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Class 9 · Tamil Nadu State Board

Samacheer Kalvi 9th Maths Probability Exercise Solutions | KalviGavani

TN Class 9 Maths Probability exercise solutions — classical definition, experiments, sample space, events, dice problems, coin problems, card problems, full step-by-step answers.

Updated 12 June 202625 min read

How to Use This Page

Work through each exercise in the Samacheer Kalvi textbook on your own first, then use the solutions below to verify your steps and answers. All questions from Chapter 7 (Probability) are solved with full working — list the sample space, count favourable outcomes, and confirm your result.

Key Concepts (Quick Recap)

  • Experiment: An operation that produces well-defined outcomes (e.g., tossing a coin, rolling a die)
  • Random experiment: An experiment whose outcome cannot be predicted in advance
  • Sample space (S): The set of all possible outcomes of an experiment
  • Event (E): A subset of the sample space — a collection of favourable outcomes
  • Classical (a priori) probability: P(E) = Number of favourable outcomes / Total number of outcomes = n(E) / n(S)
  • Range of probability: 0 ≤ P(E) ≤ 1 for any event E
  • Impossible event: P = 0 (no favourable outcomes, e.g., getting 7 on a die)
  • Sure event: P = 1 (all outcomes are favourable, e.g., getting a number ≤ 6 on a die)
  • Complementary events: P(E') = 1 − P(E), where E' is "not E"
  • Equally likely outcomes: Outcomes that have the same chance of occurring
  • Mutually exclusive events: Two events that cannot occur simultaneously (E ∩ F = empty set)
  • Playing cards: A standard deck has 52 cards — 13 each of Spades, Hearts, Diamonds, Clubs. Each suit has Ace, 2–10, Jack, Queen, King. Face cards = Jack, Queen, King (12 total). Red cards = Hearts + Diamonds = 26. Black cards = Spades + Clubs = 26.

Exercise 7.1 — Introduction to Probability

Q1. A die is rolled once. Find the probability of getting a number greater than 4.

SolutionWhen a die is rolled, the sample space is S = {1, 2, 3, 4, 5, 6} Number of outcomes n(S) = 6 Favourable outcomes (number greater than 4) = {5, 6} Number of favourable outcomes n(E) = 2

P(number greater than 4) = n(E) / n(S) = 2/6 = 1/3

Answer1/3

Q2. A coin is tossed once. Find the probability of getting a head.

SolutionWhen a coin is tossed, S = {H, T} n(S) = 2 Favourable outcome (getting head) = {H} n(E) = 1

P(getting a head) = 1/2

Answer1/2

Q3. A die is thrown once. Find the probability of getting an even number.

SolutionS = {1, 2, 3, 4, 5, 6} n(S) = 6 Favourable outcomes (even numbers) = {2, 4, 6} n(E) = 3

P(even number) = 3/6 = 1/2

Answer1/2

Q4. A die is thrown once. Find the probability of getting a number less than 7.

SolutionS = {1, 2, 3, 4, 5, 6} n(S) = 6 All outcomes on a die are less than 7. Favourable outcomes = {1, 2, 3, 4, 5, 6} n(E) = 6

P(number less than 7) = 6/6 = 1

This is a sure event.

Answer1 (sure event)

Q5. A die is rolled once. Find the probability of getting a number greater than 6.

SolutionS = {1, 2, 3, 4, 5, 6} n(S) = 6 No outcome on a die is greater than 6. Favourable outcomes = {} (empty set) n(E) = 0

P(number greater than 6) = 0/6 = 0

This is an impossible event.

Answer0 (impossible event)

Q6. A coin is tossed twice. Find the probability of getting two heads.

SolutionWhen a coin is tossed twice, the sample space is: S = {HH, HT, TH, TT} n(S) = 4 Favourable outcomes (two heads) = {HH} n(E) = 1

P(two heads) = 1/4

Answer1/4

Q7. A coin is tossed twice. Find the probability of getting at least one head.

SolutionS = {HH, HT, TH, TT} n(S) = 4 Favourable outcomes (at least one head) = {HH, HT, TH} n(E) = 3

P(at least one head) = 3/4

Answer3/4

Q8. Two coins are tossed simultaneously. Find the probability of getting no tail.

SolutionS = {HH, HT, TH, TT} n(S) = 4 No tail means both coins show heads. Favourable outcomes = {HH} n(E) = 1

P(no tail) = 1/4

Answer1/4

Q9. A die is rolled once. Find the probability of getting a prime number.

SolutionS = {1, 2, 3, 4, 5, 6} n(S) = 6 Prime numbers on a die = {2, 3, 5} n(E) = 3

P(prime number) = 3/6 = 1/2

Answer1/2

Q10. A die is thrown once. Find the probability of getting a number between 2 and 6.

SolutionS = {1, 2, 3, 4, 5, 6} n(S) = 6 Numbers between 2 and 6 = {3, 4, 5} n(E) = 3

P(number between 2 and 6) = 3/6 = 1/2

Answer1/2

Exercise 7.2 — Probability: Coins and Dice

Q1. Three coins are tossed simultaneously. Find the probability of getting exactly two heads.

SolutionWhen three coins are tossed, the sample space is: S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} n(S) = 8 Favourable outcomes (exactly two heads) = {HHT, HTH, THH} n(E) = 3

P(exactly two heads) = 3/8

Answer3/8

Q2. Three coins are tossed simultaneously. Find the probability of getting at most two heads.

SolutionS = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} n(S) = 8 At most two heads means 0, 1, or 2 heads. Favourable outcomes = {TTT, HTT, THT, TTH, HHT, HTH, THH} (All outcomes except HHH) n(E) = 7

P(at most two heads) = 7/8

Answer7/8

Q3. Three coins are tossed simultaneously. Find the probability of getting at least two heads.

SolutionS = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} n(S) = 8 At least two heads means 2 or 3 heads. Favourable outcomes = {HHT, HTH, THH, HHH} n(E) = 4

P(at least two heads) = 4/8 = 1/2

Answer1/2

Q4. Two dice are rolled simultaneously. Find the probability of getting a doublet (same number on both dice).

SolutionWhen two dice are rolled, each die has 6 outcomes. n(S) = 6 × 6 = 36

Doublets are outcomes where both dice show the same number: {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)} n(E) = 6

P(doublet) = 6/36 = 1/6

Answer1/6

Q5. Two dice are thrown simultaneously. Find the probability that the sum of numbers on the dice is 7.

Solutionn(S) = 36 Favourable outcomes (sum = 7): {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)} n(E) = 6

P(sum is 7) = 6/36 = 1/6

Answer1/6

Q6. Two dice are thrown simultaneously. Find the probability that the sum of numbers on the dice is greater than 10.

Solutionn(S) = 36 Favourable outcomes (sum > 10, i.e., sum = 11 or 12): Sum = 11: {(5,6), (6,5)} — 2 outcomes Sum = 12: {(6,6)} — 1 outcome n(E) = 3

P(sum greater than 10) = 3/36 = 1/12

Answer1/12

Q7. Two dice are rolled simultaneously. Find the probability that the product of numbers on the dice is 12.

Solutionn(S) = 36 Favourable outcomes (product = 12): {(2,6), (3,4), (4,3), (6,2)} n(E) = 4

P(product is 12) = 4/36 = 1/9

Answer1/9

Q8. A die is thrown once. What is the probability of getting a number which is a multiple of 3?

SolutionS = {1, 2, 3, 4, 5, 6} Multiples of 3 on a die = {3, 6} n(E) = 2 n(S) = 6

P(multiple of 3) = 2/6 = 1/3

Answer1/3

Q9. Two dice are rolled. Find the probability that the sum of the numbers on the dice is less than or equal to 5.

Solutionn(S) = 36 Favourable outcomes (sum ≤ 5):
  • Sum = 2: {(1,1)} — 1 outcome
  • Sum = 3: {(1,2), (2,1)} — 2 outcomes
  • Sum = 4: {(1,3), (2,2), (3,1)} — 3 outcomes
  • Sum = 5: {(1,4), (2,3), (3,2), (4,1)} — 4 outcomes

n(E) = 1 + 2 + 3 + 4 = 10

P(sum ≤ 5) = 10/36 = 5/18

Answer5/18

Q10. A pair of dice is thrown. Find the probability of getting a sum of at least 9.

Solutionn(S) = 36 Favourable outcomes (sum ≥ 9):
  • Sum = 9: {(3,6), (4,5), (5,4), (6,3)} — 4 outcomes
  • Sum = 10: {(4,6), (5,5), (6,4)} — 3 outcomes
  • Sum = 11: {(5,6), (6,5)} — 2 outcomes
  • Sum = 12: {(6,6)} — 1 outcome

n(E) = 4 + 3 + 2 + 1 = 10

P(sum ≥ 9) = 10/36 = 5/18

Answer5/18

Exercise 7.3 — Probability: Playing Cards and Other Experiments

Q1. A card is drawn at random from a well-shuffled deck of 52 playing cards. Find the probability of getting a king.

Solutionn(S) = 52 (total cards) Number of kings in a deck = 4 (King of Spades, King of Hearts, King of Diamonds, King of Clubs) n(E) = 4

P(getting a king) = 4/52 = 1/13

Answer1/13

Q2. A card is drawn from a well-shuffled deck of 52 cards. Find the probability of drawing a red card.

Solutionn(S) = 52 Number of red cards = 26 (13 Hearts + 13 Diamonds) n(E) = 26

P(red card) = 26/52 = 1/2

Answer1/2

Q3. A card is drawn from a well-shuffled pack of 52 cards. Find the probability of getting a face card.

Solutionn(S) = 52 Face cards are Jack, Queen, and King of each suit. Number of face cards = 3 per suit × 4 suits = 12 n(E) = 12

P(face card) = 12/52 = 3/13

Answer3/13

Q4. A card is drawn at random from a well-shuffled deck of 52 cards. Find the probability of getting the queen of diamonds.

Solutionn(S) = 52 There is exactly 1 queen of diamonds in the deck. n(E) = 1

P(queen of diamonds) = 1/52

Answer1/52

Q5. A card is drawn from a well-shuffled pack of 52 cards. Find the probability of getting a black king.

Solutionn(S) = 52 Black kings = King of Spades + King of Clubs = 2 n(E) = 2

P(black king) = 2/52 = 1/26

Answer1/26

Q6. A card is drawn from a pack of 52 cards. Find the probability of getting a spade.

Solutionn(S) = 52 Number of spades = 13 n(E) = 13

P(spade) = 13/52 = 1/4

Answer1/4

Q7. A card is drawn from a well-shuffled deck of 52 cards. Find the probability that it is neither a king nor a queen.

Solutionn(S) = 52 Number of kings = 4 Number of queens = 4 Cards that are either king or queen = 4 + 4 = 8 Cards that are neither king nor queen = 52 − 8 = 44 n(E) = 44

P(neither king nor queen) = 44/52 = 11/13

Answer11/13

Q8. A card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting a red face card.

Solutionn(S) = 52 Red face cards = Jack of Hearts, Queen of Hearts, King of Hearts, Jack of Diamonds, Queen of Diamonds, King of Diamonds = 6 n(E) = 6

P(red face card) = 6/52 = 3/26

Answer3/26

Q9. A card is drawn from a pack of 52 cards. Find the probability of getting a number card (2 through 10).

Solutionn(S) = 52 Number cards per suit = 9 (2, 3, 4, 5, 6, 7, 8, 9, 10) Total number cards = 9 × 4 = 36 n(E) = 36

P(number card) = 36/52 = 9/13

Answer9/13

Q10. A card is drawn from a well-shuffled pack of 52 cards. Find the probability of getting a card which is either red or a king.

Solutionn(S) = 52 Red cards = 26 Kings = 4 But red kings (King of Hearts, King of Diamonds) are counted in both, so we must subtract them once. Cards that are red or king = 26 + 4 − 2 = 28 n(E) = 28

P(red or king) = 28/52 = 7/13

Answer7/13

Exercise 7.4 — Probability: Balls in a Bag and Other Problems

Q1. A bag contains 4 red balls and 6 blue balls. A ball is drawn at random. Find the probability of drawing a red ball.

SolutionTotal number of balls = 4 + 6 = 10 n(S) = 10 Number of red balls = 4 n(E) = 4

P(red ball) = 4/10 = 2/5

Answer2/5

Q2. A bag contains 5 white balls, 3 black balls, and 2 red balls. A ball is drawn at random. Find the probability of getting a black ball.

SolutionTotal balls = 5 + 3 + 2 = 10 n(S) = 10 Number of black balls = 3 n(E) = 3

P(black ball) = 3/10

Answer3/10

Q3. A bag contains 5 white balls, 3 black balls, and 2 red balls. A ball is drawn at random. Find the probability that the ball is not white.

SolutionTotal balls = 10 n(S) = 10 White balls = 5 Not white balls = 10 − 5 = 5 n(E) = 5

P(not white) = 5/10 = 1/2

Answer1/2

Q4. A bag contains 3 red, 4 white, and 5 blue balls. A ball is drawn at random. Find the probability of getting a ball that is either red or blue.

SolutionTotal balls = 3 + 4 + 5 = 12 n(S) = 12 Red or blue balls = 3 + 5 = 8 n(E) = 8

P(red or blue) = 8/12 = 2/3

Answer2/3

Q5. A box contains 20 cards numbered 1 to 20. A card is drawn at random. Find the probability that the number on the card is a prime number.

Solutionn(S) = 20 Prime numbers from 1 to 20: {2, 3, 5, 7, 11, 13, 17, 19} (1 is not a prime number) n(E) = 8

P(prime number) = 8/20 = 2/5

Answer2/5

Q6. A box contains 20 cards numbered 1 to 20. A card is drawn at random. Find the probability that the number is divisible by 4.

Solutionn(S) = 20 Numbers divisible by 4 from 1 to 20: {4, 8, 12, 16, 20} n(E) = 5

P(divisible by 4) = 5/20 = 1/4

Answer1/4

Q7. A number is chosen at random from the numbers 1 to 50. Find the probability that the chosen number is a multiple of 5.

Solutionn(S) = 50 Multiples of 5 from 1 to 50: {5, 10, 15, 20, 25, 30, 35, 40, 45, 50} n(E) = 10

P(multiple of 5) = 10/50 = 1/5

Answer1/5

Q8. Numbers from 1 to 30 are written on cards and placed in a box. A card is drawn at random. Find the probability that the number is a perfect square.

Solutionn(S) = 30 Perfect squares from 1 to 30: {1, 4, 9, 16, 25} n(E) = 5

P(perfect square) = 5/30 = 1/6

Answer1/6

Q9. A box contains 12 balls of which x are red. If 6 more red balls are added, the probability of drawing a red ball becomes 2/3. Find x.

SolutionInitially: Total balls = 12, Red balls = x After adding 6 red balls: Total balls = 18, Red balls = x + 6

P(drawing red ball after addition) = (x + 6) / 18 = 2/3

Solving: 3(x + 6) = 2 × 18 3x + 18 = 36 3x = 18 x = 6

Verification: After adding 6 red balls, total red = 12, total = 18 P(red) = 12/18 = 2/3 ✓

Answerx = 6

Q10. A bag contains 4 red balls and some blue balls. If the probability of drawing a blue ball is thrice the probability of drawing a red ball, find the number of blue balls.

SolutionLet the number of blue balls = y Total balls = 4 + y

P(red ball) = 4 / (4 + y) P(blue ball) = y / (4 + y)

GivenP(blue) = 3 × P(red) y / (4 + y) = 3 × 4 / (4 + y) y / (4 + y) = 12 / (4 + y) y = 12
AnswerNumber of blue balls = 12

Exercise 7.5 — Complementary Events and Combined Probability

Q1. If P(E) = 3/7, find P(not E).

SolutionP(not E) = 1 − P(E) = 1 − 3/7 = 4/7
Answer4/7

Q2. If the probability of winning a game is 0.3, what is the probability of losing the game?

SolutionP(losing) = 1 − P(winning) = 1 − 0.3 = 0.7
Answer0.7

Q3. A die is thrown once. Find the probability of getting a number less than or equal to 5.

SolutionS = {1, 2, 3, 4, 5, 6} n(S) = 6 Favourable outcomes (number ≤ 5) = {1, 2, 3, 4, 5} n(E) = 5

P(number ≤ 5) = 5/6

Answer5/6

Q4. A die is thrown once. The probability of getting a number greater than 3 is p. Find the value of p.

SolutionS = {1, 2, 3, 4, 5, 6} n(S) = 6 Numbers greater than 3 = {4, 5, 6} n(E) = 3

p = P(number > 3) = 3/6 = 1/2

Answerp = 1/2

Q5. The probability of an event A is 0.6. Find the probability of the event "not A".

SolutionP(not A) = 1 − P(A) = 1 − 0.6 = 0.4
Answer0.4

Q6. In a class, 40% of students play cricket. What is the probability that a randomly chosen student does not play cricket?

SolutionP(plays cricket) = 40% = 40/100 = 2/5 = 0.4 P(does not play cricket) = 1 − 2/5 = 3/5
Answer3/5

Q7. A coin is tossed 1000 times and head appears 455 times. Find the empirical probability of getting a head.

SolutionNumber of trials = 1000 Number of heads = 455 Empirical probability of head = 455/1000 = 91/200 = 0.455
Answer91/200 (or 0.455)

Q8. A coin is tossed 500 times. Head appears 240 times and tail appears 260 times. Find the probability of getting a head and the probability of getting a tail.

SolutionTotal trials = 500 Number of heads = 240 Number of tails = 260

P(head) = 240/500 = 12/25 = 0.48 P(tail) = 260/500 = 13/25 = 0.52

Check: P(head) + P(tail) = 0.48 + 0.52 = 1 ✓

AnswerP(head) = 12/25, P(tail) = 13/25

Q9. In a survey of 200 students, 80 like Mathematics, 60 like Science, and the rest like other subjects. If a student is chosen at random, find the probability that the student likes Mathematics.

SolutionTotal students = 200 Students who like Mathematics = 80

P(likes Mathematics) = 80/200 = 2/5

Answer2/5

Q10. A die is thrown 300 times and the outcomes 1, 2, 3, 4, 5, 6 appear with frequencies 55, 48, 52, 53, 47, 45 respectively. Find the empirical probability of getting the outcome 3.

SolutionTotal trials = 300 Frequency of outcome 3 = 52

Empirical probability of getting 3 = 52/300 = 13/75

Answer13/75

Exercise 7.6 — Combined and Word Problems

Q1. Two coins are tossed simultaneously. Find the probability of getting:

(i) exactly one head (ii) at least one tail (iii) no head

SolutionS = {HH, HT, TH, TT} n(S) = 4

(i) Exactly one head: {HT, TH} P(exactly one head) = 2/4 = 1/2

(ii) At least one tail: {HT, TH, TT} (all except HH) P(at least one tail) = 3/4

(iii) No head: {TT} P(no head) = 1/4

Answer(i) 1/2, (ii) 3/4, (iii) 1/4

Q2. A bag contains 3 red, 4 green, and 5 blue balls. A ball is drawn at random. Find the probability that the ball is:

(i) red (ii) green (iii) not blue

SolutionTotal balls = 3 + 4 + 5 = 12

(i) P(red) = 3/12 = 1/4 (ii) P(green) = 4/12 = 1/3 (iii) P(not blue) = (3 + 4)/12 = 7/12

Answer(i) 1/4, (ii) 1/3, (iii) 7/12

Q3. A card is drawn from a well-shuffled deck of 52 cards. Find the probability that the card is:

(i) a black queen (ii) a red ace (iii) a diamond

Solutionn(S) = 52

(i) Black queens: Queen of Spades and Queen of Clubs = 2 P(black queen) = 2/52 = 1/26

(ii) Red aces: Ace of Hearts and Ace of Diamonds = 2 P(red ace) = 2/52 = 1/26

(iii) Diamonds: 13 cards P(diamond) = 13/52 = 1/4

Answer(i) 1/26, (ii) 1/26, (iii) 1/4

Q4. A die is thrown once. Find the probability of getting:

(i) a number greater than 2 (ii) a number less than or equal to 3 (iii) an odd number

SolutionS = {1, 2, 3, 4, 5, 6}, n(S) = 6

(i) Numbers greater than 2: {3, 4, 5, 6} P(number > 2) = 4/6 = 2/3

(ii) Numbers ≤ 3: {1, 2, 3} P(number ≤ 3) = 3/6 = 1/2

(iii) Odd numbers: {1, 3, 5} P(odd number) = 3/6 = 1/2

Answer(i) 2/3, (ii) 1/2, (iii) 1/2

Q5. In a single throw of two dice, find the probability of getting:

(i) a sum of 8 (ii) a doublet (iii) a sum greater than 9

Solutionn(S) = 36

(i) Sum = 8: {(2,6), (3,5), (4,4), (5,3), (6,2)} — 5 outcomes P(sum of 8) = 5/36

(ii) Doublets: {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)} — 6 outcomes P(doublet) = 6/36 = 1/6

(iii) Sum > 9 (i.e., sum = 10, 11, or 12): Sum = 10: {(4,6), (5,5), (6,4)} — 3 outcomes Sum = 11: {(5,6), (6,5)} — 2 outcomes Sum = 12: {(6,6)} — 1 outcome n(E) = 3 + 2 + 1 = 6 P(sum > 9) = 6/36 = 1/6

Answer(i) 5/36, (ii) 1/6, (iii) 1/6

Q6. A jar contains 24 marbles — some are green and the rest are blue. If a marble is drawn at random, the probability of drawing a green marble is 2/3. Find the number of blue marbles.

SolutionTotal marbles = 24 P(green) = 2/3 Number of green marbles = (2/3) × 24 = 16 Number of blue marbles = 24 − 16 = 8
AnswerNumber of blue marbles = 8

Q7. A number is selected at random from the first 50 natural numbers. Find the probability that the number is:

(i) an even number (ii) a perfect square (iii) a multiple of 7

Solutionn(S) = 50

(i) Even numbers from 1 to 50: 2, 4, 6, ..., 50 — that's 25 numbers P(even) = 25/50 = 1/2

(ii) Perfect squares from 1 to 50: 1, 4, 9, 16, 25, 36, 49 — that's 7 numbers P(perfect square) = 7/50

(iii) Multiples of 7 from 1 to 50: 7, 14, 21, 28, 35, 42, 49 — that's 7 numbers P(multiple of 7) = 7/50

Answer(i) 1/2, (ii) 7/50, (iii) 7/50

Q8. Two dice are rolled simultaneously. Find the probability that the difference of the numbers on the two dice is 2.

Solutionn(S) = 36

(1,3), (2,4), (3,5), (4,6), (3,1), (4,2), (5,3), (6,4) — 8 outcomes n(E) = 8

Favourable outcomes wheredie1 − die2= 2:

P(difference is 2) = 8/36 = 2/9

Answer2/9

Q9. A coin is tossed 3 times. Find the probability of getting exactly one tail.

SolutionSample space for 3 coin tosses: S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} n(S) = 8

Exactly one tail: {HHT, HTH, THH} n(E) = 3

P(exactly one tail) = 3/8

Answer3/8

Q10. If P(E) = 0.05, what is the probability of "not E"?

SolutionP(not E) = 1 − P(E) = 1 − 0.05 = 0.95
Answer0.95

Q11. A ticket is drawn at random from a bag containing tickets numbered 1 to 40. Find the probability of getting a ticket with a number that is a multiple of 5 or 7.

Solutionn(S) = 40

Multiples of 5 from 1 to 40: 5, 10, 15, 20, 25, 30, 35, 40 — 8 numbers Multiples of 7 from 1 to 40: 7, 14, 21, 28, 35 — 5 numbers Common multiples (of both 5 and 7, i.e., multiples of 35): 35 — 1 number

By inclusion-exclusion: n(multiple of 5 or 7) = 8 + 5 − 1 = 12

P(multiple of 5 or 7) = 12/40 = 3/10

Answer3/10

Q12. A card is drawn from a well-shuffled deck of 52 cards. Find the probability that the card drawn is:

(i) a black card (ii) a face card (iii) a black face card (iv) neither red nor a queen

Solutionn(S) = 52

(i) Black cards: 26 P(black card) = 26/52 = 1/2

(ii) Face cards: 12 (J, Q, K of each suit) P(face card) = 12/52 = 3/13

(iii) Black face cards: King, Queen, Jack of Spades and Clubs = 6 P(black face card) = 6/52 = 3/26

(iv) Neither red nor queen: Red cards = 26, Queens = 4 (but 2 red queens are already in red cards) Cards that are red or queen = 26 + 2 = 28 (26 red cards + 2 black queens) Cards that are neither red nor queen = 52 − 28 = 24 P(neither red nor queen) = 24/52 = 6/13

Answer(i) 1/2, (ii) 3/13, (iii) 3/26, (iv) 6/13

Common Mistakes

1. Mistake: Writing probability as greater than 1 or less than 0 — Probability always lies between 0 and 1 (inclusive). If your answer is greater than 1, you have likely swapped the numerator and denominator. Remember: P(E) = favourable outcomes / total outcomes.

2. Mistake: Not listing the complete sample space — For two dice, n(S) = 36 (not 12 or 11). For two coins, n(S) = 4 (not 3). For three coins, n(S) = 8. Always list or count the sample space first before finding favourable outcomes.

3. Mistake: Confusing "at least" with "exactly" — "At least 2 heads" means 2 or more (i.e., 2, 3, 4, ...), while "exactly 2 heads" means precisely 2. "At most 2 heads" means 2 or fewer (i.e., 0, 1, or 2). Read the question carefully.

4. Mistake: Counting (2,5) and (5,2) as the same outcome — When rolling two dice, (2,5) and (5,2) are different outcomes because they represent different results on each die. The first number is die 1 and the second is die 2.

5. Mistake: Forgetting that 1 is not a prime number — 1 is neither prime nor composite. The primes on a die are {2, 3, 5}. When counting primes from 1 to 20, the list is {2, 3, 5, 7, 11, 13, 17, 19} — that is 8 primes.

6. Mistake: Double-counting in card problems — When asked for "red or king," remember that the red kings (King of Hearts, King of Diamonds) appear in both categories. Use inclusion-exclusion: n(red or king) = n(red) + n(king) − n(red king) = 26 + 4 − 2 = 28.

7. Mistake: Not using complementary probability when easier — If the question asks for "not getting 6 on a die," it is easier to calculate P(6) = 1/6 and then P(not 6) = 1 − 1/6 = 5/6 rather than counting {1, 2, 3, 4, 5} individually. Use this approach for "at least one" or "not" type questions.

Exam Checklist

  • [ ] I can define sample space and list all outcomes for coins (1, 2, 3 coins) and dice (1 or 2)
  • [ ] I can apply P(E) = n(E)/n(S) to find probability of simple events
  • [ ] I can find probability for dice problems (single die and two dice)
  • [ ] I can find probability for coin toss problems (single, two, and three coins)
  • [ ] I can find probability for card problems (knowing the 52-card deck structure)
  • [ ] I can find probability for ball-in-bag problems
  • [ ] I can distinguish between "at least," "at most," and "exactly"
  • [ ] I can use complementary events: P(not E) = 1 − P(E)
  • [ ] I can verify that all probabilities in a sample space sum to 1
  • [ ] I can solve problems involving numbers (1–20, 1–50, etc.) for prime, square, divisible types
  • [ ] I can use inclusion-exclusion for "or" probability problems
  • [ ] I know that probability of an impossible event = 0 and a sure event = 1

Internal Links

  • [Class 9 Maths — Probability Chapter Guide](/school/tamil-nadu-state-board/class/9/maths/probability)
  • [All Class 9 Maths Solutions](/school/tamil-nadu-state-board/class/9/maths/solutions)
  • [Class 9 Maths Subject Guide](/school/tamil-nadu-state-board/class/9/maths)

Frequently Asked Questions

If an experiment has n equally likely outcomes and m of them are favourable to event E, then P(E) = m/n. The probability of any event always lies between 0 and 1 (inclusive).
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Chapter Notes

Tamil Nadu Class 9 Maths Probability — Samacheer Kalvi

Exercise Solutions

Samacheer Kalvi 9th Maths Algebra Exercise Solutions | KalviGavani

Exercise Solutions

Samacheer Kalvi 9th Maths Coordinate Geometry Exercise Solutions | KalviGavani