Exercise 14.1 — Real Life Applications
Applications of probability in real life.
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Formula to Remember
Every question in Exercise 14.1 uses one of two probability formulas. Keep these in front of you while solving:
Theoretical Probability:
P(E) = n(E) / n(S)
= Favourable outcomes / Total outcomes
Experimental (Empirical) Probability:
P(E) = Frequency of event / Total number of trials
💡 Always remember: 0 ≤ P(E) ≤ 1 | P(impossible) = 0 | P(certain) = 1 | Sum of all elementary probabilities = 1
Question 1 — Rolling a Die
Question 1
A die has six faces numbered 1 to 6. It is rolled. (a) Possible outcomes? (b) Are they equally likely? (c) Find the probability of a composite number on top.
(a) Possible outcomes: The die can land on any of its six faces.
S = {1, 2, 3, 4, 5, 6} → n(S) = 6
(b) Are they equally likely? Yes. An unbiased die has no reason to favour one face over another — each face has an equal 1/6 chance of turning up.
(c) Probability of a composite number:
Composite numbers from 1–6: 4 and 6 (note: 1 is neither prime nor composite)
Favourable event E = {4, 6} → n(E) = 2
P(composite) = n(E) / n(S) = 2/6
∴ P(composite number) = 1/3
✅ Answer: P(composite number) = 1/3
📐 Why is 1 not composite? A composite number has more than 2 factors. The number 1 has only one factor (itself), so it is neither prime nor composite. From 1 to 6: Primes = {2, 3, 5}, Composite = {4, 6}, Neither = {1}.
Question 2 — Experimental Probability with Coin
Question 2
A coin is tossed 100 times: Head = 45 times, Tail = 55 times. (a) Compute probability of each outcome. (b) Find the sum of all probabilities.
This is an experimental probability question — you use the actual recorded frequencies, not the theoretical formula with equal outcomes.
Total trials n = 100
Heads occurred f(H) = 45 times
Tails occurred f(T) = 55 times
P(Head) = 45/100 = 9/20
P(Tail) = 55/100 = 11/20
Sum of all probabilities = 9/20 + 11/20 = 20/20
= 1 ✓
✅ Answers: P(Head) = 9/20 | P(Tail) = 11/20 | Sum = 1
📌 Note: Theoretically P(Head) = 1/2 = 0.5. Here we got 0.45 — slightly less than expected. This is normal for a finite experiment. As trials increase towards infinity, the experimental probability approaches 1/2.
Question 3 — Coloured Spinner
Question 3
A spinner has four colours — Red, Blue, Green and Yellow — in different proportions. Answer five questions about likelihood.
From the spinner diagram, the sectors have different sizes: Red occupies the most area, Yellow the least, while Blue and Green are approximately equal in size.
| Part | Question | Answer |
|---|---|---|
| (a) | Most likely colour | 🔴 Red (largest sector) |
| (b) | Least likely colour | 🟡 Yellow (smallest sector) |
| (c) | Equally likely colours | 🔵 Blue & 🟢 Green (equal sectors) |
| (d) | Chance of stopping on White | 0 — No White sector exists |
| (e) | Colour where pointer certainly stops | None — no colour covers 100% |
✅ Key insight: A colour with a larger sector on a spinner is more likely because the pointer spends proportionally more angular range on it. White has probability 0 (impossible). No single colour has probability 1 (certain).
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Question 4 — Drawing Marbles from a Bag
Question 4
A bag has 5 green, 3 blue, 2 red, 2 yellow marbles. One marble is drawn. (a) Are the four colour outcomes equally likely? (b) Find P(green), P(blue), P(red), P(yellow). (c) Find the sum of all probabilities.
Visual — all 12 marbles in the bag:
🟢 Green (5) 🔵 Blue (3) 🔴 Red (2) 🟡 Yellow (2)
(a) Are they equally likely? No. The colours have different counts — 5 green but only 2 yellow — so each colour does not have the same chance of being drawn.
Total marbles n(S) = 5 + 3 + 2 + 2 = 12
P(green) = 5/12
P(blue) = 3/12 = 1/4
P(red) = 2/12 = 1/6
P(yellow) = 2/12 = 1/6
Sum = 5/12 + 3/12 + 2/12 + 2/12 = 12/12
= 1 ✓
✅ Answers: Not equally likely | P(G) = 5/12, P(B) = 3/12, P(R) = 2/12, P(Y) = 2/12 | Sum = 1
Question 5 — Letters of the English Alphabet
Question 5
A letter is chosen randomly from the English alphabet (26 letters). Find: (a) P(vowel) (b) P(letter after P) (c) P(vowel or consonant) (d) P(not a vowel)
Total letters n(S) = 26
Vowels: A, E, I, O, U → n = 5
Letters after P: Q,R,S,T,U,V,W,X,Y,Z → n = 10
Consonants = 26 − 5 = 21
| Part | Event | Favourable Letters | Count | P(E) |
|---|---|---|---|---|
| (a) | Vowel | A, E, I, O, U | 5 | 5/26 |
| (b) | Letter after P | Q, R, S, T, U, V, W, X, Y, Z | 10 | 10/26 = 5/13 |
| (c) | Vowel or consonant | All 26 letters | 26 | 26/26 = 1 |
| (d) | Not a vowel | All consonants (21) | 21 | 21/26 |
📌 Part (c) gives P = 1 because every letter in the alphabet is either a vowel or a consonant — this is a certain event. Also notice: P(vowel) + P(not a vowel) = 5/26 + 21/26 = 1. These are complementary events.
Question 6 — Wheat Flour Bags (Experimental Probability)
Question 6
Eleven flour bags marked "5 kg" actually weigh: 4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98, 5.04, 5.07, 5.00. Find P(bag contains more than 5 kg).
| Bag # | Actual Weight (kg) | More than 5 kg? |
|---|---|---|
| 1 | 4.97 | ❌ No |
| 2 | 5.05 | ✅ Yes |
| 3 | 5.08 | ✅ Yes |
| 4 | 5.03 | ✅ Yes |
| 5 | 5.00 | ❌ No (exactly 5) |
| 6 | 5.06 | ✅ Yes |
| 7 | 5.08 | ✅ Yes |
| 8 | 4.98 | ❌ No |
| 9 | 5.04 | ✅ Yes |
| 10 | 5.07 | ✅ Yes |
| 11 | 5.00 | ❌ No (exactly 5) |
Total bags n(S) = 11
Bags with weight > 5 kg: bags 2,3,4,6,7,9,10 → n(E) = 7
P(more than 5 kg) = 7/11
∴ P(bag > 5 kg) = 7/11
✅ Answer: P(more than 5 kg) = 7/11
⛔ Common trap: Bags weighing exactly 5.00 kg are NOT counted as "more than 5 kg." Only strictly greater than 5 counts. Here 2 bags weigh exactly 5.00 — they are excluded.
Question 7 — Driver Accident Data (Frequency-based Probability)
Question 7
An insurance company studied 2000 drivers. Find probabilities for three specific events using the accident data table below.
| Age Group | 0 Accidents | 1 Accident | 2 Accidents | 3 Accidents | More than 3 | Row Total |
|---|---|---|---|---|---|---|
| 18–29 years | 440 | 160 | 110 | 61 | 35 | 806 |
| 30–50 years | 505 | 125 | 60 | 22 | 18 | 730 |
| Over 50 years | 360 | 45 | 35 | 15 | 9 | 464 |
| Total | 1305 | 330 | 205 | 98 | 62 | 2000 |
Part (i) — Driver aged 18–29 with exactly 3 accidents
Total drivers n(S) = 2000
Drivers aged 18–29 with exactly 3 accidents = 61
P(age 18–29 AND exactly 3 accidents) = 61/2000
∴ P = 61/2000
Part (ii) — Driver aged 30–50 with 1 or more accidents
Drivers aged 30–50 with ≥1 accident = 125 + 60 + 22 + 18 = 225
P(age 30–50 AND ≥1 accident) = 225/2000
= 225/2000 = 9/80
∴ P = 9/80
Part (iii) — Driver with no accidents (any age)
Drivers with 0 accidents = 440 + 505 + 360 = 1305
P(no accidents) = 1305/2000
= 261/400
∴ P = 261/400
✅ Answers:
(i) P = 61/2000 |
(ii) P = 9/80 |
(iii) P = 261/400
📌 This is an experimental probability question using real survey data. There is no sample space to list — just read the required count from the table and divide by 2000.
Question 8 — Geometric Probability (Dart on a Board)
Question 8
A circle of radius 2 cm is inscribed in a square. A dart is thrown randomly at the square board. Find the probability it hits the shaded region (outside the circle). Express as a percentage. (Use π = 22/7)
Setup: The circle of radius 2 cm sits perfectly inside the square. Since the diameter = 4 cm = side of the square, the square's side is 4 cm.
r = 2 cm, Side of square a = 2r = 4 cm
Area of circle = πr²
= (22/7) × 2 × 2 = 88/7 cm²
Area of square = a² = 4 × 4 = 16 cm²
Shaded area = Square − Circle
= 16 − 88/7
= 112/7 − 88/7
= 24/7 cm²
P(shaded) = (24/7) ÷ 16
= 24 / (7 × 16) = 24/112
= 3/14
As % = (3/14) × 100 = 300/14
= 150/7 %
≈ 21.43%
✅ Answer: P(shaded region) = 3/14 ≈ 21.43%
📐 Key concept — Geometric Probability: When a dart is thrown randomly at a board, the probability of hitting any region is proportional to its area. P(event) = Area of region / Total area. This type of question connects Chapter 14 (Probability) with mensuration concepts from Chapter 10.
Quick Reference — All 8 Questions at a Glance
| Q# | Topic | Type | Key Answer |
|---|---|---|---|
| Q1 | Die — composite number | Theoretical | P = 1/3 |
| Q2 | Coin — 100 tosses | Experimental | P(H)=9/20, P(T)=11/20 |
| Q3 | Spinner — 4 colours | Conceptual | Red=most; Yellow=least; Blue=Green; White=0 |
| Q4 | Marbles — 4 colours | Theoretical | 5/12, 3/12, 2/12, 2/12 |
| Q5a | Alphabet — vowel | Theoretical | 5/26 |
| Q5b | Alphabet — after P | Theoretical | 10/26 = 5/13 |
| Q5c | Alphabet — vowel or consonant | Theoretical | 1 (certain event) |
| Q5d | Alphabet — not a vowel | Theoretical | 21/26 |
| Q6 | Flour bags — over 5 kg | Experimental | 7/11 |
| Q7i | Drivers 18–29 age, 3 accidents | Experimental | 61/2000 |
| Q7ii | Drivers 30–50, ≥1 accident | Experimental | 9/80 |
| Q7iii | Drivers — no accidents | Experimental | 261/400 |
| Q8 | Dart — shaded region | Geometric | 3/14 ≈ 21.43% |
Common Mistakes to Avoid
- Counting 1 as a composite number (Q1): The number 1 is neither prime nor composite. Composites from 1–6 are only 4 and 6, giving P = 2/6 = 1/3, not 3/6.
- Including exactly 5 kg bags as "more than 5 kg" (Q6): "More than" means strictly greater than. Bags weighing exactly 5.00 kg are not included. Count only the 7 bags with weight > 5.00.
- Adding instead of checking the table in Q7: For Part (ii), 1-or-more accidents for age 30–50 means summing columns 1, 2, 3 and "more than 3": 125 + 60 + 22 + 18 = 225. Do not include the 0-accidents column.
- Using r as the side of the square in Q8: The diameter (2r = 4 cm) is the side of the square, not the radius. Always derive the side as 2r before computing the square area.
- Forgetting to simplify fractions: In board exams, always reduce: 225/2000 = 9/80, 1305/2000 = 261/400. Unsimplified fractions may lose a mark.
- Saying spinner outcomes are equally likely (Q3): Spinner probabilities depend on sector angle/area. Different-sized sectors give different probabilities — they are not equally likely unless all sectors are the same size.
⛔ Board exam alert (Telangana & AP): Q7 (accident data table) and Q8 (geometric probability) are frequently set as 4–5 mark questions. For Q7, always write "total drivers = 2000" and identify the correct cell from the table before dividing. For Q8, show all four steps: area of circle, area of square, shaded area, then probability as a fraction and percentage.
What This Exercise Prepares You For
Exercise 14.1 is the only exercise in Class 9 Probability, but the concepts here are the direct foundation for Class 10 Probability, which introduces complementary events (P(not E) = 1 − P(E)), and problems with coloured balls and two dice. The geometric probability in Q8 connects to mensuration work from Chapter 10 Surface Areas and Volumes.
The data-based Q7 (accident table) is similar in structure to Chapter 13 Statistics problems, where frequency tables are used to calculate relative frequencies — essentially the same as experimental probability.
🎲 Chapter 14 Probability
📊 Experimental Probability
📐 Geometric Probability
🔟 Class 10 Probability
📌 Revision checklist for Exercise 14.1 (CBSE / Telangana / AP):
✔ P(E) = Favourable outcomes / Total outcomes (theoretical)
✔ P(E) = Frequency of event / Total trials (experimental)
✔ Sum of all probabilities of an experiment = 1
✔ 1 is neither prime nor composite
✔ Ace is not a face card (though not tested here, it will be in Class 10)
✔ For geometric probability: P = Area of region / Total area
✔ P(E) = Favourable outcomes / Total outcomes (theoretical)
✔ P(E) = Frequency of event / Total trials (experimental)
✔ Sum of all probabilities of an experiment = 1
✔ 1 is neither prime nor composite
✔ Ace is not a face card (though not tested here, it will be in Class 10)
✔ For geometric probability: P = Area of region / Total area