Chapter 7

Chapter 7: Introduction to Probability — Solutions

Exercise Set 7.1

1. Rank the following events on a scale from 0 (Impossible) to 1 (Certain). Label each event: Impossible, less likely, equally likely (even chance), more likely, certain. Give reasons why you gave each event its ranking.

(i) The next Monday will come after Sunday.

• Label: Certain (Value: 1)
• Reason: Monday always follows Sunday according to the calendar structure.

(ii) It will snow in Mumbai in July.

• Label: Impossible (Value: 0)
• Reason: Mumbai has a tropical climate where atmospheric conditions never drop to freezing levels necessary for snowfall.

(iii) An elephant will walk through your classroom today.

• Label: Impossible / Less Likely (Value near 0)
• Reason: Unless the school is adjacent to a wild sanctuary or a parade route, the probability is practically 0 (impossible).

(iv) You will greet at least one friend at school tomorrow.

• Label: More likely / Certain (Value near 1)
• Reason: Attending school typically involves interacting with peers and classmates.

Exercise Set 7.2

1. A teacher mixes a large bag of sweets of different colours and randomly selects a sample of 30 sweets: 10 red, 8 green, 7 yellow, 5 blue.

(i) Calculate the probability that a randomly picked sweet from the sample is green.

P(Green) = 8 / 30 = 4 / 15 ≈ 0.267 (26.7%)

(ii) If there are 600 sweets in total in the large bag, estimate how many are likely to be yellow.

P(Yellow) = 7 / 30

Estimated yellow sweets = (7 / 30) × 600 = 140 sweets

2. Survey of 40 students regarding favourite clubs: 14 Science, 11 Arts, 9 Sports, 6 Debate. Total school population = 800.

(i) What is the probability that a randomly chosen student prefers the Arts Club?

P(Arts) = 11 / 40 = 0.275 (27.5%)

(ii) Estimate how many students in the whole school are likely to prefer the Sports Club.

P(Sports) = 9 / 40

Estimated Sports Club members = (9 / 40) × 800 = 180 students

3. Toss a coin 20 times and record results.

(Note: Answers to (i)-(iii) depend on experimental data. Below is a sample outcome.)

(i) Heads count: Say, 11 times.

(ii) Tails count: Say, 9 times.

(iii) Experimental probability of getting heads: 11 / 20 = 0.55

(iv) If you toss the coin once more, what is the probability of getting tails?

Theoretical probability remains 1 / 2 = 0.5 because previous coin tosses do not affect future independent trials (avoiding Gambler's Fallacy).

4. Toss a paper cup 100 times and record landing positions (bottom, top, side).

This is an empirical activity. The experimental probability for each outcome is calculated as:

P(Outcome) = (Number of times outcome occurred) / 100

Typically, landing on the side has the highest relative frequency compared to bottom or top.

5. What is the probability of getting an even number when rolling a fair 6-sided die?

Sample Space = {1, 2, 3, 4, 5, 6} (Total = 6 outcomes)

Favourable outcomes (Even numbers) = {2, 4, 6} (Count = 3)

P(Even) = 3 / 6 = 1 / 2 = 0.5 (50%)

6. Suppose you roll a 6-sided die 12 times and get a '3' three times.

(i) Experimental probability of rolling a '3': 3 / 12 = 1 / 4 = 0.25

(ii) Theoretical probability of rolling a '3': 1 / 6 ≈ 0.167

(iii) Why might these probabilities be different? What happens as trials increase?

They differ because 12 trials is a small sample size, leading to short-term experimental variation. According to the Law of Large Numbers, as the number of rolls increases to 60, 600, or 6000, the experimental probability will get closer to the theoretical probability of 1/6.

Exercise Set 7.3

1. Total possible outcomes when a single 6-sided die is rolled?

Total outcomes = 6 (i.e., {1, 2, 3, 4, 5, 6}).

2. Write the sample space S for each experiment:

(i) Rolling a die and tossing a coin together:

S = {(1,H), (2,H), (3,H), (4,H), (5,H), (6,H), (1,T), (2,T), (3,T), (4,T), (5,T), (6,T)}

Sample size n(S) = 12

(ii) Choosing a random integer between -5 and +5:

S = {-4, -3, -2, -1, 0, 1, 2, 3, 4} (Assuming 'between' excludes endpoints); or {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5} if inclusive.

(iii) A box containing 5 green and 7 red balls. One ball is drawn:

S = {Green, Red} or {G1, G2, G3, G4, G5, R1, R2, R3, R4, R5, R6, R7}

3. Village fair options — Snacks: Samosa (S), Pakora (P), Bhaji (B); Drinks: Chai (C), Lassi (L).

(i) Sample space of combinations:

S = {(S, C), (S, L), (P, C), (P, L), (B, C), (B, L)}

(ii) Event 'Selecting Samosa as a snack':

E = {(S, C), (S, L)}

Exercise Set 7.4

1. Basket A: {Apple (A), Orange1 (O1), Orange2 (O2)}. Basket B: {Banana (B), Mango (M)}.

(i) Tree Diagram Description:

Branch 1 (Basket A) splits into A, O1, O2. From each of these 3, branch 2 (Basket B) splits into B and M, creating 6 total paths.

(ii) Sample Space:

S = {(A, B), (A, M), (O1, B), (O1, M), (O2, B), (O2, M)}

(iii) Probability of picking one apple and one banana:

Favourable outcome = {(A, B)}

P(Apple and Banana) = 1 / 6

2. Box: 3 Red, 4 Black, 2 Green pens (Total = 9). Pick a pen, replace it, then friend picks.

(i) Possible outcomes:

S = {(R,R), (R,B), (R,G), (B,R), (B,B), (B,G), (G,R), (G,B), (G,G)}

(ii) Probability that both pick the same colour:

• P(Both Red) = (3/9) × (3/9) = 9/81
• P(Both Black) = (4/9) × (4/9) = 16/81
• P(Both Green) = (2/9) × (2/9) = 4/81

P(Same Colour) = (9 + 16 + 4) / 81 = 29 / 81 ≈ 0.358

End-of-Chapter Exercises

1. Fill in the blanks:

(i) The probability of an impossible event is 0.

(ii) The set of all possible outcomes of a random experiment is called the sample space.

(iii) The probability of an event that is certain to happen is 1.

(iv) Tossing a fair coin has a probability of 1/2 (or 0.5) for getting heads.

2. In a survey of 50 students, 15 like football.

The frequency is 15, and the relative frequency is 15/50 = 3/10 = 0.3.

3. Which experiments have equally likely outcomes?

(i) Tossing a fair coin once: Yes, heads and tails are equally likely.

(ii) A driver attempts to start a car: No, whether a car starts depends on its mechanical condition, so outcomes are not equally likely.

(iii) Rolling a fair 6-sided die: Yes, each face from 1 to 6 has an equal chance.

(iv) Drawing a marble from 3 red and 7 blue marbles: No, drawing a blue marble is more likely than red.

(v) A baby is born (boy or girl): Yes, biologically male and female births are approximately equally likely.

4. Write sample space and calculate probability:

(i) Two coins tossed. P(at least one head):

S = {HH, HT, TH, TT}; Favourable = {HH, HT, TH}; P = 3 / 4

(ii) Cards 1 to 10. P(even number):

S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; Even = {2, 4, 6, 8, 10}; P = 5 / 10 = 1 / 2

(iii) Die rolled once. P(greater than 4):

S = {1, 2, 3, 4, 5, 6}; Favourable = {5, 6}; P = 2 / 6 = 1 / 3

(iv) Bag: 3 red, 2 blue, 1 green (Total 6). P(not red):

Not red = Blue + Green = 3; P = 3 / 6 = 1 / 2

(v) Three coins tossed. P(exactly two heads):

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}; Favourable = {HHT, HTH, THH}; P = 3 / 8

5. Bag with 3 candies (strawberry, lemon, mint). P(strawberry)?

P(Strawberry) = 1 / 3

6. Outfit combinations table (2 shirts: Red, Blue; 3 pants: Jeans, Khakis, Shorts):

Shirt \ Pants	Jeans (J)	Khakis (K)	Shorts (S)
Red (R)	(Red, Jeans)	(Red, Khakis)	(Red, Shorts)
Blue (B)	(Blue, Jeans)	(Blue, Khakis)	(Blue, Shorts)

Total combinations = 6

7. Tyre company dataset (Total 1000 cases):

(i) P(Less than 4000 km): 20 / 1000 = 0.02

(ii) P(Between 4000 and 14000 km): (210 + 325) / 1000 = 535 / 1000 = 0.535

(iii) P(More than 14000 km): 445 / 1000 = 0.445

8. Word 'PEACE' (Letters: P, E, A, C, E — Total = 5 letters):

(i) P(P, E or C): Favourable = {P, E, E, C} (Count = 4). P = 4 / 5

(ii) P(not an E): Favourable = {P, A, C} (Count = 3). P = 3 / 5

9. Spinner numbered 1 to 8:

(i) P(8): 1 / 8

(ii) P(Odd number): Odds = {1, 3, 5, 7}; P = 4 / 8 = 1 / 2

(iii) P(Greater than 2): {3, 4, 5, 6, 7, 8}; P = 6 / 8 = 3 / 4

(iv) P(Less than 9): {1, 2, 3, 4, 5, 6, 7, 8}; P = 8 / 8 = 1 (Certain)

(v) P(Multiple of 3): Multiples = {3, 6}; P = 2 / 8 = 1 / 4

10. 4 red and 5 blue balls (Total = 9). Draw 1 without replacement, then draw 2nd.

(i) P(Red then Blue):

P(R1) = 4 / 9; P(B2 | R1) = 5 / 8

P(Red and Blue) = (4 / 9) × (5 / 8) = 20 / 72 = 5 / 18

(ii) P(2 Blue balls):

P(B1) = 5 / 9; P(B2 | B1) = 4 / 8

P(Both Blue) = (5 / 9) × (4 / 8) = 20 / 72 = 5 / 18

11. Pair of 6-sided dice:

Event with P = 0: Getting a sum of 13 or getting a 7 on a single die.

Outcome with P = 1: Getting a total sum between 2 and 12 (inclusive).

12. Advanced probability problems:

(i) Sum is a prime number greater than 5 (i.e., Sum = 7 or 11):

Sum = 7: {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)} (6 outcomes)

Sum = 11: {(5,6), (6,5)} (2 outcomes)

Total = 8 outcomes. P = 8 / 36 = 2 / 9

(ii) 4 Red, 3 Green, 2 Blue (Total = 9). Two drawn without replacement. P(different colours):

P(Same colour) = P(RR) + P(GG) + P(BB) = [(4×3) + (3×2) + (2×1)] / (9×8) = (12 + 6 + 2) / 72 = 20 / 72

P(Different colours) = 1 - (20 / 72) = 52 / 72 = 13 / 18

(iii) 3 coins. P(first coin is H AND exactly two H in total):

Favourable = {HHT, HTH}; P = 2 / 8 = 1 / 4

(iv) Four-digit number using 1, 2, 3, 4 without repetition. P(Even):

Total numbers = 4! = 24

For even, last digit must be 2 or 4 (2 choices). Remaining digits = 3! = 6 choices.

Favourable = 2 × 6 = 12. P = 12 / 24 = 1 / 2

(v) MCQ test with 3 questions, 4 options each. P(guessing exactly 2 correct):

p = 1/4 (correct), q = 3/4 (wrong).

P(Exactly 2 correct) = 3C2 × (1/4)^2 × (3/4)^1 = 3 × (1/16) × (3/4) = 9 / 64

13. Box with 4 balls (1 to 4):

(i) With replacement: Size n(S) = 4 × 4 = 16

(ii) Without replacement: Size n(S) = 4 × 3 = 12

14. Simultaneous toss of a coin and card drawn from {1, 2, 3, 4, 5, 6}:

S = {(H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}

15. Three coins tossed, number of heads recorded. Which list is a sample space?

Correct List: (iv) {0, 1, 2, 3}

Reasons why others fail:

• (i) {1, 2, 3} misses 0 heads (TTT).
• (ii) {0, 1, 2} misses 3 heads (HHH).
• (iii) {0, 1, 2, 3, 4} includes 4, which is impossible with 3 coins.

16. Rectangular region 3 m × 2 m. Circle diameter = 1 m (Radius r = 0.5 m). P(landing inside circle)?

Area of Rectangle = 3 × 2 = 6 m²

Area of Circle = π × r² = π × (0.5)² = 0.25π m²

P(Inside Circle) = (0.25π) / 6 = π / 24 ≈ 0.1309 (or ~13.1%)