Exercise 13.2 — Probability Problems

Problems based on probability.

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Chapter 13 · Probability

Exercise 13.2 — Classical Probability

Class 10 Mathematics · CBSE · Telangana · Andhra Pradesh Board
🎲 Balls & Marbles 🃏 Card Problems 🎯 Spinner & Dice 🪙 Coin Toss 🔵 Discs & Pens 15 Questions Fully Solved

What is Probability? — Core Concepts for Exercise 13.2

Probability is the branch of mathematics that measures the likelihood of an event occurring. It is expressed as a number between 0 and 1, where 0 means the event is impossible and 1 means the event is certain. Exercise 13.2 of Class 10 Mathematics (CBSE, Telangana, and Andhra Pradesh syllabus) covers classical (theoretical) probability — calculating probabilities by counting equally likely outcomes without actually performing the experiment.

🎯 The Classical Probability Formula
P(Event A) = n(A) / n(S)
n(A) = number of favourable outcomes  |  n(S) = total number of outcomes in Sample Space

Three essential terms you must know before solving any problem in this exercise:

  • Sample Space (S) — the set of all possible outcomes of an experiment. For example, rolling one die gives S = {1, 2, 3, 4, 5, 6}, so n(S) = 6.
  • Event (A) — any specific outcome or group of outcomes you are interested in. The number of outcomes in that event is written as n(A).
  • Complementary Event (Ā) — the event that A does NOT happen. P(Ā) = 1 − P(A). If the probability of picking a red ball is 3/8, the probability of NOT picking a red ball is 1 − 3/8 = 5/8.
Key identities:   0 ≤ P(A) ≤ 1  |  P(A) + P(Ā) = 1  |  P(certain event) = 1  |  P(impossible event) = 0
💡 Quick number sense for exams: When tossing n coins → 2ⁿ total outcomes. When rolling n dice → 6ⁿ total outcomes. Two dice rolled = 6² = 36 outcomes. Three coins tossed = 2³ = 8 outcomes.

Question 1 — Red and Black Balls in a Bag

Q1
A bag has 3 red balls and 5 black balls. Find the probability of drawing: (i) a red ball (ii) not a red ball
Total balls in bag = 3 (Red) + 5 (Black) = 8  →  n(S) = 8
(i) Red ball Let A = event of red ball n(A) = 3 P(A) = n(A)/n(S) = 3/8 P(red) = 3/8
(ii) Not red ball (Ā) P(Ā) = 1 − P(A) = 1 − 3/8 = (8−3)/8 P(not red) = 5/8
Key insight: The "not red" event is the complementary event of "red." Since there are 5 black balls, we can also verify directly: 5/8. Both methods give the same answer.

Question 2 — Red, White and Green Marbles

Q2
A box has 5 red, 8 white, and 4 green marbles. One marble is picked at random. Find the probability it is: (i) red (ii) white (iii) not green
Total marbles = 5 + 8 + 4 = 17  →  n(S) = 17
(i) Red n(A) = 5 P(A) = 5/17 = 5/17
(ii) White n(B) = 8 P(B) = 8/17 = 8/17
(iii) Not green n(C) = 5+8 = 13 P(C) = 13/17 = 13/17
📌 "Not green" = red + white = 5 + 8 = 13. Alternatively, P(not green) = 1 − P(green) = 1 − 4/17 = 13/17. Both approaches give the same answer — use whichever is faster!

Question 3 — Coins Falling from a Kiddy Bank

Q3
A kiddy bank has 100 fifty-paise coins, 50 one-rupee coins, 20 two-rupee coins, and 10 five-rupee coins. Find P(50p coin) and P(not Rs. 5 coin).
Total coins = 100 + 50 + 20 + 10 = 180  →  n(S) = 180
Coin TypeCountFractionSimplified
50 Paise (50p)100100/1805/9
Rs. 15050/1805/18
Rs. 22020/1801/9
Rs. 51010/1801/18
Total180180/1801
(i) P(50p coin) n(A) = 100 P(A) = 100/180 = 5/9
(ii) P(not Rs.5 coin) n(B) = 180−10 = 170 P(B) = 170/180 = 17/18

Question 4 — Fish from an Aquarium Tank

Q4
A tank has 5 male fish and 8 female fish. One fish is taken at random. What is the probability it is a male fish?
Total fish = 5 (male) + 8 (female) = 13  →  n(S) = 13 Let A = event that the fish is male  →  n(A) = 5 P(A) = n(A)/n(S) = 5/13 P(male fish) = 5/13
Answer: 5/13 ≈ 0.385 (about 38.5% chance)

Question 5 — The Spinning Arrow (Numbers 1 to 8)

A spinning arrow can point to any of the numbers 1, 2, 3, 4, 5, 6, 7, 8 with equal probability. The sample space has n(S) = 8 equally likely outcomes.

1 2 3 4 5 6 7 8
n(S) = 8 (numbers 1 through 8) (i) Arrow points at 8 n(A) = 1  →  P = 1/8 (ii) Odd number {1,3,5,7} n(B) = 4  →  P = 4/8 = 1/2 (iii) Number > 2 {3,4,5,6,7,8} n(C) = 6  →  P = 6/8 = 3/4 (iv) Number < 9 {1,2,...,8} n(D) = 8  →  P = 8/8 = 1 (certain!)
📌 Part (iv) is a certain event. Since all 8 numbers on the wheel are less than 9, the arrow is guaranteed to point at a number less than 9. Probability = 1. This is a great example of a sure event.
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Question 6 — A Deck of 52 Playing Cards

This is one of the most important and most-examined topics in Probability for Class 10 board exams. You must know the structure of a standard deck of 52 cards perfectly. Here is a complete breakdown:

Hearts (Red)
13 cards:
A,2,3…10,J,Q,K
Face cards: J,Q,K (3)
Diamonds (Red)
13 cards:
A,2,3…10,J,Q,K
Face cards: J,Q,K (3)
Spades (Black)
13 cards:
A,2,3…10,J,Q,K
Face cards: J,Q,K (3)
Clubs (Black)
13 cards:
A,2,3…10,J,Q,K
Face cards: J,Q,K (3)
CategoryWhich Cards?CountProbability
Total cardsAll 5252
Face cardsJack, Queen, King (×4 suits)1212/52 = 3/13
Red cardsAll Hearts + All Diamonds2626/52 = 1/2
Black cardsAll Spades + All Clubs2626/52 = 1/2
KingsK♥, K♦, K♠, K♣44/52 = 1/13
Red KingsK♥, K♦22/52 = 1/26
Red face cardsJ♥,Q♥,K♥,J♦,Q♦,K♦66/52 = 3/26
Jack of HeartsJ♥ only11/52
SpadesAll 13 spades1313/52 = 1/4
Queen of DiamondsQ♦ only11/52
Q6 — All Parts
One card selected from 52-card deck. n(S) = 52
(i) King of red colour A = {K♥, K♦} n(A) = 2 P = 2/52 = 1/26
(ii) Face card J, Q, K in 4 suits n(B) = 12 P = 12/52 = 3/13
(iii) Red face card ♥ and ♦ face cards n(C) = 6 P = 6/52 = 3/26
(iv) Jack of Hearts Only one J♥ n(D) = 1 P = 1/52
(v) A Spade 13 spades in deck n(E) = 13 P = 13/52 = 1/4
(vi) Queen of Diamonds Only one Q♦ n(F) = 1 P = 1/52

Question 7 — Five Diamond Cards (Without Replacement)

This question introduces an important concept: probability without replacement. When a card is removed and NOT put back, the sample space for the next draw changes.

Q7
Five cards — 10, J, Q, K, A of diamonds — face down. (i) P(Queen) (ii) After Queen is removed without replacement: P(Ace) and P(Queen)
(i) P(Queen) — 5 cards n(S) = 5 n(A) = 1 (one Queen) P(A) = 1/5 P(Queen) = 1/5
(ii) After Queen removed Remaining cards = 4 n(S) = 4 (a) P(Ace): n = 1 → 1/4 (b) P(Queen): n = 0 → 0 (impossible!)
Part (ii)(b) is an impossible event. The Queen has already been removed, so P(Queen) = 0/4 = 0. This is a great example of a null/impossible event.

Questions 8 & 9 — Defective Pens and Bulbs

Q8
12 defective pens mixed with 132 good ones (total 144). P(pen is good)?
n(S) = 144 total pens n(A) = 132 (good pens) P(good pen) = 132/144 = 11/12
Answer: P(good pen) = 11/12 ≈ 0.917 (about 91.7% chance)
Q9
20 bulbs with 4 defective. (i) P(defective) (ii) A good bulb is removed WITHOUT replacement. P(next is not defective)?
(i) First draw n(S) = 20 n(A) = 4 (defective) P(defective) = 4/20 = 1/5
(ii) Second draw (no replacement) Good bulb removed → 19 left Good bulbs remaining = 16−1 = 15 n(B) = 15, n(S) = 19 P(not defective) = 15/19
📌 Without replacement — why n(S) changes: Originally 20 bulbs (16 good + 4 defective). After removing one good bulb (not replacing it), we have 19 bulbs left: 15 good + 4 defective. Always update both n(S) and n(A) in such problems.

Question 10 — Numbered Discs from 1 to 90

Q10
90 discs numbered 1–90. Find P(i) two-digit number (ii) perfect square (iii) divisible by 5
n(S) = 90. This question requires careful counting of the favourable outcomes.
(i) Two-digit numbers From 10 to 90 Count = 90−9 = 81 n(A) = 81 P = 81/90 = 9/10
(ii) Perfect squares {1,4,9,16,25, 36,49,64,81} n(B) = 9 P = 9/90 = 1/10
(iii) Divisible by 5 {5,10,15,...,90} Count = 90÷5 = 18 n(C) = 18 P = 18/90 = 1/5
💡 Counting tip for perfect squares up to 90: Find √90 ≈ 9.49, so there are 9 perfect squares (1² through 9²). This trick works for any range — find the floor of the square root of the maximum number.

Question 11 — Geometric Probability (Rectangle and Circle)

This question introduces geometric probability — where the probability is the ratio of areas rather than the count of outcomes. This is a unique type that frequently appears in board exams.

Q11
A die is dropped randomly on a 3m × 2m rectangle. A circle of diameter 1m is drawn inside it. P(die lands inside the circle)?
Area of rectangle = 3 × 2 = 6 m² n(S) = 6 m² Area of circle diameter = 1m → radius = 0.5m = π × r² = (22/7) × (1/2)² = 22/7 × 1/4 = 22/28 m²
P(inside circle) = Area of circle / Area of rectangle = (22/28) / 6 = 22 / (28 × 6) = 22 / 168 = 11/84 P = 11/84
📌 Geometric probability formula: P(event) = (Favourable Area) / (Total Area). This concept extends naturally to problems involving sectors of circles, triangles inside rectangles, etc.

Question 12 — Sudha Buying a Pen

Q12
144 ball pens (20 defective, 124 good). Shopkeeper gives one to Sudha. P(she buys it) and P(she does not buy it)?
Sudha will only buy a good pen. Total pens = 144. Good pens = 144 − 20 = 124.
(i) P(Sudha buys — good pen) n(A) = 124 P(A) = 124/144 = 31/36
(ii) P(Sudha does not buy — defective) n(B) = 20 P(B) = 20/144 = 5/36
Verification: P(buy) + P(not buy) = 31/36 + 5/36 = 36/36 = 1 ✓. The two probabilities always add up to 1 — a reliable self-check!

Question 13 — Rolling Two Dice: Sum Probability Table

When two dice are rolled simultaneously, the total number of outcomes is 6 × 6 = 36. The possible sums range from 2 (both dice show 1) to 12 (both dice show 6). The number of ways to achieve each sum follows a symmetric pattern — increasing from sum 2 to sum 7, then decreasing back.

Sum 2
(1,1) — 1 way
1/36
Sum 3
(1,2),(2,1) — 2 ways
2/36
Sum 4
(1,3),(2,2),(3,1) — 3 ways
3/36
Sum 5
(1,4),(2,3),(3,2),(4,1) — 4 ways
4/36
Sum 6
5 ways
5/36
Sum 7
6 ways — PEAK!
6/36 = 1/6
Sum 8
5 ways
5/36
Sum 9
4 ways
4/36
Sum 10
3 ways
3/36
Sum 11
2 ways
2/36
Sum 12
(6,6) — 1 way
1/36
Sum23456789101112
Ways 12345654321
Probability 1/362/363/364/365/366/365/364/363/362/361/36
Q13 (ii) — Critical Thinking
A student says: "There are 11 possible sums (2 to 12), so each has probability 1/11." Is this correct?
❌ The student's argument is WRONG. The sums 2, 3, 4, ... 12 are NOT outcomes — they are derived values. The actual outcomes are ordered pairs like (1,1), (1,2), ..., (6,6). Total outcomes = 36 (not 11), and each ordered pair is equally likely. Different sums arise from different numbers of pairs, so they do NOT all have equal probability. Sum 7 has probability 6/36 = 1/6, but sum 2 has only 1/36.
This is a classic board exam concept check! The 11 possible sums are NOT equally likely outcomes. Probability requires that the basic outcomes in the sample space be equally likely — and those are the 36 ordered pairs, not the 11 sums.

Question 14 — Three Coin Tosses: Deekshitha's Game

When 3 coins are tossed, the total outcomes = 2³ = 8. Deekshitha wins only if all three tosses match (HHH or TTT). Otherwise she loses.

HHH ✓ WIN
HHT ✗ LOSE
HTH ✗ LOSE
THH ✗ LOSE
TTT ✓ WIN
TTH ✗ LOSE
THT ✗ LOSE
HTT ✗ LOSE
Q14
Three coins tossed. Deekshitha wins if all results are same (HHH or TTT). Find P(she loses).
n(S) = 2³ = 8 Winning outcomes (all same): {HHH, TTT} → n(win) = 2 Losing outcomes: {HHT, HTH, THH, TTH, THT, HTT} → n(E) = 6 P(loses) = n(E)/n(S) = 6/8 = 3/4 Verify: P(wins) = 2/8 = 1/4  |  P(wins) + P(loses) = 1/4 + 3/4 = 1 ✓
P(Deekshitha loses) = 3/4 (she loses 75% of the time!)

Question 15 — A Die Thrown Twice: Getting (or Not Getting) 5

Throwing a die twice is the same experiment as rolling two dice simultaneously — the sample space has n(S) = 36 equally likely outcomes.

Q15
(i) P(5 does not appear either time) (ii) P(5 appears at least once)
(i) 5 does NOT appear either time Each die can show {1,2,3,4,6} — 5 choices Outcomes without 5 = 5×5 = 25 n(E) = 25, n(S) = 36 P = 25/36
(ii) 5 appears AT LEAST ONCE P(at least one 5) = 1 − P(no 5) = 1 − 25/36 = 11/36 P = 11/36
Direct count for (ii) — outcomes with at least one 5: {(1,5),(2,5),(3,5),(4,5),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,5)} n(A) = 11 → P = 11/36 ✓ (same answer!)
💡 "At least once" shortcut: P(at least once) = 1 − P(never). This complementary approach is almost always faster than listing all favourable outcomes directly. Use it in exams!

Quick Reference — All Answers at a Glance

Q#SituationAnswer(s)
1(i)Red ball from 3R+5B bag3/8
1(ii)Not red ball5/8
2(i)Red marble (5R+8W+4G)5/17
2(ii)White marble8/17
2(iii)Not green marble13/17
3(i)50p coin from kiddy bank5/9
3(ii)Not Rs. 5 coin17/18
4Male fish (5M+8F)5/13
5(i-iv)Spinner 1–81/8 · 1/2 · 3/4 · 1
6(i-vi)52-card deck1/26 · 3/13 · 3/26 · 1/52 · 1/4 · 1/52
7(i)Queen from 5 diamond cards1/5
7(ii-a)Ace after Queen removed1/4
7(ii-b)Queen after Queen removed0 (impossible)
8Good pen (144 total, 12 defective)11/12
9(i)Defective bulb (20 total, 4 defective)1/5
9(ii)Good bulb after good removed (19 left)15/19
10(i)Two-digit number on disc (1–90)9/10
10(ii)Perfect square disc1/10
10(iii)Divisible by 5 disc1/5
11Die landing in circle (rectangle 3×2)11/84
12(i)Sudha buys pen (144 total, 20 defective)31/36
12(ii)Sudha does not buy pen5/36
13Two-dice sum table (complete)See table above
14Deekshitha loses 3-coin game3/4
15(i)5 never appears in two throws25/36
15(ii)5 appears at least once11/36

Common Mistakes to Avoid in Board Exams

  • Forgetting the "without replacement" update: In Q7 and Q9, once a card or bulb is removed without being put back, both n(S) and n(A) must be recalculated. Students who forget this score zero on the second part.
  • Wrong count of face cards: There are 12 face cards (3 per suit × 4 suits = J, Q, K of each suit), NOT 16. Ace is NOT a face card in standard probability convention.
  • Confusing "at least once" with "exactly once": "At least once" = 1 or more times. Always use the complementary approach: P(at least once) = 1 − P(never). This avoids long enumeration.
  • Assuming 11 dice-sum outcomes are equally likely (Q13 trap): They are NOT. The 36 ordered pairs are the equally likely outcomes. Sums are derived values with different frequencies.
  • Not simplifying fractions: Board examiners expect fractions in their lowest terms. 100/180 must be simplified to 5/9. 132/144 must be simplified to 11/12.
  • Using diameter instead of radius for circle area: In Q11, diameter = 1m, so radius = 0.5m. Area = π(0.5)² = π/4. Using diameter directly in the formula will give a wrong answer.
High-risk exam trap — Q13: The argument that "11 possible sums → probability 1/11 each" is a very common distractor in board papers. Always remember: probability is calculated from the sample space of equally likely outcomes, which for two dice is the 36 ordered pairs — not the 11 derived sums.

What This Exercise Prepares You For

Exercise 13.2 covers the bulk of the marks-earning problems in Chapter 13. The concepts here — complementary events, without-replacement problems, deck of cards, and two-dice outcomes — appear in nearly every Telangana, Andhra Pradesh, and CBSE Class 10 board paper in the 2–4 mark question section.

Mastering this exercise also builds intuition for the Exercise 13.1 (Introduction to Probability) concepts, and creates a strong foundation for advanced probability topics encountered in Class 11 and 12 (such as conditional probability and Bayes' theorem). Students preparing for competitive exams like JEE Foundation will find the two-dice and cards problems particularly relevant.

📐 Telangana & AP Board Exam Tip: In recent years, Q6 (cards), Q13 (two dice table + critical thinking), Q14 (coin toss game), and Q15 (at least one 5) have been the most frequently examined problems from this exercise. For CBSE boards, Q7 (without replacement), Q10 (discs numbered 1–90), and Q11 (geometric probability) are perennial favourites. Practise all of them until the method is second nature.
🎲 Classical Probability ✅ Complementary Events 🃏 Cards & Suits 🎯 Dice & Spinners 🔄 Without Replacement 📐 Geometric Probability
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