Chapter 13: Probability — Comprehensive Solved Notes & Solution Manual
Mastery Guide: Probability — Sample Space, Events, Complementary Events & Tree Diagrams
Comprehensive student mastery guide, formulas, and step-by-step solved textbook manual for Class 8 Mathematics.
📖 Unit Overview & Target Learning Outcomes
By mastering this unit, Class 8 students will be able to:
- Understand Statistical Experiments & Outcomes: Define an experiment whose results depend on chance and identify all individual possible outcomes.
- Define and Construct Sample Spaces ($S$): List the universal set of all possible outcomes for single and multi-stage experiments and state $n(S)$.
- Analyze Events ($E$) & Calculate Theoretical Probability: Define an event as a subset of the sample space ($E \subseteq S$) and calculate: $$P(E) = \frac{n(E)}{n(S)} = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$$
- Interpret the Probability Continuum: Recognize that $0 \le P(E) \le 1$, where $P=0$ denotes an impossible event and $P=1$ denotes a certain event.
- Identify Mutually Exclusive Events: Determine when two events cannot occur simultaneously ($A \cap B = \emptyset \implies P(A \text{ or } B) = P(A) + P(B)$).
- Calculate Complementary Probabilities: Apply the complement rule $P(\text{not } E) = 1 - P(E)$ to solve probability problems.
- Construct & Interpret Tree Diagrams: Track and display compound multi-stage events with branching paths.
💡 Kid-Friendly Tips & Memory Hooks
Probability can NEVER be negative and can NEVER be greater than 1 (or 100%). If your answer is $-0.2$ or $1.5$, re-check your calculations!
When asked for "not getting something", just calculate the chance of getting it first, then subtract from 1: $$P(\text{Not } E) = 1 - P(E)$$
To find total outcomes of multi-stage decisions, multiply branch choices: $(n_1 \text{ options}) \times (n_2 \text{ options})$.
🌍 Real-World Connections
- Meteorology & Daily Forecasts: A "70% chance of rain" means that under identical atmospheric conditions, historical data shows precipitation in 70 out of 100 days.
- Board Games & Game Theory (Ludo, Monopoly): Rolling a single 6 on a fair die occurs $\frac{1}{6} \approx 16.7\%$ of the time.
- Quality Assurance & Manufacturing: Factories test random batches of microchips to compute the probability of defective products before shipping.
- Medical Diagnostic Testing: Epidemiologists use probability models and sample spaces to evaluate disease incidence and diagnostic screening accuracy.
🔑 Study Cues & Essential Inquiries
- If you flip a fair coin 5 times and get 5 Heads in a row, is the 6th flip more likely to be Tails? (No! Coin flips are independent; $P(\text{Head}) = 0.5$ always).
- Why can two mutually exclusive events never happen at the same time?
- How does replacing an item (with replacement) vs. keeping it (without replacement) alter branch probabilities in a tree diagram?
- What is the relationship between an event $E$ and the sample space $S$? (An event is always a subset: $E \subseteq S$).
🌟 Section-by-Section Conceptual Breakdown
13.1 Experiments, Sample Space & The Probability Scale
An experiment is any planned activity or trial that yields one or more unpredictable outcomes governed by chance.
- Sample Space ($S$): The complete set of all possible outcomes. E.g., for a fair 6-sided die: $$S = \{1, 2, 3, 4, 5, 6\}, \quad n(S) = 6$$
- Event ($E$): Any subset of the sample space representing a specific outcome of interest. E.g., rolling an odd number: $$E = \{1, 3, 5\}, \quad n(E) = 3 \implies P(E) = \frac{3}{6} = \frac{1}{2}$$
13.2 Mutually Exclusive & Complementary Events
- Mutually Exclusive Events: Two events $A$ and $B$ cannot happen together ($A \cap B = \emptyset$). E.g., rolling an odd and an even number on a single die. $$P(A \text{ or } B) = P(A) + P(B)$$
- Complementary Events: $P(\text{not } E) = 1 - P(E)$.
13.3 Multi-Stage Combined Events & Tree Diagrams
A Tree Diagram tracks every branch and compound outcome across multiple stages:
🎯 Unit Synthesis Summary
Unit 13 bridges foundational set theory and real-world predictive modeling. Every probability question begins with identifying the random experiment, establishing the universal sample space ($S$), and isolating the subset of favorable outcomes ($E$). Probabilities are bounded between $0$ and $1$. The complement rule ($P(\text{not } E) = 1 - P(E)$) and mutually exclusive addition rule ($P(A \cup B) = P(A) + P(B)$) simplify complex multi-condition calculations, while tree diagrams provide systematic visualization of compound random trials.
📝 Solved Textbook Exercises — Step-by-Step Manual
Exercise 13.1 • Sample Spaces, Single Events & Complementary Rules
Q1. Jar containing 12 cards numbered 1 to 12:
• (i) Sample space $S$: $\mathbf{S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}, \quad n(S) = 12}$.
• (ii) Multiples of 4 ($E$): $\mathbf{E = \{4, 8, 12\}, \quad n(E) = 3}$.
Q2. Box with 1 red, 2 green, 2 white marbles:
• (i) Sample space $S$: $\mathbf{S = \{R, G_1, G_2, W_1, W_2\}, \quad n(S) = 5}$.
• (ii) Event not white ($E$): $\mathbf{E = \{R, G_1, G_2\}, \quad n(E) = 3}$.
Q3. School buses: 2 red, 2 blue, 4 white:
• (i) Outcomes: $\mathbf{S = \{R_1, R_2, B_1, B_2, W_1, W_2, W_3, W_4\}}$.
• (ii) Total: $\mathbf{n(S) = 8}$.
Q4. 18 jellybeans (8 red, 1 white, 3 yellow, 2 green, 2 black, 2 orange):
$$P(\text{Red}) = \frac{8}{18} = \mathbf{\frac{4}{9}}$$
Q5. Ayesha chooses from 4 hats:
• (i) Purple hat: $P = \mathbf{\frac{1}{4} = 0.25}$.
• (ii) Hat with black bow: $P = \mathbf{\frac{1}{4} = 0.25}$.
Q6. Letters of the word 'CHANCE' ($n = 6$):
• (i) Letter 'H': $P = \mathbf{\frac{1}{6}}$.
• (ii) Letter 'C': $P = \frac{2}{6} = \mathbf{\frac{1}{3}}$.
• (iii) Consonant (C, H, N, C): $P = \frac{4}{6} = \mathbf{\frac{2}{3}}$.
Q7. 8-sector spinner (1 to 8):
• (i) Any sector: $P = \frac{8}{8} = \mathbf{1}$.
• (ii) Even sector ($2, 4, 6, 8$): $P = \frac{4}{8} = \mathbf{\frac{1}{2}}$.
• (iii) Sector $> 5$ ($6, 7, 8$): $P = \mathbf{\frac{3}{8}}$.
Q8. Standard pack of 52 playing cards:
• (i) Queen of hearts: $P = \mathbf{\frac{1}{52}}$.
• (ii) A heart: $P = \frac{13}{52} = \mathbf{\frac{1}{4}}$.
• (iii) A red card: $P = \frac{26}{52} = \mathbf{\frac{1}{2}}$.
• (iv) A picture card (12 face cards): $P = \frac{12}{52} = \mathbf{\frac{3}{13}}$.
Q9. Number chosen from {1, 2, ..., 20}:
• (i) Multiple of 6 ($6, 12, 18$): $P = \mathbf{\frac{3}{20}}$.
• (ii) Prime number ($2, 3, 5, 7, 11, 13, 17, 19$): $P = \frac{8}{20} = \mathbf{\frac{2}{5}}$.
• (iii) Perfect cube ($1, 8$): $P = \frac{2}{20} = \mathbf{\frac{1}{10}}$.
Q10. 300 lottery tickets (Nahyan has 6):
• (i) Winning ticket: $P = \frac{6}{300} = \mathbf{\frac{1}{50} = 0.02}$.
• (ii) Not winning ticket: $P = 1 - \frac{1}{50} = \mathbf{\frac{49}{50} = 0.98}$.
Q11. Fair 6-sided dice:
• (i) Getting a 3: $P(3) = \mathbf{\frac{1}{6}}$.
• (ii) Not getting a 3: $P(\text{not } 3) = 1 - \frac{1}{6} = \mathbf{\frac{5}{6}}$.
Q12. Fair 6-sided dice with P(6) = 1/6:
• (i) Not getting a 6: $P(\text{not } 6) = \mathbf{\frac{5}{6}}$.
• (ii) Either 6 or not 6: $P = \frac{1}{6} + \frac{5}{6} = \mathbf{1}$ (Certain event).
Q13. 2 red, 3 green, 4 yellow balls (draw with replacement):
• (i) $X$ (both red) & $Y$ (at least one red): Not mutually exclusive (Outcome $(R, R)$ belongs to both).
• (ii) $X$ (both red) & $Z$ (at least one yellow): Mutually exclusive ($X \cap Z = \emptyset$).
• (iii) $Y$ (at least one red) & $Z$ (at least one yellow): Not mutually exclusive (Outcome $(R, Y)$ belongs to both).
Exercise 13.2 • Multi-Stage Combined Events & Tree Diagrams
Q1. Tree diagrams & sample spaces:
• (i) Soup/Salad with Beef/Chicken/Fish/Pasta: $\mathbf{2 \times 4 = 8\text{ outcomes}}$.
• (ii) Basketball/Park/Concert on Friday/Saturday: $\mathbf{3 \times 2 = 6\text{ outcomes}}$.
Q2. Outfits (Jeans/Khaki pants with Yellow/White/Blue shirt):
$\mathbf{2 \times 3 = 6\text{ outfits}}: \{(J,Y), (J,W), (J,B), (K,Y), (K,W), (K,B)\}$.
Q3. Rahat, Ghazala, Lalarukh line up permutations:
$\mathbf{3! = 6\text{ ways}}: \{(R,G,L), (R,L,G), (G,R,L), (G,L,R), (L,R,G), (L,G,R)\}$.
Q4. Burger and Drink combinations:
$\mathbf{2 \times 3 = 6\text{ combinations}}: \{(C,S), (C,W), (C,J), (F,S), (F,W), (F,J)\}$.
Q5. Red, blue, green socks (two picks):
$\mathbf{3 \times 3 = 9\text{ outcomes}}: \{(R,R), (R,B), (R,G), (B,R), (B,B), (B,G), (G,R), (G,B), (G,G)\}$.
Q6. 1 black, 1 white ball (draw, replace, draw again):
$\mathbf{2 \times 2 = 4\text{ outcomes}}: \{(B,B), (B,W), (W,B), (W,W)\}$.
Q7. Bag 1 (Red, White, Yellow) and Bag 2 (Red, Blue):
$\mathbf{3 \times 2 = 6\text{ outcomes}}: \{(R,R), (R,B), (W,R), (W,B), (Y,R), (Y,B)\}$.
Q8. Ice cream choices (Vanilla, Chocolate for 2 buyers):
$\mathbf{2 \times 2 = 4\text{ outcomes}}: \{(V,V), (V,C), (C,V), (C,C)\}$.
Review Exercise 13 • Official Solutions
- Probability 0 $\implies$ (a) Impossible
- Probability 1 $\implies$ (d) certain to happen
- Set of all possible outcomes $\implies$ (b) Sample space
- Trial with two combined events $\implies$ (d) Tree diagram
- Chance of happening $\implies$ (d) likelihood
- Low chance of happening $\implies$ (d) unlikely
- Queen from 52 cards $\implies$ (a) 1/13 ($\frac{4}{52} = \frac{1}{13}$)
- Odd number on dice $\implies$ (c) 3/6 (or 1/2)
- Prime number on dice $\implies$ (c) 1/2
- Head on coin toss $\implies$ (b) 1/2
- Q2 (Dice): (i) Multiple of 3: $\mathbf{\frac{1}{3}}$, (ii) Less than 7: $\mathbf{1}$.
- Q3 (36 Discs): (i) Marked ($17$): $\mathbf{\frac{17}{36}}$, (ii) Unmarked ($19$): $\mathbf{\frac{19}{36}}$.
- Q4 (52 Cards): (i) Picture card ($12$): $\mathbf{\frac{3}{13}}$, (ii) Not a picture card: $\mathbf{\frac{10}{13}}$.
- Q5 (24 Red, $x$ White, $P(W) = \frac{3}{7}$): (i) $P(\text{Red}) = \mathbf{\frac{4}{7}}$, (ii) $\mathbf{x = 18\text{ white balls}}$.
- Q6 (30 Apples): (i) $P(\text{Green}) = \mathbf{\frac{8}{15}}$, (ii) Red apples removed for $P = \frac{2}{3}$: $\mathbf{6\text{ red apples}}$.
- Q7 (52 Cards): (i) Black card: $\mathbf{\frac{1}{2}}$, (ii) King: $\mathbf{\frac{1}{13}}$, (iii) Multiple of 3 card ($3,6,9$): $\mathbf{\frac{3}{13}}$, (iv) Ace/Jack/Queen/King: $\mathbf{\frac{4}{13}}$.
- Q8 (10-sided dice): (i) $1$: $\mathbf{\frac{1}{10}}$, (ii) Even: $\mathbf{\frac{1}{2}}$, (iii) $<4$ or $>7$: $\mathbf{\frac{3}{5}}$.
- Q9 (Hexagon spinner 1-6): (i) Odd: $\mathbf{\frac{1}{2}}$, (ii) Multiple of 3: $\mathbf{\frac{1}{3}}$, (iii) $<6$: $\mathbf{\frac{5}{6}}$, (iv) $>6$: $\mathbf{0}$, (v) $<7$: $\mathbf{1}$.
- Q10 (3 Flavors, 2 buyers): $\mathbf{9\text{ outcomes}}: \{(V,V), (V,C), (V,S), (C,V), (C,C), (C,S), (S,V), (S,C), (S,S)\}$.
- Q11 (40 Students: 20 Math, 25 Stats): Not mutually exclusive ($20 + 25 = 45 > 40 \implies \text{at least } 5\text{ students take both}$).
- Q12 (52 Cards Union): (i) King or Queen: $\mathbf{\frac{2}{13}}$, (ii) Queen or Spade: $\frac{4 + 13 - 1}{52} = \mathbf{\frac{4}{13}}$.
❓ Frequently Asked Questions — Class 8 Mathematics Chapter 13
More Chapter Notes for Class 8 (FBISE)
MathematicsTest Your Knowledge on Chapter 13: Chapter 13: Probability — Comprehensive Solved Notes & Solution Manual
Practice textbook-aligned solved MCQs with instant answer feedback, step-by-step solutions, and timed test simulation.
Class 8 Mathematics - Ch 13: Probability Chapter Mock Test
Test your complete conceptual mastery across all chapters under real board exam conditions with official timer, anti-cheat surveillance, and instant grading.