Probability Foundations, Scale & Complements
Probability measures the likelihood of an event occurring on a real scale between $0$ (impossible) and $1$ (certain).
- Theoretical Probability (Equally Likely Outcomes): $$P(A) = \frac{n(A)}{n(U)} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes in sample space } U}$$
- Experimental Probability (Relative Frequency): $$\text{Relative Frequency} = \frac{\text{Number of times event observed}}{\text{Total number of experimental trials}}$$
- Law of Large Numbers: As the number of trials increases indefinitely ($n \to \infty$), the experimental relative frequency converges toward the true theoretical probability.
- Complementary Events Rule: The probability of event $A$ not occurring ($A'$): $$P(A') = 1 - P(A) \iff P(A) + P(A') = 1$$
- Expected Frequency: If an experiment is repeated $n$ times: $$\text{Expected Occurrences} = n \times P(A)$$
Combined Events: Addition Rules & Mutual Exclusivity
When calculating the probability that event $A$ OR event $B$ occurs ($P(A \cup B)$), the existence of an overlap determines the formula.
Events that cannot happen simultaneously ($A \cap B = \emptyset \implies P(A \cap B) = 0$): $$P(A \cup B) = P(A) + P(B)$$ Example: Rolling a die and getting an odd number ($1, 3, 5$) OR an even number ($2, 4, 6$).
When events can happen at the same time, the intersection is counted twice and must be subtracted once: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ Example: Drawing a card from a deck that is an Ace ($A$) OR a Heart ($B$): $$P(\text{Ace} \cup \text{Heart}) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$$
Independent Events & Multiplication Rules
Two events $A$ and $B$ are independent if the occurrence of one event has no effect whatsoever on the probability of the other event.
- With Replacement: The selected item is returned before the next draw. Trials remain identical and independent ($P$ remains constant).
- Without Replacement: The selected item is NOT returned. The sample space shrinks and remaining counts change, creating dependent events.
Probability Modeling: Sample Grids & Tree Diagrams
Visual probability structures systematically map multi-stage random experiments.
- Multiplication Along Branches: To find the probability of a specific sequence of consecutive events, multiply probabilities along the path: $$P(\text{Path}) = P(E_1) \times P(E_2)$$
- Addition Across Branches: If an outcome can occur via multiple distinct paths, add their path probabilities together.
- Conservation of Probability: The sum of probabilities branching out from any single node must always equal $1$. The sum of all final outcome paths must equal $1$.
Two-stage tree diagram: Multiply probabilities along branches, sum probabilities of valid paths.
Worked Example 1: Venn Diagram & Addition Rule
In a group of 40 students, 24 study French ($F$), 18 study Spanish ($S$), and 8 study neither language.
a) Complete a Venn diagram representing this information.
b) Find the probability that a randomly chosen student studies both languages: $P(F \cap S)$.
c) Find the probability that a student studies French or Spanish: $P(F \cup S)$.
Step-by-step Solution:
Part a & b: Both Languages $P(F \cap S)$
Total students $n(U) = 40$. Neither $= 8 \implies n(F \cup S) = 40 - 8 = 32$. [M1]
Using the addition rule: $n(F \cup S) = n(F) + n(S) - n(F \cap S)$
$32 = 24 + 18 - n(F \cap S) \implies 32 = 42 - n(F \cap S) \implies n(F \cap S) = 10$. [A1]
Therefore, $P(F \cap S) = \frac{10}{40} = \frac{1}{4} = 0.25$. [A1]
French only: $24 - 10 = 14$; Spanish only: $18 - 10 = 8$.
Part c: French or Spanish $P(F \cup S)$
$P(F \cup S) = \frac{n(F \cup S)}{n(U)} = \frac{32}{40} = \frac{4}{5} = 0.80$. [Criterion A & C: Knowledge & Communication]
Worked Example 2: Tree Diagram Without Replacement
A bag contains 5 red marbles and 3 blue marbles (total 8 marbles). Two marbles are drawn at random one after another without replacement.
a) Draw a probability tree diagram showing all probabilities.
b) Calculate the probability that both marbles drawn are the same color.
c) Calculate the probability that at least one blue marble is drawn.
Step-by-step Solution:
Part a: Branch Probabilities:
1st draw: $P(R_1) = \frac{5}{8}$, $P(B_1) = \frac{3}{8}$. [M1]
2nd draw if Red first: $P(R_2|R_1) = \frac{4}{7}$, $P(B_2|R_1) = \frac{3}{7}$.
2nd draw if Blue first: $P(R_2|B_1) = \frac{5}{7}$, $P(B_2|B_1) = \frac{2}{7}$. [M1]
Part b: Both same color ($RR$ or $BB$):
$P(RR) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56}$.
$P(BB) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56}$.
$P(\text{Same Color}) = \frac{20}{56} + \frac{6}{56} = \frac{26}{56} = \frac{13}{28} \approx 0.464$. [A1]
Part c: At least one Blue:
Use the complement rule: $P(\text{At least one Blue}) = 1 - P(\text{No Blue}) = 1 - P(RR)$ [M1]
$P(\text{At least one Blue}) = 1 - \frac{20}{56} = \frac{36}{56} = \frac{9}{14} \approx 0.643$. [Criterion A: Conceptual Efficiency]
Tiered Practice Exercises
10 problems advancing from single-stage theoretical events and complementary calculations to overlapping Venn additions, two-stage tree diagrams, and sampling without replacement.
A fair 12-sided die numbered 1 to 12 is rolled once. Find the probability of rolling:
a) A prime number
b) A multiple of 4
c) A number greater than 8
View Solution & Markscheme
Method & Marks:
Total outcomes $n(U) = 12$.
a) Primes: $\{2, 3, 5, 7, 11\}$ (5 numbers) $\implies P(\text{Prime}) = \frac{5}{12}$. [A1]
b) Multiples of 4: $\{4, 8, 12\}$ (3 numbers) $\implies P(\text{Multiple of 4}) = \frac{3}{12} = \frac{1}{4}$. [A1]
c) Greater than 8: $\{9, 10, 11, 12\}$ (4 numbers) $\implies P(>8) = \frac{4}{12} = \frac{1}{3}$. [A1]
The probability that a bus arrives on time is $0.78$.
a) What is the probability that the bus is late?
b) In a month with 50 bus trips, how many trips are expected to be on time?
View Solution & Markscheme
Method & Marks:
a) Complement: $P(\text{Late}) = 1 - P(\text{On time}) = 1 - 0.78 = 0.22$. [A1]
b) Expected frequency $= n \times P = 50 \times 0.78 = 39\text{ trips}$. [A1]
Events $A$ and $B$ are mutually exclusive. Given $P(A) = 0.35$ and $P(B) = 0.45$, find:
a) $P(A \cap B)$
b) $P(A \cup B)$
c) $P(A' \cap B')$
View Solution & Markscheme
Method & Marks:
a) For mutually exclusive events, $P(A \cap B) = 0$. [A1]
b) $P(A \cup B) = P(A) + P(B) = 0.35 + 0.45 = 0.80$. [A1]
c) Neither occurs $= 1 - P(A \cup B) = 1 - 0.80 = 0.20$. [A1]
Two fair six-sided dice are rolled.
a) Draw a sample space grid showing all 36 outcomes of the sum of the two dice.
b) Find the probability of obtaining a sum of 7 or 11.
c) Find the probability of obtaining a sum greater than 9.
View Solution & Markscheme
Method & Marks:
Total possible outcomes $= 6 \times 6 = 36$. [M1]
b) Sum of 7: $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$ (6 outcomes).
Sum of 11: $(5,6),(6,5)$ (2 outcomes).
$P(7 \text{ or } 11) = \frac{6 + 2}{36} = \frac{8}{36} = \frac{2}{9}$. [A1]
c) Sum $> 9$: Sum 10: $(4,6),(5,5),(6,4)$ (3 outcomes); Sum 11: 2 outcomes; Sum 12: $(6,6)$ (1 outcome).
Total $= 3 + 2 + 1 = 6 \implies P(>9) = \frac{6}{36} = \frac{1}{6}$. [A1]
Given two independent events $A$ and $B$ with $P(A) = 0.6$ and $P(B) = 0.4$:
a) Calculate $P(A \cap B)$.
b) Calculate $P(A \cup B)$.
c) Calculate $P(A \cap B')$ (event $A$ occurs but event $B$ does not).
View Solution & Markscheme
Method & Marks:
a) Independent multiplication: $P(A \cap B) = P(A) \times P(B) = 0.6 \times 0.4 = 0.24$. [A1]
b) General addition rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.6 + 0.4 - 0.24 = 0.76$. [A1]
c) $P(A \cap B') = P(A) \times P(B') = 0.6 \times (1 - 0.4) = 0.6 \times 0.6 = 0.36$. [A1]
A card is drawn from a standard deck of 52 cards. Let $K$ be the event of drawing a King, and $D$ be the event of drawing a Diamond.
a) Show mathematically whether events $K$ and $D$ are independent.
b) Find $P(K \cup D)$.
View Solution & Markscheme
Method & Marks:
a) Independence test:
$P(K) = \frac{4}{52} = \frac{1}{13}$. $P(D) = \frac{13}{52} = \frac{1}{4}$. [M1]
$P(K \cap D) = P(\text{King of Diamonds}) = \frac{1}{52}$.
$P(K) \times P(D) = \frac{1}{13} \times \frac{1}{4} = \frac{1}{52}$.
Since $P(K \cap D) = P(K)P(D)$, events $K$ and $D$ are independent. [A1][R1]
b) $P(K \cup D) = P(K) + P(D) - P(K \cap D) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$. [A1]
A box contains 6 red pens and 4 blue pens. A student picks two pens at random without replacement. Find the probability that the student picks one pen of each color.
View Solution & Markscheme
Method & Marks:
Two possible paths: Red then Blue ($RB$) OR Blue then Red ($BR$): [M1]
$P(RB) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90}$.
$P(BR) = \frac{4}{10} \times \frac{6}{9} = \frac{24}{90}$. [M1]
$P(\text{One of each}) = \frac{24}{90} + \frac{24}{90} = \frac{48}{90} = \frac{8}{15} \approx 0.533$. [A1]
In a competition, athlete A has a probability of $0.7$ of hitting the target, and athlete B has a probability of $0.8$ of hitting the target. Both shoot simultaneously and independently.
a) Find the probability that both hit the target.
b) Find the probability that exactly one hits the target.
c) Find the probability that neither hits the target.
View Solution & Markscheme
Method & Marks:
a) Both hit: $P(A \cap B) = 0.7 \times 0.8 = 0.56$. [A1]
b) Exactly one hits: ($A$ hits and $B$ misses) OR ($A$ misses and $B$ hits): [M1]
$P(A \cap B') + P(A' \cap B) = (0.7 \times 0.2) + (0.3 \times 0.8) = 0.14 + 0.24 = 0.38$. [A1]
c) Neither hits: $P(A' \cap B') = 0.3 \times 0.2 = 0.06$. [A1]
(Check: $0.56 + 0.38 + 0.06 = 1.00$).
A bag contains $n$ red counters and $4$ yellow counters. Two counters are drawn at random without replacement. If the probability of drawing two red counters is $\frac{1}{3}$, determine the value of $n$.
View Solution & Markscheme
Method & Marks:
Total counters $= n + 4$.
$P(RR) = \frac{n}{n + 4} \times \frac{n - 1}{n + 3} = \frac{1}{3}$. [M1]
$\frac{n(n - 1)}{(n + 4)(n + 3)} = \frac{1}{3} \implies 3n(n - 1) = (n + 4)(n + 3)$ [M1]
$3n^2 - 3n = n^2 + 7n + 12$
$2n^2 - 10n - 12 = 0 \implies n^2 - 5n - 6 = 0$. [M1]
Factoring: $(n - 6)(n + 1) = 0$.
Since count of counters must be positive: $n = 6$. [A1][R1]
Criterion B & D Gambler's Fallacy Investigation: In a roulette game, the ball has landed on Red 6 consecutive times. A gambler bets heavily on Black, reasoning that "Black is due to appear because probability must balance out."
a) Use the mathematical definition of independent events to evaluate the gambler's claim.
b) Calculate the probability that the 7th spin lands on Black (assuming a fair wheel with 18 Red, 18 Black, 1 Green).
c) Distinguish between the theoretical probability of a sequence $P(7 \text{ Reds in a row})$ before any spins occur versus the conditional probability of the 7th spin given that the first 6 were Red.
View Solution & Markscheme
Method & Marks:
Part a: Evaluation: The roulette wheel has no physical memory of previous spins. Each spin is an independent event. The gambler's claim is the classic Gambler's Fallacy, confusing long-term law of large numbers with short-term self-correction. [R1: Criterion C]
Part b: 7th spin probability:
$P(\text{Black}) = \frac{18}{37} \approx 0.4865$ ($48.65\%$), exactly the same as on any other spin! [A1]
Part c: Sequence vs. Single Step:
- Prior to the experiment, the sequence of 7 Reds is extremely rare: $\left(\frac{18}{37}\right)^7 \approx 0.0064$ ($0.64\%$).
- However, once the first 6 Reds have already occurred, their probability is $1$. The probability of the 7th being Red is simply $\frac{18}{37}$. The past spins have no impact on the future outcome. [A1][R1: Criterion D Reflection]
Probability Laws & Formulas Sheet
Probability Scale
$$0 \le P(A) \le 1$$
Impossible $0$ to Certain $1$.Complement
$$P(A') = 1 - P(A)$$
Probability of event NOT occurring.Mutually Exclusive
$$P(A \cup B) = P(A) + P(B)$$
When $P(A \cap B) = 0$.General Addition Rule
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
Subtract overlapping intersection.Independent Events
$$P(A \cap B) = P(A) \times P(B)$$
Multiplication test for independence.Expected Frequency
$$\text{Exp} = n \times P(A)$$
Expected counts over $n$ trials.Top 5 Common MYP Probability Pitfalls
- Double Counting in the Addition Rule: When events can occur together (e.g. Ace and Heart), never write $P(A \cup B) = P(A) + P(B)$. You MUST subtract $P(A \cap B)$ so the overlap is not counted twice!
- Assuming Real-World Events are Independent: Do not multiply probabilities unless you have verified that the events are independent. If sampling is done without replacement, probabilities change on the second draw!
- The Gambler's Fallacy: Past independent outcomes do NOT influence future outcomes. Flipping 5 heads in a row does not make the next flip "due" to be tails.
- Tree Diagram Path Confusion: Multiply along branches of the same path; ADD the results of different separate paths.
- Probability Greater than 1 or Negative: A probability can NEVER be less than 0 or greater than 1. If your calculation yields $1.25$ or $-0.4$, you made an algebraic error.
Self-Grading Mini Test
5 quiz questions on addition laws, independence, and probability trees.
If the probability of rain tomorrow is $\frac{3}{7}$, what is the probability that it does NOT rain?
Explanation: $P(\text{No rain}) = 1 - P(\text{Rain}) = 1 - \frac{3}{7} = \frac{4}{7}$.
If $P(A) = 0.5$, $P(B) = 0.4$, and $P(A \cap B) = 0.2$, what is $P(A \cup B)$?
Explanation: $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.5 + 0.4 - 0.2 = 0.7$.
Two coins are flipped. What is the probability of getting two Heads?
Explanation: $P(H \cap H) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.
A box has 3 red and 2 green balls. Two are drawn without replacement. What is $P(\text{Both Red})$?
Explanation: $P(R_1) = \frac{3}{5}$. After 1 red is removed, 2 red remain out of 4 total: $P(R_2|R_1) = \frac{2}{4}$. $P(RR) = \frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = \frac{3}{10}$.
If events $A$ and $B$ are mutually exclusive, what must $P(A \cap B)$ equal?
Explanation: Mutually exclusive events cannot occur at the same time, so their intersection has zero probability ($P(A \cap B) = 0$).