3. Probability

# 3. Probability ## Objectifs et prérequis By the end of this chapter, students will be able to: - Define key probability terms such as experiment, outcome, sample space, event, and probability. - Construct sample spaces for simple and compound experiments using lists, tables, and tree diagrams. - Calculate the probability of an event using the classical formula and interpret it on the probability scale [0,1]. - Determine the probability of complementary events and apply the complement rule. - Calculate probabilities of compound events using addition and multiplication rules. - Distinguish between mutually exclusive and independent events and apply the correct rules accordingly. - Compute conditional probability using the definition and interpret it in context. - Construct and use probability tree diagrams to model successive events and calculate combined probabilities. - Solve multi-step problems involving successive events with and without replacement. Prerequisites: - Proficiency in fractions, decimals, and percentages. - Understanding of set notation (union, intersection, complement). - Ability to interpret and draw tree diagrams. - Familiarity with basic counting principles (multiplication principle). --- ## I. Probability vocabulary and sample space ### Definition of key terms An **experiment** is a process that leads to well-defined results called **outcomes**. The **sample space** (denoted $S$) is the set of all possible outcomes of an experiment. An **event** is a subset of the sample space. > **To remember**: The sample space $S$ contains all possible outcomes; an event $E \subseteq S$ is any collection of outcomes. ### Examples of sample spaces - Tossing a fair coin once: $S = \{H, T\}$ - Rolling a fair six-sided die: $S = \{1, 2, 3, 4, 5, 6\}$ - Tossing two coins: $S = \{HH, HT, TH, TT\}$ - Drawing a card from a standard deck: $S$ has 52 elements (e.g., $A\spadesuit, 2\spadesuit, \dots, K\heartsuit$) ### Worked example: Sample space for two dice When two fair dice are rolled, the sample space consists of all ordered pairs $(i,j)$ where $i$ is the result of the first die and $j$ the second. $$ S = \{(1,1), (1,2), \dots, (1,6), (2,1), \dots, (6,6)\} $$ Total outcomes: $6 \times 6 = 36$. Let $E$ be the event “sum is 7”. Then: $$ E = \{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\} \Rightarrow |E| = 6 $$ ### Exercises with solutions **Exercise 1**: A coin is tossed and a die is rolled. List the sample space and find the number of outcomes. **Solution**: $S = \{H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6\}$ $|S| = 2 \times 6 = 12$ **Exercise 2**: From the sample space of two dice, list the outcomes for the event “both dice show the same number”. **Solution**: $\{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\}$ → 6 outcomes --- ## II. The probability of an event and the probability scale ### Classical probability formula If all outcomes in the sample space are equally likely, the probability of an event $E$ is: $$ P(E) = \frac{\text{Number of outcomes in } E}{\text{Total number of outcomes in } S} = \frac{|E|}{|S|} $$ > **To remember**: Probability is a number between 0 and 1 inclusive: $0 \leq P(E) \leq 1$. > $P(E) = 0$ means impossible event; $P(E) = 1$ means certain event. ### Probability scale The probability scale ranges from 0 (impossible) to 1 (certain). Events can be described as: - Impossible: $P = 0$ - Unlikely: $0 < P < 0.5$ - Even chance: $P = 0.5$ - Likely: $0.5 < P < 1$ - Certain: $P = 1$ ### Worked example: Probability with dice Using the two-dice sample space ($|S| = 36$): - $P(\text{sum} = 7) = \frac{6}{36} = \frac{1}{6}$ - $P(\text{sum} > 10) = P(\text{sum}=11 \text{ or } 12) = \frac{2 + 1}{36} = \frac{3}{36} = \frac{1}{12}$ - $P(\text{double}) = \frac{6}{36} = \frac{1}{6}$ ### Exercises with solutions **Exercise 1**: A fair coin is tossed three times. Find the probability of getting exactly two heads. **Solution**: $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$ → $|S| = 8$ $E = \{HHT, HTH, THH\}$ → $|E| = 3$ $P(E) = \frac{3}{8}$ **Exercise 2**: A bag contains 5 red, 3 blue, and 2 green balls. One ball is drawn at random. Find $P(\text{not green})$. **Solution**: Total balls = $5+3+2=10$ Not green = red or blue = $5+3=8$ $P(\text{not green}) = \frac{8}{10} = \frac{4}{5}$ --- ## III. Complementary and compound events ### Complementary events The complement of an event $E$, denoted $E'$ or $\overline{E}$, consists of all outcomes in $S$ not in $E$. $$ P(E') = 1 - P(E) $$ > **To remember**: $P(E) + P(E') = 1$ — this is the complement rule. ### Compound events A compound event is formed by combining two or more simple events using “and” (intersection) or “or” (union). - **Union** ($E \cup F$): outcomes in $E$ or $F$ or both. - **Intersection** ($E \cap F$): outcomes in both $E$ and $F$. ### Addition rule For any two events $E$ and $F$: $$ P(E \cup F) = P(E) + P(F) - P(E \cap F) $$ If $E$ and $F$ are mutually exclusive, then $P(E \cap F) = 0$, so: $$ P(E \cup F) = P(E) + P(F) $$ > **To remember**: Subtract the overlap when using “or” to avoid double-counting. ### Worked example: Cards From a standard deck of 52 cards: - Let $A$ = drawing a heart, $B$ = drawing a king $P(A) = \frac{13}{52} = \frac{1}{4}$, $P(B) = \frac{4}{52} = \frac{1}{13}$, $P(A \cap B) = \frac{1}{52}$ (king of hearts) $P(A \cup B) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$ ### Exercises with solutions **Exercise 1**: In a class of 30 students, 18 play football, 12 play basketball, and 5 play both. Find the probability that a randomly selected student plays football or basketball. **Solution**: $P(F) = \frac{18}{30}$, $P(B) = \frac{12}{30}$, $P(F \cap B) = \frac{5}{30}$ $P(F \cup B) = \frac{18}{30} + \frac{12}{30} - \frac{5}{30} = \frac{25}{30} = \frac{5}{6}$ **Exercise 2**: A die is rolled. Find $P(\text{odd or greater than 4})$. **Solution**: Odd: $\{1,3,5\}$ → $P = \frac{3}{6}$ >4: $\{5,6\}$ → $P = \frac{2}{6}$ Intersection: $\{5\}$ → $P = \frac{1}{6}$ $P = \frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$ --- ## IV. Mutually exclusive and independent events ### Mutually exclusive events Two events are **mutually exclusive** (or disjoint) if they cannot occur at the same time: $E \cap F = \emptyset$, so $P(E \cap F) = 0$. > **To remember**: For mutually exclusive events, $P(E \cup F) = P(E) + P(F)$. ### Independent events Two events $E$ and $F$ are **independent** if the occurrence of one does not affect the probability of the other: $$ P(E \cap F) = P(E) \cdot P(F) $$ Equivalently, $P(E|F) = P(E)$ and $P(F|E) = P(F)$. > **To remember**: For independent events, use multiplication rule: $P(E \text{ and } F) = P(E) \times P(F)$. ### Key distinction - Mutually exclusive: $P(E \cap F) = 0$ → they cannot happen together. - Independent: $P(E \cap F) = P(E)P(F)$ → they can happen together, and one doesn’t influence the other. Note: Mutually exclusive events with $P>0$ are **not** independent (since $P(E \cap F)=0 \neq P(E)P(F)$). ### Worked example: Coins and dice Toss a coin and roll a die. Let $E$ = head on coin, $F$ = six on die. $P(E) = \frac{1}{2}$, $P(F) = \frac{1}{6}$, $P(E \cap F) = \frac{1}{12}$ Since $\frac{1}{12} = \frac{1}{2} \times \frac{1}{6}$, $E$ and $F$ are independent. Are they mutually exclusive? No, because $(H,6)$ is a valid outcome → $P(E \cap F) > 0$. ### Exercises with solutions **Exercise 1**: A box contains 4 red and 6 blue balls. Two balls are drawn with replacement. Let $R_1$ = first ball red, $R_2$ = second ball red. Are $R_1$ and $R_2$ independent? **Solution**: $P(R_1) = \frac{4}{10} = 0.4$, $P(R_2) = 0.4$ (same due to replacement) $P(R_1 \cap R_2) = \frac{4}{10} \times \frac{4}{10} = 0.16$ Since $0.16 = 0.4 \times 0.4$, they are independent. **Exercise 2**: From the same box (without replacement), are $R_1$ and $R_2$ independent? **Solution**: $P(R_1) = 0.4$ $P(R_2 | R_1) = \frac{3}{9} = \frac{1}{3} \approx 0.333 \neq 0.4$ So $P(R_1 \cap R_2) = P(R_1) \cdot P(R_2|R_1) = 0.4 \times \frac{1}{3} = \frac{2}{15}$ But $P(R_1) \cdot P(R_2) = 0.4 \times 0.4 = 0.16 = \frac{8}{50} \neq \frac{2}{15}$ Thus, not independent. --- ## V. Conditional probability ### Definition The conditional probability of event $E$ given that $F$ has occurred is: $$ P(E|F) = \frac{P(E \cap F)}{P(F)}, \quad \text{provided } P(F) > 0 $$ > **To remember**: $P(E|F)$ is the probability of $E$ assuming $F$ has already happened. ### Interpretation - $P(E|F)$ focuses on the reduced sample space $F$. - If $E$ and $F$ are independent, then $P(E|F) = P(E)$. ### Worked example: Weather and traffic Suppose: - $P(\text{rain}) = 0.3$ - $P(\text{traffic jam}) = 0.4$ - $P(\text{rain and traffic jam}) = 0.12$ Then: $$ P(\text{traffic jam} | \text{rain}) = \frac{0.12}{0.3} = 0.4 $$ Since this equals $P(\text{traffic jam})$, the events are independent. Now suppose $P(\text{rain and traffic jam}) = 0.09$: $$ P(\text{traffic jam} | \text{rain}) = \frac{0.09}{0.3} = 0.3 \neq 0.4 $$ So they are not independent. ### Exercises with solutions **Exercise 1**: In a school, 60% of students take Mathematics, 40% take Physics, and 25% take both. Find the probability that a student takes Physics given that they take Mathematics. **Solution**: Let $M$ = Maths, $P$ = Physics $P(P|M) = \frac{P(M \cap P)}{P(M)} = \frac{0.25}{0.60} = \frac{5}{12} \approx 0.4167$ **Exercise 2**: A bag has 3 black and 5 white balls. Two balls are drawn without replacement. Find the probability that the second ball is white given that the first was black. **Solution**: After removing one black, 2 black and 5 white remain → total 7 balls $P(\text{second white} | \text{first black}) = \frac{5}{7}$ --- ## VI. Probability tree diagram and successive events ### Tree diagrams A probability tree diagram represents successive events. Each branch is labeled with an outcome and its probability. Probabilities along a path are multiplied to get the probability of that sequence. > **To remember**: > - Multiply probabilities along branches (for “and”). > - Add probabilities of end branches (for “or”). ### Successive events with and without replacement - **With replacement**: probabilities remain constant across trials. - **Without replacement**: probabilities change after each draw (dependent events). ### Worked example: Drawing balls without replacement Urn: 3 red (R), 2 blue (B). Draw two balls without replacement. **Tree diagram**: - First draw: - $P(R) = \frac{3}{5}$ → then state: 2R, 2B - $P(B) = \frac{2}{5}$ → then state: 3R, 1B - Second draw from each: - From (2R,2B): - $P(R) = \frac{2}{4} = \frac{1}{2}$, $P(B) = \frac{2}{4} = \frac{1}{2}$ - From (3R,1B): - $P(R) = \frac{3}{4}$, $P(B) = \frac{1}{4}$ **Calculate $P(\text{one red and one blue})$**: Two paths: 1. R then B: $\frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = \frac{3}{10}$ 2. B then R: $\frac{2}{5} \times \frac{3}{4} = \frac{6}{20} = \frac{3}{10}$ Total: $\frac{3}{10} + \frac{3}{10} = \frac{6}{10} = \frac{3}{5}$ ### Worked example: With replacement Same urn, but with replacement. - $P(R) = \frac{3}{5}$, $P(B) = \frac{2}{5}$ each time, independent. $P(\text{one red and one blue})$: - R then B: $\frac{3}{5} \times \frac{2}{5} = \frac{6}{25}$ - B then R: $\frac{2}{5} \times \frac{3}{5} = \frac{6}{25}$ Total: $\frac{12}{25}$ ### Exercises with solutions **Exercise 1**: A box contains 4 defective and 6 good bulbs. Two bulbs are drawn at random without replacement. Find the probability that both are good. **Solution**: - First good: $\frac{6}{10}$ - Second good: $\frac{5}{9}$ - $P = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}$ **Exercise 2**: A fair coin is tossed and then a die is rolled. Use a tree diagram to find $P(\text{head and even number})$. **Solution**: - $P(H) = \frac{1}{2}$ - From H, $P(\text{even}) = P(2,4,6) = \frac{3}{6} = \frac{1}{2}$ - $P(H \text{ and even}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ --- ## Synthèse du chapitre - **Vocabulary and sample space**: Understand experiment, outcome, sample space $S$, event $E \subseteq S$. List outcomes systematically. - **Probability of an event**: Use $P(E) = \frac{|E|}{|S|}$ for equally likely outcomes; interpret on scale $[0,1]$. - **Complementary events**: Apply $P(E') = 1 - P(E)$. - **Compound events**: Use addition rule $P(E \cup F) = P(E) + P(F) - P(E \cap F)$; subtract overlap for “or”. - **Mutually exclusive events**: $P(E \cap F) = 0$ → $P(E \cup F) = P(E) + P(F)$. - **Independent events**: $P(E \cap F) = P(E)P(F)$ → use multiplication rule for “and”. - **Conditional probability**: $P(E|F) = \frac{P(E \cap F)}{P(F)}$; update sample space to $F$. - **Tree diagrams**: Multiply along branches for sequences; add end probabilities for alternatives; adapt for replacement/no replacement. --- ## Glossaire - **Experiment**: A process that leads to well-defined outcomes. - **Outcome**: A possible result of an experiment. - **Sample space ($S$)**: The set of all possible outcomes of an experiment. - **Event**: A subset of the sample space; a collection of outcomes. - **Probability**: A measure of the likelihood of an event, ranging from 0 to 1. - **Complementary event ($E'$)**: The event that $E$ does not occur; $P(E') = 1 - P(E)$. - **Mutually exclusive events**: Events that cannot occur together; $P(E \cap F) = 0$. - **Independent events**: Events where one does not affect the probability of the other; $P(E \cap F) = P(E)P(F)$. - **Conditional probability ($P(E|F)$)**: The probability of $E$ given that $F$ has occurred. - **Tree diagram**: A visual tool to represent successive events and compute combined probabilities. - **Replacement**: Returning an item to the sample space after selection, keeping probabilities constant. - **Without replacement**: Not returning an item, causing probabilities to change on subsequent draws.