Surprisingly, YES — switching wins with probability 2/3 (NOT 1/2 as most people intuit). Research the mathematical explanation and why it seems so counterintuitive. This problem generated enormous controversy when published in 1990.
Investigate the Birthday Problem — in a room of n randomly chosen people, what is the probability that at least two share a birthday? Surprisingly, with just 23 people, P(at least one shared birthday) > 50%! With 70 people, P > 99.9%. Research the derivation using the complementary probability approach: P(no shared birthday) = (365/365) × (364/365) × (363/365) × ... × (365−n+1)/365. The Birthday Problem appears in cryptography (hash collisions) and is a key intuition-builder for probability.
Explore Bayesian probability — a fundamentally different interpretation from classical probability. Classical: P(event) = fraction of equally likely outcomes. Bayesian: P(event) = degree of belief based on evidence, updated by new information using Bayes' Theorem: P(A|B) = P(B|A) × P(A) / P(B). Bayesian reasoning is used in spam filters (is this email spam given these words?), medical diagnosis (does this test result mean the patient has the disease?), and AI classifiers. Research Bayes' Theorem and one real application.
Research the Law of Large Numbers and its limits — as the number of trials increases, empirical probability approaches theoretical probability. But HOW FAST? The Central Limit Theorem says the standard deviation of sample means decreases as 1/√n — so quadrupling the number of trials halves the error. This explains why casinos are always profitable: individual gambler outcomes vary widely, but with thousands of customers, the casino's actual take closely tracks the theoretical probability. Research how the LLN is used in actuarial science and financial risk management.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

AP Board SSC (Class 10)Medium — Probability appears for 2–4 marks in AP SSC; card, dice, and ball-drawing problems are most common

JEE Main and Advanced (Probability)Very High — Probability and statistics is one of the most tested topics in JEE Mathematics, building extensively on Class 10 foundations

NTSE (Mathematics)Medium — Basic probability appears in NTSE Stage I

AP EAPCET (Mathematics)High — Probability (events, conditional, Bayes, random variables) is a full chapter in EAPCET Mathematics; Class 10 is the starting point

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Questions students ask

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For two dice, the outcomes are ORDERED PAIRS — meaning the first die's result and the second die's result are counted separately. (2, 3) and (3, 2) are DIFFERENT outcomes because different dice show different faces. There are 6 possible values for the first die and 6 for the second → 6 × 6 = 36. If you used 21 (treating (2,3) and (3,2) as the same), the outcomes would NOT be equally likely — some unordered pairs have more ways to occur than others. For example, (3,4) has two ordered arrangements [(3,4) and (4,3)] while (3,3) has only one [(3,3)]. Since probability requires EQUALLY LIKELY outcomes, we must use ordered pairs: n(S) = 36.

Diamonds: 13 cards. Kings: 4 cards. King of Diamonds: 1 card (counted in BOTH groups). Since diamond and king are NOT mutually exclusive (king of diamonds is both): P(diamond OR king) = P(diamond) + P(king) − P(diamond AND king) = 13/52 + 4/52 − 1/52 = 16/52 = 4/13. This is a non-mutually exclusive event problem. ALWAYS check: 'Is there overlap?' If yes, subtract the overlap. The king of diamonds is drawn in 1 way — subtract it once to avoid double-counting.

Two-digit numbers: 10 to 99. Total = 90 numbers. n(S) = 90. Multiples of 3 in range 10–99: smallest = 12 (first multiple ≥ 10), largest = 99. Count = (99−12)/3 + 1 = 87/3 + 1 = 29 + 1 = 30. P(multiple of 3) = 30/90 = 1/3. Alternative: multiples of 3 are uniformly distributed → every 3rd number → 90/3 = 30. P = 30/90 = 1/3. This type of number-based probability is tested in AP SSC.

Without replacement means after taking the first ball, it is NOT put back. After first draw: 9 balls remain. P(first ball is red) = 4/10 = 2/5. P(second ball is red | first was red) = 3/9 = 1/3 (only 3 red balls remain out of 9). P(both red) = P(first red) × P(second red | first red) = 4/10 × 3/9 = 12/90 = 2/15. This uses conditional probability (Class 12 topic) but appears in AP SSC as combinatorics: n(S) = 10C2 = 45, n(both red) = 4C2 = 6, P = 6/45 = 2/15 ✓.

EQUALLY LIKELY: Each outcome has the same chance of occurring — no outcome is favoured. This condition is REQUIRED for the classical formula P(E) = n(E)/n(S). Examples of equally likely: tossing a fair coin (P(H) = P(T) = 1/2). Rolling a fair die (each face = 1/6). Choosing a random card. EXAMPLES NOT EQUALLY LIKELY: A loaded die (one face is heavier, comes up more often). Drawing a ball from a bag where some balls are larger (larger balls might be grabbed more easily). In AP SSC problems, 'at random' always means equally likely. If a probability problem specifies a 'biased' or 'loaded' coin/die, the equally likely assumption does NOT apply.

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