Probability | IB - Wyatt's Notes

Probability Fundamentals

Sample Space and Events

The sample space $\Omega$ (or $S$ ) is the set of all possible outcomes of an experiment. An event $A$ is a subset of the sample space.

Probability Axioms (Kolmogorov)

$P(A) \ge 0$ for every event $A$ .
$P(\Omega) = 1$ .
If $A_1, A_2, \ldots$ are mutually exclusive events, then:

P\!\left(\bigcup_{i=1}^{\infty} A_i\right) = \sum_{i=1}^{\infty} P(A_i)

Complementary Events

P(A") = 1 - P(A)

Where $A'$ (or $\bar{A}$ ) is the complement of $A$ .

Venn Diagrams

Venn diagrams visually represent events and their relationships:

Union: $A \cup B$ (elements in $A$ or $B$ or both)
Intersection: $A \cap B$ (elements in both $A$ and $B$ )
Disjoint (mutually exclusive): $A \cap B = \emptyset$

Addition Rule

For any two events $A$ and $B$ :

P(A \cup B) = P(A) + P(B) - P(A \cap B)

For mutually exclusive events ( $A \cap B = \emptyset$ ):

P(A \cup B) = P(A) + P(B)

Example In a class of 40 students, 25 play football, 18 play basketball, and 8 play both. Find the Probability that a randomly selected student plays at least one sport.

P(F \cup B) = P(F) + P(B) - P(F \cap B) = \frac{25}{40} + \frac{18}{40} - \frac{8}{40} = \frac{35}{40} = \frac{7}{8}

Multiplication Rule

Independent Events

Two events $A$ and $B$ are independent if and only if:

P(A \cap B) = P(A) \cdot P(B)

For independent events:

P(A \cap B) = P(A) \cdot P(B)

Dependent Events

For dependent events, the conditional probability is needed:

P(A \cap B) = P(A) \cdot P(B|A) = P(B) \cdot P(A|B)

Example A box contains 4 defective and 6 non-defective items. Two items are drawn without replacement. Find The probability that exactly one is defective.

Paths giving exactly one defective:

First defective, second non-defective: $\dfrac{4}{10} \times \dfrac{6}{9} = \dfrac{24}{90}$
First non-defective, second defective: $\dfrac{6}{10} \times \dfrac{4}{9} = \dfrac{24}{90}$

P(\mathrm{exactly one defective}) = \frac{24}{90} + \frac{24}{90} = \frac{48}{90} = \frac{8}{15}

Bayes’ Theorem

Theorem

For events $A$ and $B$ with $P(B) \neq 0$ :

P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}

Extended Form

If $A_1, A_2, \ldots, A_n$ form a partition of $\Omega$ :

P(A_k|B) = \frac{P(B|A_k) \cdot P(A_k)}{\displaystyle\sum_{i=1}^{n} P(B|A_i) \cdot P(A_i)}

The denominator $P(B) = \displaystyle\sum_{i=1}^{n} P(B|A_i) \cdot P(A_i)$ is the law of total Probability.

Example A factory has three machines producing items. Machine A produces 50% of items with 2% defect rate. Machine B produces 30% with 3% defect rate. Machine C produces 20% with 1% defect rate. An item Is found to be defective. What is the probability it came from machine B?

P(\mathrm{defective}) = 0.5 \times 0.02 + 0.3 \times 0.03 + 0.2 \times 0.01 = 0.01 + 0.009 + 0.002 = 0.021

P(B|\mathrm{defective}) = \frac{0.3 \times 0.03}{0.021} = \frac{0.009}{0.021} = \frac{3}{7}

Medical Testing Example

A disease affects 1% of a population. A test has 99% sensitivity ( $P(\mathrm{positive}|\mathrm{disease}) = 0.99$ ) and 95% specificity ( $P(\mathrm{negative}|\mathrm{no disease}) = 0.95$ ). What is $P(\mathrm{disease}|\mathrm{positive})$ ?

P(\mathrm{positive}) = 0.01 \times 0.99 + 0.99 \times 0.05 = 0.0099 + 0.0495 = 0.0594

P(\mathrm{disease}|\mathrm{positive}) = \frac{0.01 \times 0.99}{0.0594} = \frac{0.0099}{0.0594} \approx 0.167

Exam Tip This result (approximately 16.7%) is counterintuitively low. Always work through Bayes’ theorem Carefully rather than relying on intuition for conditional probability questions.

Discrete Random Variables

Definition

A discrete random variable $X$ takes a countable set of values $x_1, x_2, \ldots$ with Probabilities $P(X = x_i) = p_i$ .

Probability Distribution

A probability distribution satisfies:

$p_i \ge 0$ for all $i$
$\displaystyle\sum_{\mathrm{all } i} p_i = 1$

Expectation (Mean)

E(X) = \mu = \sum x_i \cdot p_i

Variance

\mathrm{Var}(X) = \sigma^2 = E(X^2) - [E(X)]^2 = \sum x_i^2 p_i - \mu^2

Standard Deviation

\sigma = \sqrt{\mathrm{Var}(X)}

Properties of Expectation and Variance

For any constant $a$ and $b$ :

E(aX + b) = aE(X) + b

\mathrm{Var}(aX + b) = a^2 \mathrm{Var}(X)

For independent random variables $X$ and $Y$ :

E(X + Y) = E(X) + E(Y)

\mathrm{Var}(X + Y) = \mathrm{Var}(X) + \mathrm{Var}(Y)

Example A random variable $X$ has the following probability distribution:

$x$	0	1	2	3
$P(X = x)$	0.1	0.4	0.3	0.2

E(X) = 0(0.1) + 1(0.4) + 2(0.3) + 3(0.2) = 0 + 0.4 + 0.6 + 0.6 = 1.6

E(X^2) = 0(0.1) + 1(0.4) + 4(0.3) + 9(0.2) = 0 + 0.4 + 1.2 + 1.8 = 3.4

\mathrm{Var}(X) = 3.4 - 1.6^2 = 3.4 - 2.56 = 0.84

\sigma = \sqrt{0.84} \approx 0.917

The Binomial Distribution

Conditions

A random variable $X$ follows a binomial distribution $X \sim B(n, p)$ if:

There are a fixed number $n$ of trials.
Each trial has exactly two outcomes (success/failure).
The probability of success $p$ is constant for each trial.
Trials are independent.

Probability Mass Function

P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}

For $x = 0, 1, 2, \ldots, n$ .

Mean and Variance

E(X) = np

\mathrm{Var}(X) = np(1-p)

\sigma = \sqrt{np(1-p)}

Exam Tip For binomial probability questions, always state the distribution : ” $X \sim B(n, p)$ Where…”. Use your GDC for calculations involving large $n$ or cumulative probabilities.

The Normal Distribution

Properties

A continuous random variable $X$ follows a normal distribution $X \sim N(\mu, \sigma^2)$ .

The curve is bell-shaped and symmetric about $\mu$ .
The mean, median, and mode are all equal to $\mu$ .
Approximately 68% of data lies within $\mu \pm \sigma$ .
Approximately 95% of data lies within $\mu \pm 2\sigma$ .
Approximately 99.7% of data lies within $\mu \pm 3\sigma$ .

Standardisation

To find probabilities, convert to the standard normal $Z \sim N(0, 1)$ :

Z = \frac{X - \mu}{\sigma}

P(X \lt a) = P\!\left(Z \lt \frac{a - \mu}{\sigma}\right)

Example Heights of a population follow $N(170, 64)$ (in cm). Find the height that is at the 90th percentile.

$\mu = 170$ , $\sigma = 8$ .

P(X \lt x) = 0.90 \implies \frac{x - 170}{8} = z_{0.90} = 1.282

X = 170 + 1.282 \times 8 = 170 + 10.26 = 180.26 \mathrm{ cm}

Combining Normal Variables

For independent normal variables $X \sim N(\mu_X, \sigma_X^2)$ and $Y \sim N(\mu_Y, \sigma_Y^2)$ :

X + Y \sim N(\mu_X + \mu_Y, \sigma_X^2 + \sigma_Y^2)

X - Y \sim N(\mu_X - \mu_Y, \sigma_X^2 + \sigma_Y^2)

AX + b \sim N(a\mu_X + b, a^2\sigma_X^2)

Example The weight of a parcel is $X \sim N(2, 0.04)$ kg. The packaging adds $Y \sim N(0.3, 0.01)$ kg. Find The probability that the total exceeds 2.5 kg.

X + Y \sim N(2.3, 0.05)

P(X + Y \gt 2.5) = P\!\left(Z \gt \frac{2.5 - 2.3}{\sqrt{0.05}}\right) = P(Z \gt 0.894) = 1 - 0.814 = 0.186

Continuous Random Variables

Probability Density Functions (PDF)

A function $f(x)$ is a PDF if:

$f(x) \ge 0$ for all $x$ .
$\displaystyle\int_{-\infty}^{\infty} f(x)\,dx = 1$ .

Probabilities from a PDF

P(a \le X \le b) = \int_a^b f(x)\,dx

Mean and Variance

E(X) = \int_{-\infty}^{\infty} x f(x)\,dx

E(X^2) = \int_{-\infty}^{\infty} x^2 f(x)\,dx

\mathrm{Var}(X) = E(X^2) - [E(X)]^2

Median

The median $m$ satisfies:

\int_{-\infty}^{m} f(x)\,dx = 0.5

Example A continuous random variable $X$ has PDF $f(x) = 2x$ for $0 \le x \le 1$ .

Verify it is a valid PDF:

\int_0^1 2x\,dx = [x^2]_0^1 = 1

Find $P(X \lt 0.5)$ :

P(X \lt 0.5) = \int_0^{0.5} 2x\,dx = [x^2]_0^{0.5} = 0.25

Find $E(X)$ :

E(X) = \int_0^1 x \cdot 2x\,dx = \int_0^1 2x^2\,dx = \left[\frac{2x^3}{3}\right]_0^1 = \frac{2}{3}

Find the median:

\int_0^m 2x\,dx = 0.5 \implies m^2 = 0.5 \implies m = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}

IB Exam-Style Questions

Question 1 (Paper 1 style)

Events A and B are such that $P(A) = 0.6$ , $P(B) = 0.4$ And $P(A|B) = 0.3$ .

(a) Find $P(A \cap B)$ .

P(A \cap B) = P(A|B) \cdot P(B) = 0.3 \times 0.4 = 0.12

(b) Determine whether A and B are independent.

$P(A) \cdot P(B) = 0.6 \times 0.4 = 0.24 \neq 0.12 = P(A \cap B)$ .

Not independent.

(c) Find $P(A \cup B)$ .

P(A \cup B) = 0.6 + 0.4 - 0.12 = 0.88

Question 2 (Paper 2 style)

A bag contains 7 red and 5 blue marbles. Three marbles are drawn without replacement.

(a) Find the probability that all three are red.

P = \frac{7}{12} \times \frac{6}{11} \times \frac{5}{10} = \frac{210}{1320} = \frac{7}{44}

(b) Find the probability that exactly two are red.

P = \binom{3}{2} \times \frac{7}{12} \times \frac{6}{11} \times \frac{5}{10} = 3 \times \frac{210}{1320} = \frac{21}{44}

Wait, let me recalculate using a tree diagram approach:

RRB: $\dfrac{7}{12} \times \dfrac{6}{11} \times \dfrac{5}{10} = \dfrac{7}{44}$
RBR: $\dfrac{7}{12} \times \dfrac{5}{11} \times \dfrac{6}{10} = \dfrac{7}{44}$
BRR: $\dfrac{5}{12} \times \dfrac{7}{11} \times \dfrac{6}{10} = \dfrac{7}{44}$

P = \frac{7}{44} + \frac{7}{44} + \frac{7}{44} = \frac{21}{44}

Question 3 (Paper 2 style)

The time taken to complete a task follows a normal distribution with mean 45 minutes and standard Deviation 8 minutes.

(a) Find the probability that a randomly selected person takes between 40 and 50 minutes.

P(40 \lt X \lt 50) = P\!\left(\frac{40-45}{8} \lt Z \lt \frac{50-45}{8}\right) = P(-0.625 \lt Z \lt 0.625)

\approx 2\Phi(0.625) - 1 \approx 2(0.734) - 1 = 0.468

(b) The fastest 10% of people receive a certificate. Find the maximum time to qualify.

P(X \lt x) = 0.10 \implies \frac{x - 45}{8} = -1.282

X = 45 - 1.282 \times 8 = 45 - 10.26 = 34.74 \mathrm{ minutes}

Question 4 (Paper 1 style)

$X \sim B(15, 0.3)$ . Find $P(X = 5)$ .

P(X = 5) = \binom{15}{5}(0.3)^5(0.7)^{10}

Using a GDC: $P(X = 5) \approx 0.206$ .

Question 5 (Paper 2 style)

A continuous random variable $X$ has PDF $f(x) = \dfrac{3x^2}{8}$ for $0 \le x \le 2$ .

(a) Verify that $f(x)$ is a valid PDF.

\int_0^2 \frac{3x^2}{8}\,dx = \frac{3}{8}\left[\frac{x^3}{3}\right]_0^2 = \frac{3}{8} \cdot \frac{8}{3} = 1

(b) Find $E(X)$ .

E(X) = \int_0^2 x \cdot \frac{3x^2}{8}\,dx = \int_0^2 \frac{3x^3}{8}\,dx = \frac{3}{8}\left[\frac{x^4}{4}\right]_0^2 = \frac{3}{8} \times 4 = \frac{3}{2}

(c) Find the mode.

Since $f(x) = \dfrac{3x^2}{8}$ is increasing on $[0, 2]$ The mode is $x = 2$ .

flowchart TD
    A[1_Probability] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Summary

Distribution	Notation	$E(X)$	$\mathrm{Var}(X)$
Binomial	$B(n, p)$	$np$	$np(1-p)$
Normal	$N(\mu, \sigma^2)$	$\mu$	$\sigma^2$

Key Formula	Expression
Addition rule	$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Conditional probability	$P(A\\|B) = \dfrac{P(A \cap B)}{P(B)}$
Bayes’ theorem	$P(A\\|B) = \dfrac{P(B\\|A) \cdot P(A)}{P(B)}$
Standardisation	$Z = \dfrac{X - \mu}{\sigma}$

Exam Strategy Always define your random variable at the start of probability questions. For normal Distribution problems, draw a sketch of the bell curve and shade the relevant area. For binomial Problems, verify the four conditions before applying the formula.

Probability Distributions: Additional Topics

Poisson Distribution

The Poisson distribution models the number of events occurring in a fixed interval of time or space.

$X \sim \mathrm{Po}(\lambda)$ where $\lambda$ is the mean number of events.

P(X = x) = \frac{e^{-\lambda}\lambda^x}{x!}, \quad x = 0, 1, 2, \ldots

E(X) = \lambda, \quad \mathrm{Var}(X) = \lambda

Example A call centre receives an average of 4 calls per minute. Find the probability of receiving exactly 6 Calls in a minute.

P(X = 6) = \frac{e^{-4} \cdot 4^6}{6!} = \frac{e^{-4} \cdot 4096}{720} = \frac{4096}{720 \times 54.60} \approx 0.104

Geometric Distribution

Models the number of trials until the first success.

$X \sim \mathrm{Geo}(p)$ where $p$ is the probability of success on each trial.

P(X = x) = (1-p)^{x-1}p, \quad x = 1, 2, 3, \ldots

E(X) = \frac{1}{p}, \quad \mathrm{Var}(X) = \frac{1-p}{p^2}

Example A die is rolled until a 6 appears. Find the probability that it takes exactly 4 rolls.

P(X = 4) = \left(\frac{5}{6}\right)^3 \times \frac{1}{6} = \frac{125}{1296} \approx 0.0965

Combinations and Permutations

Factorial

N! = n \times (n-1) \times \cdots \times 2 \times 1, \quad 0! = 1

Permutations

The number of ways to arrange $r$ objects from $n$ distinct objects (order matters):

^nP_r = \frac{n!}{(n-r)!}

Combinations

The number of ways to choose $r$ objects from $n$ distinct objects (order does not matter):

^nC_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}

Additional Exam-Style Questions

Question 6 (Paper 2 style)

A bag contains 4 red and 6 blue marbles. Marbles are drawn one at a time without replacement until a Red marble is drawn.

(a) Find the probability that exactly 3 draws are needed.

P = \frac{6}{10} \times \frac{5}{9} \times \frac{4}{8} = \frac{120}{720} = \frac{1}{6}

(b) Find the expected number of draws.

Let $X$ be the number of draws. We need $E(X)$ .

$P(X=1) = \dfrac{4}{10} = 0.4$

$P(X=2) = \dfrac{6}{10} \times \dfrac{4}{9} = \dfrac{24}{90} = \dfrac{4}{15}$

$P(X=3) = \dfrac{6}{10} \times \dfrac{5}{9} \times \dfrac{4}{8} = \dfrac{1}{6}$

$P(X=4) = \dfrac{6}{10} \times \dfrac{5}{9} \times \dfrac{4}{8} \times \dfrac{4}{7} = \dfrac{4}{42}$

$P(X=5) = \dfrac{6}{10} \times \dfrac{5}{9} \times \dfrac{4}{8} \times \dfrac{3}{7} \times \dfrac{4}{6} = \dfrac{3}{42}$

$P(X=6) = \dfrac{6}{10} \times \dfrac{5}{9} \times \dfrac{4}{8} \times \dfrac{3}{7} \times \dfrac{2}{6} \times \dfrac{4}{5} = \dfrac{2}{105}$

$P(X=7) = \dfrac{6}{10} \times \dfrac{5}{9} \times \dfrac{4}{8} \times \dfrac{3}{7} \times \dfrac{2}{6} \times \dfrac{1}{5} \times \dfrac{4}{4} = \dfrac{1}{210}$

E(X) = 1(0.4) + 2\!\left(\frac{4}{15}\right) + 3\!\left(\frac{1}{6}\right) + 4\!\left(\frac{4}{42}\right) + 5\!\left(\frac{3}{42}\right) + 6\!\left(\frac{2}{105}\right) + 7\!\left(\frac{1}{210}\right)

= 0.4 + 0.533 + 0.5 + 0.381 + 0.357 + 0.114 + 0.033 = 2.318

Question 7 (Paper 1 style)

$X \sim B(12, 0.25)$ . Find $P(X \le 2)$ .

P(X \le 2) = P(X=0) + P(X=1) + P(X=2)

= \binom{12}{0}(0.25)^0(0.75)^{12} + \binom{12}{1}(0.25)^1(0.75)^{11} + \binom{12}{2}(0.25)^2(0.75)^{10}

= 0.0317 + 0.1267 + 0.2323 = 0.3907

Question 8 (Paper 2 style)

The heights of Year 12 students follow a normal distribution with mean $165\mathrm{ cm}$ and Standard deviation $8\mathrm{ cm}$ .

(a) What percentage of students are taller than $180\mathrm{ cm}$ ?

P(X \gt 180) = P\!\left(Z \gt \frac{180-165}{8}\right) = P(Z \gt 1.875) = 1 - 0.9696 = 0.0304

About $3.0\%$ .

(b) The school needs to order desks for the middle 90% of students. What height range should the Desks accommodate?

Middle 90% means 5th to 95th percentile.

5th percentile: $\dfrac{h - 165}{8} = -1.645 \implies h = 165 - 13.16 = 151.8\mathrm{ cm}$ .

95th percentile: $\dfrac{h - 165}{8} = 1.645 \implies h = 165 + 13.16 = 178.2\mathrm{ cm}$ .

Desks should accommodate heights from about $152\mathrm{ cm}$ to $178\mathrm{ cm}$ .

the IB specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Probability with other IB mathematics topics to test synthesis under exam conditions.

See for instructions on self-marking and building a personal test matrix.

Common Pitfalls

Dropping negative signs during algebraic manipulation. Substitute back to verify your answer.
Confusing the domain and range of functions, or not considering restrictions (e.g., denominator cannot be zero).
Misreading the question, particularly with ‘hence’ vs ‘hence or otherwise’. The former requires using previous work.
Confusing $P(A|B)$ with $P(B|A)$ . These are related by Bayes’ theorem but are not equal in general.

Cross-References

Topic	Site	Link
[Probability]	A-Level	View
[Probability]	IB	View
[Probability]	DSE	View
[Probability]	University	View

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Intuition

Probability is the mathematics of uncertainty. It quantifies how likely events are, ranging from 0 (impossible) to 1 (certain). The axioms — non-negativity, total probability equals 1, and additivity for mutually exclusive events — are the minimal rules that make probability consistent.

Think of probability as measuring the “size” of the favorable outcomes relative to all possible outcomes. Venn diagrams make this visual: the probability of an event is the area of its region divided by the total area. The addition rule corrects for double-counting when events overlap. Conditional probability is like zooming in — you restrict your sample space to event B, then ask what fraction of B is also in A. Bayes’ theorem flips the conditioning, which is why it’s so powerful (and counterintuitive).

Common Mistakes

Confusing mutually exclusive with independent. Mutually exclusive events cannot occur together ( $P(A \cap B) = 0$ ), which actually makes them dependent — if you know $A$ occurred, you know $B$ did not. Independent events satisfy $P(A \cap B) = P(A)P(B)$ , which is nonzero when both probabilities are nonzero.
Applying the multiplication rule to dependent events without conditioning. $P(A \cap B) = P(A)P(B)$ only holds for independent events. For dependent events, you must use $P(A \cap B) = P(A) \cdot P(B|A)$ . Forgetting this leads to over- or under-estimating joint probabilities.
Reversing conditional probabilities (ignoring Bayes’ theorem). $P(A|B) \neq P(B|A)$ as a rule. A common error is assuming that because $B$ is likely given $A$ , $A$ must also be likely given $B$ . Bayes’ theorem corrects this by incorporating the prior probabilities of both events.
Double-counting outcomes in sample spaces. When listing outcomes, ensure each is equally likely before dividing by the total. If outcomes are not equally likely, use weighted probabilities instead of simple fractions.
Assuming the addition rule applies to all events. $P(A \cup B) = P(A) + P(B)$ is only valid for mutually exclusive events. For overlapping events, you must subtract the intersection: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ .