What is a probability distribution in the context of IB DP Mathematics?

In IB DP Mathematics, a probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes in an experiment. It describes how the probabilities are distributed over the values of the random variable. Understanding probability distributions is crucial for analyzing random phenomena and making predictions based on data.

How do discrete and continuous probability distributions differ?

Discrete probability distributions apply to scenarios where the set of possible outcomes is countable, such as rolling a die or flipping a coin. Examples include the binomial and Poisson distributions. Continuous probability distributions, on the other hand, are used when the set of possible outcomes is uncountable, such as measuring time or temperature. The normal distribution is a common example of a continuous distribution. In IB DP Mathematics, students learn to distinguish between these types and apply them appropriately.

What is the significance of the normal distribution in probability?

The normal distribution, also known as the Gaussian distribution, is significant because it describes many natural phenomena and is characterized by its bell-shaped curve. It is defined by its mean and standard deviation. In IB DP Mathematics, students explore how the normal distribution is used to model data and understand concepts like the empirical rule, which states that approximately 68%, 95%, and 99.7% of data fall within one, two, and three standard deviations from the mean, respectively.

How can you calculate the expected value of a probability distribution?

The expected value of a probability distribution, often referred to as the mean, is calculated by multiplying each possible outcome by its probability and summing all these products. For discrete distributions, this involves summing over all possible values. For continuous distributions, it involves integrating the product of the variable and its probability density function over the range of possible values. In IB DP Mathematics, understanding expected value helps in making informed predictions and decisions based on probabilistic models.

What role does the binomial distribution play in probability theory?

The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent Bernoulli trials, each with the same probability of success. It is characterized by two parameters: the number of trials (n) and the probability of success (p). In IB DP Mathematics, students use the binomial distribution to solve problems involving scenarios like flipping coins or quality control testing, where outcomes are binary (success/failure).

Why is "Applying binomial without checking independence or constant $p$." a common mistake in Probability Distribution?

Why it happens: Recognising 'success/failure' but not the assumptions. How to avoid it: Run the four checks each time: fixed n, two outcomes, constant p, independent trials. State them in AO3 communication.

Why is "Using binomCdf when binomPdf is needed (or vice versa)." a common mistake in Probability Distribution?

Why it happens: GDC menu confusion. How to avoid it: Pdf = P(X = k), exact value. Cdf = P(X k), cumulative. For 'at least k' use 1 - binomCdf(, k - 1).

Why is "Treating the second parameter of $N(\mu, \sigma^2)$ as the standard deviation." a common mistake in Probability Distribution?

Why it happens: and ^2 look similar. How to avoid it: N(60, 144) means ^2 = 144, so = 12. The GDC needs , not ^2.

Why is "Computing GDC values without writing 'Let $X \sim B(n, p)$'." a common mistake in Probability Distribution?

Why it happens: Treating Paper 2 as 'plug into GDC'. How to avoid it: Always state the distribution and parameters before computing. AO3 communication marks reward this.

ProbabilityIB Maths AA SL

Probability Distribution

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Probability Distribution — IB Maths AA SL: discrete random variables, binomial $B(n, p)$ and normal $N(\mu, \sigma^2)$

AA SL examines two named distributions: BINOMIAL (discrete, $n$ trials with constant probability $p$ ) and NORMAL (continuous, bell curve $N(\mu, \sigma^2)$ ). This note covers expected value, the binomial pmf, and using the GDC for binomial and normal probabilities.

What you’ll learn

Mapped to the IB DP Maths AA SL subject guide (2021 onwards (applies to 2026 exams)).

AO1 — State the binomial pmf and expected value.
AO1 — Identify a normal distribution from $\mu$ and $\sigma$ .
AO2 — Use the GDC to compute binomial $P(X = k)$ , $P(X \le k)$ .
AO2 — Use the GDC to compute normal $P(a \le X \le b)$ and inverse normal (find $x$ given a probability).
AO3 — Interpret probabilities in context with units.

Discrete random variables

Probabilities sum to 1; expected value is a weighted mean.

A discrete random variable $X$ takes values $x_1, x_2, \ldots$ with probabilities $p_1, p_2, \ldots$ summing to 1.

Probability distribution table:

$x$	$x_1$	$x_2$	$\ldots$
$P(X = x)$	$p_1$	$p_2$	$\ldots$

Expected value (mean): $E(X) = \sum_i x_i p_i.$

This is a weighted average. For a fair die, $E(X) = \sum_{i=1}^{6} i \cdot \dfrac{1}{6} = \dfrac{21}{6} = 3.5$ .

Worked example. $X$ takes values $0, 1, 2, 3$ with probabilities $0.1, 0.3, 0.4, 0.2$ . Find $E(X)$ and verify the probabilities sum to 1.

Sum: $0.1 + 0.3 + 0.4 + 0.2 = 1.0$ ✓.

$E(X) = 0(0.1) + 1(0.3) + 2(0.4) + 3(0.2) = 0 + 0.3 + 0.8 + 0.6 = 1.7$ .

Worked example (find unknown). $X$ takes values $1, 2, 3$ with probabilities $k, 2k, 3k$ . Find $k$ and $E(X)$ .

$k + 2k + 3k = 1 \Rightarrow k = 1/6$ .

$E(X) = 1 \cdot 1/6 + 2 \cdot 2/6 + 3 \cdot 3/6 = (1 + 4 + 9)/6 = 14/6 = 7/3$ .

Probabilities must sum to 1.
$E(X) = \sum x_i p_i$ — weighted mean.
If a probability is unknown ( $k$ ), use the sum-to-1 constraint.

Binomial distribution $B(n, p)$

$n$ independent trials, fixed success probability $p$ .

$X \sim B(n, p)$ models the number of SUCCESSES in $n$ independent trials, each with success probability $p$ .

Probability mass function (formula booklet): $P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}, \quad k = 0, 1, \ldots, n.$

Mean and variance:

$E(X) = np$ .
$\text{Var}(X) = np(1 - p)$ .

Worked example. A coin biased so $P(H) = 0.6$ is flipped 10 times. Find $P(\text{exactly 7 heads})$ .

$X \sim B(10, 0.6)$ : $P(X = 7) = \binom{10}{7}(0.6)^7(0.4)^3 = 120 \cdot 0.02799 \cdot 0.064 \approx 0.215.$

GDC: binomPdf(10, 0.6, 7) → 0.2150.

Worked example. With same coin, find $P(\text{at least 7 heads})$ .

$P(X \ge 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$ .

GDC: $1 -$ binomCdf(10, 0.6, 6) → about $0.382$ .

Mean and 'expected number'. For $B(10, 0.6)$ : $E(X) = 6$ , $\text{Var}(X) = 2.4$ .

Conditions for a binomial. Four checks:

Fixed number of trials $n$ .
Each trial has two outcomes (success / failure).
Constant probability $p$ .
Independent trials.

When in doubt, restate these in your AO3 communication.

Binomial pmf: in the formula booklet.
Mean $np$ , variance $np(1-p)$ .
GDC: binomPdf for exact $k$ ; binomCdf for $\le k$ ; complement for $\ge k$ .
Always check the four binomial conditions.

Normal distribution $N(\mu, \sigma^2)$

The bell curve. GDC normCdf for everything.

$X \sim N(\mu, \sigma^2)$ — continuous bell-shaped distribution with mean $\mu$ and variance $\sigma^2$ (standard deviation $\sigma$ ).

Key facts:

Symmetric about $\mu$ .
$P(\mu - \sigma \le X \le \mu + \sigma) \approx 0.68$ .
$P(\mu - 2\sigma \le X \le \mu + 2\sigma) \approx 0.95$ .
$P(\mu - 3\sigma \le X \le \mu + 3\sigma) \approx 0.997$ .

Standardisation. $Z = \dfrac{X - \mu}{\sigma}$ converts to $N(0, 1)$ . Z-tables and the $68$ - $95$ - $99.7$ rule live in the standard normal.

GDC commands.

$P(a \le X \le b)$ : normCdf $(a, b, \mu, \sigma)$ .
$P(X \le b)$ : normCdf $(-10^{99}, b, \mu, \sigma)$ (or just very negative).
$P(X \ge a)$ : normCdf $(a, 10^{99}, \mu, \sigma)$ .
Inverse: invNorm $(p, \mu, \sigma)$ — gives the $x$ value such that $P(X \le x) = p$ .

Worked example. Adult heights are approximately $N(170, 8^2)$ cm. Find the probability a randomly chosen adult is between 165 and 180 cm.

GDC normCdf(165, 180, 170, 8) ≈ $0.628$ .

Worked example (inverse). What height is exceeded by only 5% of adults?

We need $x$ with $P(X > x) = 0.05$ , equivalently $P(X \le x) = 0.95$ .

GDC invNorm(0.95, 170, 8) ≈ $183.2$ cm.

The 68-95-99.7 rule. About 68% of values lie within $\pm \sigma$ of the mean; 95% within $\pm 2\sigma$; 99.7% within $\pm 3\sigma$.

$N(\mu, \sigma^2)$ — note the variance not the SD in the notation.
Symmetric about $\mu$ ; 68-95-99.7 rule.
normCdf for probabilities, invNorm for values.
Always state $\mu$ and $\sigma$ used.

How it’s examined

Paper 1: rare — non-calc binomial only for small $n$ . Paper 2 is where binomial and normal questions live, with full GDC reliance. Examiner reports flag students who don't state the distribution and parameters (' $X \sim B(20, 0.3)$ ') before using the GDC.

Sources: IB Diploma Programme Mathematics: Analysis and Approaches subject guide (IBO, official). Last reviewed 2026-05-25.

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Step-by-step worked examples — Probability Distribution

Step-by-step solutions to past-paper-style questions on probability distribution, written exactly the way a tutor would explain them at the board.

1Find $k$ and $E(X)$
Getting started• discrete
Question
$X$ takes values 0, 1, 2 with $P(X=0) = 0.2$ , $P(X=1) = k$ , $P(X=2) = 0.5$ . Find $k$ and $E(X)$ .
Step-by-step solution
1. Step 1
  Sum = 1: $0.2 + k + 0.5 = 1 \Rightarrow k = 0.3$ .
2. Step 2
  $E(X) = 0(0.2) + 1(0.3) + 2(0.5) = 1.3$ .
Answer
$k = 0.3$ , $E(X) = 1.3$ .
2Binomial $P(X = k)$
Building confidence• binomial
Question
$X \sim B(8, 0.25)$ . Find $P(X = 3)$ exactly.
Step-by-step solution
1. Step 1
  $P(X = 3) = \binom{8}{3}(0.25)^3(0.75)^5$ .
2. Step 2
  $\binom{8}{3} = 56$ . $(0.25)^3 = 1/64$ . $(0.75)^5 \approx 0.2373$ .
3. Step 3
  $\approx 56 \cdot 0.01563 \cdot 0.2373 \approx 0.208$ .
Answer
$\approx 0.208$ .
3Binomial 'at least one'
Building confidence• binomial
Question
$X \sim B(20, 0.1)$ . Find $P(X \ge 1)$ .
Step-by-step solution
1. Step 1
  Complement: $P(X \ge 1) = 1 - P(X = 0)$ .
2. Step 2
  $P(X = 0) = (0.9)^{20} \approx 0.1216$ .
3. Step 3
  $P(X \ge 1) \approx 1 - 0.1216 = 0.878$ .
Answer
$\approx 0.878$ .
4Normal probability — interval
Stretch• normal
Question
Test marks are approximately $N(60, 12^2)$ . Find $P(50 \le X \le 75)$ .
Step-by-step solution
1. Step 1
  GDC: normCdf(50, 75, 60, 12).
2. Step 2
  Result $\approx 0.692$ .
Answer
$\approx 0.692$ .
5Inverse normal
Stretch• inverse normal
Question
Marks are $N(60, 12^2)$ . The top 10% receive an A. Find the lowest A-grade mark.
Step-by-step solution
1. Step 1
  $P(X \le a) = 0.90$ , so $a =$ invNorm(0.9, 60, 12).
2. Step 2
  $a \approx 75.4$ .
Answer
$\approx 75.4$ marks.

Key Definitions and Keywords — Probability Distribution

Definitions to memorise and the exact keywords mark schemes credit for probability distribution answers — sharpened from recent examiner reports for the 2026 IB DP Maths AA SL sitting.

Expected value $E(X)$
Examiner keyword
$E(X) = \sum x_i p_i$ — the long-run average of a random variable.
Binomial $B(n, p)$
Examiner keyword
Distribution of the number of successes in $n$ independent trials with success probability $p$ . $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$ .
Normal $N(\mu, \sigma^2)$
Examiner keyword
Continuous symmetric bell-shaped distribution with mean $\mu$ and variance $\sigma^2$ .
Standardisation
$Z = \dfrac{X - \mu}{\sigma}$ — converts $N(\mu, \sigma^2)$ to $N(0, 1)$ .

Common Mistakes and Misconceptions — Probability Distribution

The traps other students keep falling into on probability distribution questions — taken from recent IB DP Maths AA SL examiner reports and mark schemes — and how to avoid them.

✕Applying binomial without checking independence or constant $p$ .
Why it happens
Recognising 'success/failure' but not the assumptions.
How to avoid it
Run the four checks each time: fixed $n$ , two outcomes, constant $p$ , independent trials. State them in AO3 communication.
✕Using binomCdf when binomPdf is needed (or vice versa).
Why it happens
GDC menu confusion.
How to avoid it
Pdf = $P(X = k)$ , exact value. Cdf = $P(X \le k)$ , cumulative. For 'at least $k$ ' use $1 -$ binomCdf $(\ldots, k - 1)$ .
✕Treating the second parameter of $N(\mu, \sigma^2)$ as the standard deviation.
Why it happens
$\sigma$ and $\sigma^2$ look similar.
How to avoid it
$N(60, 144)$ means $\sigma^2 = 144$ , so $\sigma = 12$ . The GDC needs $\sigma$ , not $\sigma^2$ .
✕Computing GDC values without writing 'Let $X \sim B(n, p)$ '.
Why it happens
Treating Paper 2 as 'plug into GDC'.
How to avoid it
Always state the distribution and parameters before computing. AO3 communication marks reward this.

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Probability Distribution — frequently asked questions

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Probability Distribution

Short Study Notes in the form of Slides

Short Study Notes — Probability Distribution

Detailed Notes

What you’ll learn

How it’s examined

1Find kkk and E(X)E(X)E(X)

2Binomial P(X=k)P(X = k)P(X=k)

3Binomial 'at least one'

4Normal probability — interval

5Inverse normal

Expected value E(X)E(X)E(X)

Binomial B(n,p)B(n, p)B(n,p)

Normal N(μ,σ2)N(\mu, \sigma^2)N(μ,σ2)

Standardisation

✕Applying binomial without checking independence or constant ppp.

✕Using binomCdf when binomPdf is needed (or vice versa).

✕Treating the second parameter of N(μ,σ2)N(\mu, \sigma^2)N(μ,σ2) as the standard deviation.

✕Computing GDC values without writing 'Let X∼B(n,p)X \sim B(n, p)X∼B(n,p)'.

Past paper style quiz

Probability Distribution — frequently asked questions

1Find $k$ and $E(X)$

2Binomial $P(X = k)$

Expected value $E(X)$

Binomial $B(n, p)$

Normal $N(\mu, \sigma^2)$

✕Applying binomial without checking independence or constant $p$ .

✕Treating the second parameter of $N(\mu, \sigma^2)$ as the standard deviation.

✕Computing GDC values without writing 'Let $X \sim B(n, p)$ '.