End-of-Chapter Exercises
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Overview
This page provides comprehensive Ch 7: The Mathematics of Maybe: Introduction to Probability - End-of-Chapter Exercises. Practice solving mixed problems on probability rules, equally likely events, tree diagrams without replacement, multiple-choice guesswork, and geometric area probability with step-by-step solutions.
Predicting Probabilities: Advanced Coin, Dice, Cards & Geometric Fields
Q1: Fill in the Blanks
Fill in the blanks:
(i) The probability of an impossible event is _______.
(ii) The set of all possible outcomes of a random experiment is called the __________.
(iii) The probability of an event that is certain to happen is _______.
(iv) Tossing a fair coin has a probability of ______ for getting heads.
(i) The probability of an impossible event is _______.
(ii) The set of all possible outcomes of a random experiment is called the __________.
(iii) The probability of an event that is certain to happen is _______.
(iv) Tossing a fair coin has a probability of ______ for getting heads.
(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.
(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.
(i) 0 (ii) sample space (iii) 1 (iv) 1/2
Q2: Relative Frequency
In a survey of $50\text{ students}$, $15\text{ students}$ said they liked football. The number of students who like football is $15$, and the ________ (frequency/relative frequency) is __________ (fill in the fraction or decimal).
The observed count ($15$) is the absolute **frequency**.
The proportion of students who like football out of total is the **relative frequency**:
$$\text{Relative Frequency} = \frac{15}{50} = \mathbf{\frac{3}{10} = 0.3}$$
The proportion of students who like football out of total is the **relative frequency**:
$$\text{Relative Frequency} = \frac{15}{50} = \mathbf{\frac{3}{10} = 0.3}$$
relative frequency; 3/10 (or 0.3)
Q3: Equally Likely Outcomes
Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) Tossing a fair coin once.
(iii) Rolling a fair 6-sided die.
(iv) Choosing a marble randomly from a bag that contains 3 red marbles and 7 blue marbles.
(v) A baby is born. It is a boy or a girl.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) Tossing a fair coin once.
(iii) Rolling a fair 6-sided die.
(iv) Choosing a marble randomly from a bag that contains 3 red marbles and 7 blue marbles.
(v) A baby is born. It is a boy or a girl.
An experiment has equally likely outcomes if each outcome has the exact same theoretical probability.
• **(i) Starting a car:** **Not equally likely**. The outcome depends heavily on variables like fuel levels, engine health, battery condition, and weather.
• **(ii) Tossing a fair coin:** **Equally likely**. Both Heads and Tails have an identical chance ($P=1/2$).
• **(iii) Rolling a fair die:** **Equally likely**. All six faces $\{1, 2, 3, 4, 5, 6\}$ have the same chance ($P=1/6$).
• **(iv) Selecting a marble (3 red, 7 blue):** **Not equally likely**. Since there are more blue marbles, picking a blue marble is more likely ($7/10$) than a red one ($3/10$).
• **(v) A baby born:** **Equally likely**. Biologically, the chance of a baby being a boy or a girl is approximately equal (roughly 50% each).
• **(ii) Tossing a fair coin:** **Equally likely**. Both Heads and Tails have an identical chance ($P=1/2$).
• **(iii) Rolling a fair die:** **Equally likely**. All six faces $\{1, 2, 3, 4, 5, 6\}$ have the same chance ($P=1/6$).
• **(iv) Selecting a marble (3 red, 7 blue):** **Not equally likely**. Since there are more blue marbles, picking a blue marble is more likely ($7/10$) than a red one ($3/10$).
• **(v) A baby born:** **Equally likely**. Biologically, the chance of a baby being a boy or a girl is approximately equal (roughly 50% each).
Equally likely: (ii), (iii), and (v)
Q4: Classic Probability Calculations
Write the sample space and calculate the probability based on the given information:
(i) Two coins are tossed at the same time. What is the probability of getting at least one head?
(ii) Ten identical cards numbered 1 to 10 are placed in a box. One card is drawn at random. What is the probability of drawing a card with an even number?
(iii) A die is rolled once. What is the probability of getting a number greater than 4?
(iv) A bag contains 3 red balls, 2 blue balls, and 1 green ball. One ball is picked at random. What is the probability that it is not red?
(v) Three coins are tossed simultaneously. What is the probability of getting exactly two heads?
(i) Two coins are tossed at the same time. What is the probability of getting at least one head?
(ii) Ten identical cards numbered 1 to 10 are placed in a box. One card is drawn at random. What is the probability of drawing a card with an even number?
(iii) A die is rolled once. What is the probability of getting a number greater than 4?
(iv) A bag contains 3 red balls, 2 blue balls, and 1 green ball. One ball is picked at random. What is the probability that it is not red?
(v) Three coins are tossed simultaneously. What is the probability of getting exactly two heads?
(i) At least 1 Head (2 coins):
Sample Space $S = \{HH, HT, TH, TT\} \implies n(S) = 4$.
Favourable outcomes $E = \{HH, HT, TH\} \implies n(E) = 3$.
$$P(E) = \mathbf{\frac{3}{4} = 0.75}$$
Sample Space $S = \{HH, HT, TH, TT\} \implies n(S) = 4$.
Favourable outcomes $E = \{HH, HT, TH\} \implies n(E) = 3$.
$$P(E) = \mathbf{\frac{3}{4} = 0.75}$$
(ii) Even Number Card (1 to 10):
Sample Space $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \implies n(S) = 10$.
Even cards $E = \{2, 4, 6, 8, 10\} \implies n(E) = 5$.
$$P(E) = \frac{5}{10} = \mathbf{\frac{1}{2} = 0.50}$$
Sample Space $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \implies n(S) = 10$.
Even cards $E = \{2, 4, 6, 8, 10\} \implies n(E) = 5$.
$$P(E) = \frac{5}{10} = \mathbf{\frac{1}{2} = 0.50}$$
(iii) Die Roll greater than 4:
Sample Space $S = \{1, 2, 3, 4, 5, 6\} \implies n(S) = 6$.
Outcomes $> 4$ is $E = \{5, 6\} \implies n(E) = 2$.
$$P(E) = \frac{2}{6} = \mathbf{\frac{1}{3} \approx 0.333}$$
Sample Space $S = \{1, 2, 3, 4, 5, 6\} \implies n(S) = 6$.
Outcomes $> 4$ is $E = \{5, 6\} \implies n(E) = 2$.
$$P(E) = \frac{2}{6} = \mathbf{\frac{1}{3} \approx 0.333}$$
(iv) Not Red Ball:
Total balls $N = 3 + 2 + 1 = 6$. Red balls count $= 3$.
Not red balls (Blue + Green) $= 2 + 1 = 3$.
$$P(\text{not Red}) = \frac{3}{6} = \mathbf{\frac{1}{2} = 0.50}$$
Total balls $N = 3 + 2 + 1 = 6$. Red balls count $= 3$.
Not red balls (Blue + Green) $= 2 + 1 = 3$.
$$P(\text{not Red}) = \frac{3}{6} = \mathbf{\frac{1}{2} = 0.50}$$
(v) Exactly 2 Heads (3 coins):
Sample Space $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\} \implies n(S) = 8$.
Outcomes with exactly 2 Heads $E = \{HHT, HTH, THH\} \implies n(E) = 3$.
$$P(E) = \mathbf{\frac{3}{8} = 0.375}$$
Sample Space $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\} \implies n(S) = 8$.
Outcomes with exactly 2 Heads $E = \{HHT, HTH, THH\} \implies n(E) = 3$.
$$P(E) = \mathbf{\frac{3}{8} = 0.375}$$
(i) 3/4 (ii) 1/2 (iii) 1/3 (iv) 1/2 (v) 3/8
Q5: Strawberry Candy
A bag has 3 candies: strawberry, lemon, and mint. One is picked at random. What is the probability of picking a strawberry candy?
Sample Space $S = \{\text{Strawberry, Lemon, Mint}\} \implies n(S) = 3$.
Favourable outcome is $\{\text{Strawberry}\} \implies n(E) = 1$.
$$P(\text{Strawberry}) = \mathbf{\frac{1}{3} \approx 0.333}$$
Favourable outcome is $\{\text{Strawberry}\} \implies n(E) = 1$.
$$P(\text{Strawberry}) = \mathbf{\frac{1}{3} \approx 0.333}$$
P(Strawberry) = 1/3
Q6: Outfit Combinations
A child has 2 shirts (one red and one blue) and 3 types of pants (jeans, khakis, and shorts). List all the possible combinations of outfits consisting of one shirt and one pair of pants. Display your answer in a table format.
Total possible combinations $= 2\text{ shirts} \times 3\text{ pants} = \mathbf{6\text{ outfits}}$.
Table of Combinations:
| Outfit # | Shirt Color | Pant Type |
|---|---|---|
| 1 | Red | Jeans |
| 2 | Red | Khakis |
| 3 | Red | Shorts |
| 4 | Blue | Jeans |
| 5 | Blue | Khakis |
| 6 | Blue | Shorts |
6 possible outfits (listed in the table)
Q7: Tyre Durability Survey
A tyre company records distances before replacement in $1000\text{ cases}$:
Find the probability that a randomly chosen tyre lasts:
(i) Less than 4000 km.
(ii) Between 4000 and 14000 km.
(iii) More than 14000 km.
| Distance (km) | Less than 4000 | 4001 to 9000 | 9001 to 14000 | More than 14000 |
|---|---|---|---|---|
| **Number of cases** | 20 | 210 | 325 | 445 |
Find the probability that a randomly chosen tyre lasts:
(i) Less than 4000 km.
(ii) Between 4000 and 14000 km.
(iii) More than 14000 km.
Total cases $N = 1000$.
(i) Less than 4000 km:
Cases $= 20$.
$$P(<4000) = \frac{20}{1000} = \mathbf{0.02}$$
Cases $= 20$.
$$P(<4000) = \frac{20}{1000} = \mathbf{0.02}$$
(ii) Between 4000 and 14000 km (i.e. 4001 to 14000 km):
Cases $= 210 + 325 = 535$.
$$P(4000 \text{ to } 14000) = \frac{535}{1000} = \mathbf{0.535}$$
Cases $= 210 + 325 = 535$.
$$P(4000 \text{ to } 14000) = \frac{535}{1000} = \mathbf{0.535}$$
(iii) More than 14000 km:
Cases $= 445$.
$$P(>14000) = \frac{445}{1000} = \mathbf{0.445}$$
Cases $= 445$.
$$P(>14000) = \frac{445}{1000} = \mathbf{0.445}$$
(i) 0.02 (ii) 0.535 (iii) 0.445
Q8: Word Card PEACE
The letters of the word ‘PEACE’ are placed on cards. Leela draws a card without looking.
(i) What is the probability that it is a P, E, or C?
(ii) What is the probability that it is not an E?
(i) What is the probability that it is a P, E, or C?
(ii) What is the probability that it is not an E?
The letters are $\{P, E, A, C, E\}$. Total letters $N = 5$.
Letter frequencies: $P: 1$, $E: 2$, $A: 1$, $C: 1$.
Letter frequencies: $P: 1$, $E: 2$, $A: 1$, $C: 1$.
(i) P(P, E, or C):
Favourable cards $= 1 + 2 + 1 = 4$.
$$P(\text{P, E, or C}) = \mathbf{\frac{4}{5} = 0.80}$$
Favourable cards $= 1 + 2 + 1 = 4$.
$$P(\text{P, E, or C}) = \mathbf{\frac{4}{5} = 0.80}$$
(ii) P(not E):
Favourable cards (P, A, C) $= 3$.
$$P(\text{not E}) = \mathbf{\frac{3}{5} = 0.60}$$
Favourable cards (P, A, C) $= 3$.
$$P(\text{not E}) = \mathbf{\frac{3}{5} = 0.60}$$
(i) 4/5 (0.80) (ii) 3/5 (0.60)
Q9: Spinner Arrow
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers $1, 2, 3, 4, 5, 6, 7, 8$, and these are equally likely outcomes. What is the probability that it will point at:
(i) 8?
(ii) An odd number?
(iii) A number greater than 2?
(iv) A number less than 9?
(v) A multiple of 3?
(i) 8?
(ii) An odd number?
(iii) A number greater than 2?
(iv) A number less than 9?
(v) A multiple of 3?
Sample Space $S = \{1, 2, 3, 4, 5, 6, 7, 8\} \implies n(S) = 8$.
(i) P(8): Favourable outcome $\{8\} \implies \mathbf{1/8}$.
(ii) P(odd): Odd numbers $\{1, 3, 5, 7\} \implies 4/8 = \mathbf{1/2}$.
(iii) P(>2): Numbers $\{3, 4, 5, 6, 7, 8\} \implies 6/8 = \mathbf{3/4}$.
(iv) P(<9): Numbers $\{1, 2, \dots, 8\} \implies 8/8 = \mathbf{1}$ (Certain).
(v) P(multiple of 3): Multiples $\{3, 6\} \implies 2/8 = \mathbf{1/4}$.
(ii) P(odd): Odd numbers $\{1, 3, 5, 7\} \implies 4/8 = \mathbf{1/2}$.
(iii) P(>2): Numbers $\{3, 4, 5, 6, 7, 8\} \implies 6/8 = \mathbf{3/4}$.
(iv) P(<9): Numbers $\{1, 2, \dots, 8\} \implies 8/8 = \mathbf{1}$ (Certain).
(v) P(multiple of 3): Multiples $\{3, 6\} \implies 2/8 = \mathbf{1/4}$.
(i) 1/8 (ii) 1/2 (iii) 3/4 (iv) 1 (v) 1/4
Q10: Ball Draw Without Replacement
A basket contains **4 red balls** and **5 blue balls**. One ball is drawn and laid aside, and a second ball is drawn. Draw a tree diagram to represent the possible outcomes and probabilities. Use the tree diagram to answer:
(i) What is the probability of drawing a red ball and then a blue ball?
(ii) What is the probability of drawing 2 blue balls?
(i) What is the probability of drawing a red ball and then a blue ball?
(ii) What is the probability of drawing 2 blue balls?
Total balls originally $= 9$ ($4\text{ Red}, 5\text{ Blue}$).
First Pick:
• $P(R) = 4/9$
• $P(B) = 5/9$
First Pick:
• $P(R) = 4/9$
• $P(B) = 5/9$
Second Pick (without replacement, total balls left $= 8$):
• If Red was drawn first: Red remaining $= 3$, Blue $= 5$.
— $P(R|R) = 3/8$
— $P(B|R) = 5/8$
• If Blue was drawn first: Red remaining $= 4$, Blue $= 4$.
— $P(R|B) = 4/8 = 1/2$
— $P(B|B) = 4/8 = 1/2$
• If Red was drawn first: Red remaining $= 3$, Blue $= 5$.
— $P(R|R) = 3/8$
— $P(B|R) = 5/8$
• If Blue was drawn first: Red remaining $= 4$, Blue $= 4$.
— $P(R|B) = 4/8 = 1/2$
— $P(B|B) = 4/8 = 1/2$
Calculations:
**(i) P(Red then Blue):**
$$P(R \text{ then } B) = P(R) \times P(B|R) = \frac{4}{9} \times \frac{5}{8} = \frac{20}{72} = \mathbf{\frac{5}{18} \approx 0.278}$$
**(ii) P(2 Blue balls):**
$$P(B \text{ then } B) = P(B) \times P(B|B) = \frac{5}{9} \times \frac{4}{8} = \frac{20}{72} = \mathbf{\frac{5}{18} \approx 0.278}$$
**(i) P(Red then Blue):**
$$P(R \text{ then } B) = P(R) \times P(B|R) = \frac{4}{9} \times \frac{5}{8} = \frac{20}{72} = \mathbf{\frac{5}{18} \approx 0.278}$$
**(ii) P(2 Blue balls):**
$$P(B \text{ then } B) = P(B) \times P(B|B) = \frac{5}{9} \times \frac{4}{8} = \frac{20}{72} = \mathbf{\frac{5}{18} \approx 0.278}$$
(i) 5/18 (ii) 5/18
Q11: Sum of Two Dice Boundaries
I throw a pair of 6-sided dice. Write down an event that has a probability of $0$ and an outcome that has a probability of $1$.
The sum of numbers on a pair of 6-sided dice ranges from $2$ (minimum $1+1$) to $12$ (maximum $6+6$).
• **Event with $P=0$ (Impossible):** Obtaining a sum of **$13$** (or "getting a sum greater than 12").
• **Event with $P=1$ (Certain):** Obtaining a sum **between $2$ and $12$ inclusive** (or "getting a sum less than 15").
• **Event with $P=1$ (Certain):** Obtaining a sum **between $2$ and $12$ inclusive** (or "getting a sum less than 15").
P=0: Sum is 13; P=1: Sum is between 2 and 12
Q12: Advanced Probability Calculations
Write the sample space and calculate the probability based on the given information:
(i) Two dice are rolled. What is the probability that the sum is a prime number greater than 5?
(ii) A bag contains 4 red, 3 green, and 2 blue balls. Two balls are drawn without replacement. What is the probability that both are of different colours?
(iii) Three coins are tossed. What is the probability that the first coin shows heads and exactly two heads occur in total?
(iv) A four-digit number is formed using the digits 1, 2, 3, and 4 with no repetition. What is the probability that the number is even?
(v) A student takes a multiple-choice test with 3 questions, each having 4 options (A, B, C, D), with only one correct answer. What is the probability that the student guesses and gets exactly 2 answers correct?
(i) Two dice are rolled. What is the probability that the sum is a prime number greater than 5?
(ii) A bag contains 4 red, 3 green, and 2 blue balls. Two balls are drawn without replacement. What is the probability that both are of different colours?
(iii) Three coins are tossed. What is the probability that the first coin shows heads and exactly two heads occur in total?
(iv) A four-digit number is formed using the digits 1, 2, 3, and 4 with no repetition. What is the probability that the number is even?
(v) A student takes a multiple-choice test with 3 questions, each having 4 options (A, B, C, D), with only one correct answer. What is the probability that the student guesses and gets exactly 2 answers correct?
(i) Sum of 2 dice is prime $>5$:
Total outcomes $n(S) = 36$. Primes $> 5$ are $7$ and $11$.
• 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 favourable $= 6 + 2 = 8$.
$$P = \frac{8}{36} = \mathbf{\frac{2}{9} \approx 0.222}$$
Total outcomes $n(S) = 36$. Primes $> 5$ are $7$ and $11$.
• 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 favourable $= 6 + 2 = 8$.
$$P = \frac{8}{36} = \mathbf{\frac{2}{9} \approx 0.222}$$
(ii) Drawn 2 balls of different colors (4 R, 3 G, 2 B):
Total balls $= 9$.
$$P(\text{diff colors}) = 1 - P(\text{same color}) = 1 - [P(RR) + P(GG) + P(BB)]$$
$$P(RR) = \frac{4}{9} \times \frac{3}{8} = \frac{12}{72}, \quad P(GG) = \frac{3}{9} \times \frac{2}{8} = \frac{6}{72}, \quad P(BB) = \frac{2}{9} \times \frac{1}{8} = \frac{2}{72}$$
$$P(\text{same color}) = \frac{12 + 6 + 2}{72} = \frac{20}{72}$$
$$P(\text{diff colors}) = 1 - \frac{20}{72} = \frac{52}{72} = \mathbf{\frac{13}{18} \approx 0.722}$$
Total balls $= 9$.
$$P(\text{diff colors}) = 1 - P(\text{same color}) = 1 - [P(RR) + P(GG) + P(BB)]$$
$$P(RR) = \frac{4}{9} \times \frac{3}{8} = \frac{12}{72}, \quad P(GG) = \frac{3}{9} \times \frac{2}{8} = \frac{6}{72}, \quad P(BB) = \frac{2}{9} \times \frac{1}{8} = \frac{2}{72}$$
$$P(\text{same color}) = \frac{12 + 6 + 2}{72} = \frac{20}{72}$$
$$P(\text{diff colors}) = 1 - \frac{20}{72} = \frac{52}{72} = \mathbf{\frac{13}{18} \approx 0.722}$$
(iii) 3 coins: 1st shows Heads and exactly 2 Heads in total:
Sample Space $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$.
Favourable outcomes (starts with H, exactly two H's): $\{HHT, HTH\} \implies n(E) = 2$.
$$P = \frac{2}{8} = \mathbf{\frac{1}{4} = 0.25}$$
Sample Space $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$.
Favourable outcomes (starts with H, exactly two H's): $\{HHT, HTH\} \implies n(E) = 2$.
$$P = \frac{2}{8} = \mathbf{\frac{1}{4} = 0.25}$$
(iv) Even 4-digit number using 1, 2, 3, 4 without repetition:
Total permutations $= 4! = 24$.
Even number must end in $2$ or $4$.
• If ends in 2: remaining 3 digits can be arranged in $3! = 6$ ways.
• If ends in 4: remaining 3 digits can be arranged in $3! = 6$ ways.
Total even numbers $= 6 + 6 = 12$.
$$P = \frac{12}{24} = \mathbf{\frac{1}{2} = 0.50}$$
Total permutations $= 4! = 24$.
Even number must end in $2$ or $4$.
• If ends in 2: remaining 3 digits can be arranged in $3! = 6$ ways.
• If ends in 4: remaining 3 digits can be arranged in $3! = 6$ ways.
Total even numbers $= 6 + 6 = 12$.
$$P = \frac{12}{24} = \mathbf{\frac{1}{2} = 0.50}$$
(v) Guesses on 3 MCQs (each 4 options), exactly 2 correct:
For each question: $P(\text{correct}) = 1/4$, $P(\text{incorrect}) = 3/4$.
Ways to get exactly 2 correct out of 3 is $\binom{3}{2} = 3$ (CCI, CIC, ICC).
$$P = 3 \times \left(\frac{1}{4}\right)^2 \times \frac{3}{4} = 3 \times \frac{1}{16} \times \frac{3}{4} = \mathbf{\frac{9}{64} \approx 0.141}$$
For each question: $P(\text{correct}) = 1/4$, $P(\text{incorrect}) = 3/4$.
Ways to get exactly 2 correct out of 3 is $\binom{3}{2} = 3$ (CCI, CIC, ICC).
$$P = 3 \times \left(\frac{1}{4}\right)^2 \times \frac{3}{4} = 3 \times \frac{1}{16} \times \frac{3}{4} = \mathbf{\frac{9}{64} \approx 0.141}$$
(i) 2/9 (ii) 13/18 (iii) 1/4 (iv) 1/2 (v) 9/64
Q13: Double Ball Draw Sample Space
A box contains 4 balls numbered 1 to 4. Record a sample space using a tree diagram for:
(i) A ball is drawn, and the number is recorded. Then the ball is returned, and a second ball is drawn and recorded.
(ii) A ball is drawn and recorded. Without replacing the first ball, the experimenter draws and records a second ball.
(iii) What are the sizes of these two sample spaces?
(i) A ball is drawn, and the number is recorded. Then the ball is returned, and a second ball is drawn and recorded.
(ii) A ball is drawn and recorded. Without replacing the first ball, the experimenter draws and records a second ball.
(iii) What are the sizes of these two sample spaces?
(i) With Replacement:
Since the ball is returned, the second draw also has outcomes $1, 2, 3, 4$.
$$S_1 = \{(1,1), (1,2), (1,3), (1,4), (2,1), \dots, (4,4)\}$$
Size $= \mathbf{16}$.
Since the ball is returned, the second draw also has outcomes $1, 2, 3, 4$.
$$S_1 = \{(1,1), (1,2), (1,3), (1,4), (2,1), \dots, (4,4)\}$$
Size $= \mathbf{16}$.
(ii) Without Replacement:
Since the first ball is not returned, duplicate pairs (e.g. $(1,1)$) are impossible.
$$S_2 = \{(1,2), (1,3), (1,4), (2,1), (2,3), (2,4), (3,1), (3,2), (3,4), (4,1), (4,2), (4,3)\}$$
Size $= \mathbf{12}$.
Since the first ball is not returned, duplicate pairs (e.g. $(1,1)$) are impossible.
$$S_2 = \{(1,2), (1,3), (1,4), (2,1), (2,3), (2,4), (3,1), (3,2), (3,4), (4,1), (4,2), (4,3)\}$$
Size $= \mathbf{12}$.
(iii) Sizes:
• Size of Sample Space (i) $= \mathbf{16}$.
• Size of Sample Space (ii) $= \mathbf{12}$.
• Size of Sample Space (i) $= \mathbf{16}$.
• Size of Sample Space (ii) $= \mathbf{12}$.
Sizes: (i) 16, (ii) 12
Q14: Coin & Card Drawing
List the elements of a sample space for the simultaneous tossing of a coin and drawing of a card from a set of 6 cards numbered 1 through 6.
Coin outcomes: $\{H, T\}$. Card outcomes: $\{1, 2, 3, 4, 5, 6\}$.
$$S = \mathbf{\{(H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)\}}$$
Size $= 2 \times 6 = 12$.
$$S = \mathbf{\{(H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)\}}$$
Size $= 2 \times 6 = 12$.
S = {(H,1)...(H,6), (T,1)...(T,6)} (size 12)
Q15: Head Count Sample Space
Three coins are tossed, and the number of heads is recorded. Which of the following lists is a sample space for this experiment? Why do the other lists fail to qualify as a sample space?
(i) $\{1, 2, 3\}$
(ii) $\{0, 1, 2\}$
(iii) $\{0, 1, 2, 3, 4\}$
(iv) $\{0, 1, 2, 3\}$
(i) $\{1, 2, 3\}$
(ii) $\{0, 1, 2\}$
(iii) $\{0, 1, 2, 3, 4\}$
(iv) $\{0, 1, 2, 3\}$
When tossing three coins, the number of heads obtained can be $0$ (all Tails), $1$, $2$, or $3$ (all Heads).
Therefore, the set of all possible outcomes is **$\{0, 1, 2, 3\}$** (list iv).
Therefore, the set of all possible outcomes is **$\{0, 1, 2, 3\}$** (list iv).
Why others fail:
• **(i) $\{1, 2, 3\}$** fails because it omits the possible outcome of $0$ heads ($TTT$).
• **(ii) $\{0, 1, 2\}$** fails because it omits the possible outcome of $3$ heads ($HHH$).
• **(iii) $\{0, 1, 2, 3, 4\}$** fails because it includes $4$, which is impossible when only tossing $3$ coins.
• **(i) $\{1, 2, 3\}$** fails because it omits the possible outcome of $0$ heads ($TTT$).
• **(ii) $\{0, 1, 2\}$** fails because it omits the possible outcome of $3$ heads ($HHH$).
• **(iii) $\{0, 1, 2, 3, 4\}$** fails because it includes $4$, which is impossible when only tossing $3$ coins.
List (iv) {0, 1, 2, 3} is the correct sample space.
Q16: Geometric Area Probability
Suppose you drop a die at random on the rectangular region of dimensions $3\text{ m} \times 2\text{ m}$. What is the probability that it will land inside the circle with a diameter of $1\text{ m}$?
Step 1: Calculate Area of Rectangular Region (Total Area):
$$\text{Area of Rectangle} = \text{Length} \times \text{Width} = 3\text{ m} \times 2\text{ m} = 6\text{ sq. m}$$
$$\text{Area of Rectangle} = \text{Length} \times \text{Width} = 3\text{ m} \times 2\text{ m} = 6\text{ sq. m}$$
Step 2: Calculate Area of Circle (Favourable Area):
Diameter $D = 1\text{ m} \implies \text{Radius } r = 0.5\text{ m}$.
$$\text{Area of Circle} = \pi r^2 = \pi (0.5)^2 = 0.25\pi\text{ sq. m}$$
Diameter $D = 1\text{ m} \implies \text{Radius } r = 0.5\text{ m}$.
$$\text{Area of Circle} = \pi r^2 = \pi (0.5)^2 = 0.25\pi\text{ sq. m}$$
Step 3: Calculate Probability:
$$P = \frac{\text{Area of Circle}}{\text{Area of Rectangle}} = \frac{0.25\pi}{6} = \frac{\pi}{24} \approx \mathbf{0.131 \text{ (or 13.1\%)}}$$
$$P = \frac{\text{Area of Circle}}{\text{Area of Rectangle}} = \frac{0.25\pi}{6} = \frac{\pi}{24} \approx \mathbf{0.131 \text{ (or 13.1\%)}}$$
P = $\pi / 24 \approx 0.131$