Probability distributions are where the Statistics & Probability strand of IB Mathematics Applications & Interpretation HL turns from describing data to modelling randomness. AI treats each distribution as a ready-made model for a real situation — a fixed number of pass/fail trials (binomial), events arriving at a steady rate (Poisson), or a continuous measurement clustered around a mean (normal). A calculator is always allowed, so almost every probability drops out of a single GDC command; the marks live in choosing the right distribution, entering the correct parameters, and reading the answer in context — not in hand-grinding factorials.
This cheatsheet condenses SL 4.7–4.9 and AHL 4.17 onto one page — valid distributions and expected value, the binomial and its mean $np$, normal probabilities and the inverse normal, and the HL Poisson model with its additive rate $\lambda$ — and flags the traps that quietly cost marks (pdf vs cdf, which tail invNorm wants, dropping the $(1-p)$ in the variance). The print-ready PDF is at the bottom, free to download.
§1 — Discrete random variables & valid distributions SL 4.7
A discrete random variable $X$ takes separate values, each with its own probability; the list of value–probability pairs is its probability distribution, usually given as a table.
Each probability:$0\le P(X=x)\le 1$ — never negative, never above $1$.
Valid distribution:$\sum P(X=x)=1$ — every entry in the table must add to one.
Missing entry:Set the sum equal to $1$ and solve for the unknown $k$.
TrickAdd every probability you are given, then subtract from $1$ to fill a gap — that missing value is one of the most commonly dropped marks on the whole topic.
TrapYour answer for $k$ must itself be a valid probability. If solving gives $k>1$ or $k<0$, re-check the algebra — a probability can never sit outside $[0,1]$.
§2 — Expected value & fair games SL 4.7
The expected value (mean) is the long-run average outcome if the experiment were repeated many, many times.
Expected value:$E(X)=\sum x\,P(X=x)$ — weight each value by its probability, then add.
Net gain (game):$E(\text{gain})=E(\text{payout})-\text{cost to play}$.
Fair game:Fair when $E(\text{gain})=0$; positive favours the player, negative favours the house.
TrickFind any missing $k$ from $\sum P(X=x)=1$ before you touch $E(X)$ — the two steps almost always appear in the same question, and $E(X)$ is wrong if $k$ is.
Trap$E(\text{payout})$ on its own is not the verdict on a game — always subtract the stake first. An expected payout smaller than the ticket price is still a losing game.
Note$E(X)$ need not be an attainable value — a fair die has mean $3.5$. It is a balance point of the distribution, not a prediction of any single outcome.
§3 — The binomial distribution: exact probabilities SL 4.8
Use $X\sim B(n,p)$ when you count the successes in a fixed number $n$ of independent trials, each with the same success probability $p$ and only two outcomes.
Model:$X\sim B(n,p)$ — $n$ independent trials, constant $p$, success/failure each time.
Exactly $k$:$P(X=k)=\binom{n}{k}p^{k}(1-p)^{n-k}$.
On the GDC:binompdf$(n,p,k)$ returns $P(X=k)$ directly — no need to expand the formula by hand.
Trick"Exactly", "precisely" or "equal to" $k$ successes $\Rightarrow$ binompdf. Read the wording before you type a single number.
TrapThe binomial needs a fixed $n$, a constant $p$ and independent trials. Sampling without replacement changes $p$ each draw, so the binomial no longer applies.
§4 — Binomial 'at least' & 'at most' SL 4.8
Most exam parts ask for a range of successes rather than an exact count — that is the cumulative binomial, binomcdf.
At most $k$:$P(X\le k)=$ binomcdf$(n,p,k)$ — the calculator sums from $0$ up to $k$.
At least $k$:$P(X\ge k)=1-P(X\le k-1)=1-$ binomcdf$(n,p,k-1)$.
Between:$P(a\le X\le b)=$ binomcdf$(n,p,b)-$ binomcdf$(n,p,a-1)$.
TrapFor "at least $k$" you subtract $P(X\le k-1)$, not $P(X\le k)$ — "at least $2$" excludes only $0$ and $1$. The off-by-one $k-1$ is the classic slip here.
TrickTranslate the words first: "fewer than $k$" $=P(X\le k-1)$; "more than $k$" $=1-P(X\le k)$. A quick sketch of which counts are included stops every mix-up.
§5 — Binomial mean & standard deviation SL 4.8
The shape of $B(n,p)$ is fixed entirely by $n$ and $p$, so its centre and spread come straight from them — no GDC needed.
Mean:$E(X)=np$ — expected number of successes.
Variance:$\mathrm{Var}(X)=np(1-p)$.
Standard deviation:$\sigma=\sqrt{np(1-p)}$.
TrapThe variance carries the extra $(1-p)$ factor; the mean does not. Dropping $(1-p)$ — writing the spread as $np$ — is the single most common error in this sub-topic.
TrickCompute $np(1-p)$ in full first, then take the square root once at the very end for $\sigma$. Rounding the variance before rooting throws away accuracy.
§6 — The normal distribution: probabilities SL 4.9
A continuous measurement that clusters symmetrically around a mean is modelled by $X\sim N(\mu,\sigma^{2})$ — the bell curve. A probability is an area under that curve, read off with normalcdf.
The normal curve N(μ,σ²) is symmetric about the mean μ. The shaded area between a and b is the probability P(a < X < b) = normalcdf(a, b, μ, σ) — always sketch and shade before reaching for the GDC.
Area (probability):$P(a<X<b)=$ normalcdf(lower, upper, $\mu$, $\sigma$).
One-sided tail:Use $-1\text{E}99$ or $1\text{E}99$ for the open end, e.g. $P(X>a)=$ normalcdf$(a, 1\text{E}99, \mu, \sigma)$.
Interval rule:$P(a<X<b)=\text{cdf}(b)-\text{cdf}(a)$ — subtract the two areas; don't stop at one bound.
TrickSketch the bell and shade the region before touching the GDC. The picture tells you which bounds to enter and instantly settles "greater-than vs less-than" confusion.
TrapEnter the standard deviation $\sigma$, not the variance $\sigma^{2}$. If the question quotes a variance, take its square root first — a "variance of $16$" means you type $\sigma=4$.
NoteFor a continuous normal, $P(X=a)=0$, so $<$ and $\le$ give the same answer — the area over a single point is zero.
§7 — Inverse normal & finding an unknown parameter SL 4.9
The inverse normal runs the process backwards: given an area, it returns the value $x$ that cuts it off. It is the key to "find the mark that the top 10% exceed" style questions.
Value from area:$x=$ invNorm(area to the LEFT, $\mu$, $\sigma$).
Standardise:$z=\dfrac{x-\mu}{\sigma}$, so $x=\mu+z\sigma$.
Unknown $\mu$ or $\sigma$:Read $z$ from invNorm(area, $0$, $1$), then solve $x=\mu+z\sigma$ for the missing parameter.
TrapinvNorm always expects the area to the LEFT of $x$. For a "top $5$%" cut-off, enter $0.95$; for the "bottom $5$%", enter $0.05$. Feeding the wrong tail is the number-one inverse-normal error.
TrickSketch first so the sign of $z$ is unmistakable: a below-average cut-off gives a negative $z$, so $x=\mu+z\sigma$ subtracts $|z|\sigma$ from the mean.
§8 — The Poisson distribution AHL 4.17
Use $X\sim\text{Po}(\lambda)$ to count independent events that occur at a steady average rate $\lambda$ over a fixed interval of time, length or area — with no fixed upper limit. This is HL only in IB AI.
Probability:$P(X=k)=\dfrac{e^{-\lambda}\lambda^{k}}{k!}$.
On the GDC:poissonpdf$(\lambda,k)$ for an exact count; poissoncdf$(\lambda,k)$ for $P(X\le k)$.
Mean & variance:Both equal $\lambda$ — a Poisson's spread is fixed by its mean.
Additive rate:Independent $X\sim\text{Po}(\lambda_1)$ and $Y\sim\text{Po}(\lambda_2)\Rightarrow X+Y\sim\text{Po}(\lambda_1+\lambda_2)$.
TrapRescale $\lambda$ to match the interval in the question. If the rate is $6$ per hour and you want a $20$-minute window, use $\lambda=2$ — scale before computing, not after.
NoteMean $\approx$ variance is the fingerprint of a Poisson model. A binomial with large $n$ and small $p$ is well approximated by $\text{Po}(np)$ — handy when $n$ is too big to be practical.
§9 — Sums & averages of independent variables AHL 4.17
HL adds rules for combining several independent random variables into one — the toolkit behind totals, differences and sample averages.
Combine variance:$\mathrm{Var}(aX\pm bY)=a^{2}\mathrm{Var}(X)+b^{2}\mathrm{Var}(Y)$ (independent).
Poisson sum:Independent Poissons add their rates: $X+Y\sim\text{Po}(\lambda_1+\lambda_2)$.
Sample mean:$\bar{X}\sim N\!\left(\mu,\ \dfrac{\sigma^{2}}{n}\right)$, so the standard error is $\sigma_{\bar{X}}=\dfrac{\sigma}{\sqrt{n}}$.
TrapWhen combining, square every coefficient and add the variances — never subtract a variance, even for a difference $aX-bY$. Two errors live here: forgetting to square, and subtracting.
NoteThe sample-mean result — an average of $n$ readings clusters $\sqrt{n}$ times more tightly than a single reading — is the Central Limit Theorem behind confidence intervals (developed in the
Estimation cheatsheet, AHL 4.14). Standardise with $z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}$, dividing $\sigma$ by $\sqrt{n}$ first —
not $\sigma$.
§10 — Exam attack plan All sections
| Question cue | What to do | Watch for |
|---|
| "Table of probabilities, find $k$" | Set $\sum P(X=x)=1$ and solve for $k$ | Every entry in $[0,1]$; do this before $E(X)$ |
| "Find $E(X)$ / expected winnings" | $\sum x\,P(X=x)$; for a game subtract the stake | Fair $\Leftrightarrow$ net $0$; $E(X)$ need not be attainable |
| "Exactly $k$ successes in $n$ trials" | binompdf$(n,p,k)$ | "Exactly" = pdf, never cdf |
| "At least / at most $k$" (binomial) | binomcdf; $P(X\ge k)=1-$ binomcdf$(n,p,k-1)$ | The $k-1$ shift for "at least" |
| "Mean / sd of a binomial" | $np$ and $\sqrt{np(1-p)}$ | Variance keeps $(1-p)$; root only at the end |
| "$P(a<X<b)$ for a normal" | normalcdf$(a,b,\mu,\sigma)$; sketch & shade | Enter $\sigma$ not $\sigma^{2}$; $\pm1\text{E}99$ for open tails |
| "Find $x$, $\mu$ or $\sigma$ from a normal probability" | invNorm(area to left) $\to z$, then $x=\mu+z\sigma$ | Feed the LEFT area; "top $p$%" $\to$ use $1-p$; sign of $z$ |
| "Events at rate $\lambda$ per interval" | Poisson poissonpdf / poissoncdf; scale $\lambda$ to the interval | $\lambda$ is mean AND variance; independent sums add $\lambda$ |
Worked Example — Choosing & Applying the Right Distribution
Question (AI HL Paper 2 style — 10 marks)
A company manufactures LED bulbs. Independently, $4$% of all bulbs produced are defective.
(a) An inspector selects a box of $25$ bulbs at random. Find the probability that the box contains
(i) exactly one defective bulb;
(ii) at least two defective bulbs.
(b) Find the mean and the standard deviation of the number of defective bulbs in a box of $25$.
(c) The lifetime $L$ hours of a bulb is modelled by $L\sim N(1200,\ 150^{2})$. Find $P(L>1000)$.
(d) The manufacturer replaces free of charge the shortest-lived $5$% of bulbs. Find the lifetime below which a bulb is replaced.
Solution
- (a)(i) The count of defectives is $X\sim B(25,0.04)$ — a fixed $25$ trials with constant $p=0.04$. $P(X=1)=$ binompdf$(25,0.04,1)=0.375$ (3 s.f.). (M1)(A1)
- (a)(ii) $P(X\ge 2)=1-P(X\le 1)=1-$ binomcdf$(25,0.04,1)=1-0.736=0.264$ (3 s.f.). (M1)(A1) ("At least $2$" removes only the $0$ and $1$ cases — hence binomcdf at $k-1=1$.)
- (b) Mean $E(X)=np=25\times0.04=1$ defective. (A1) Standard deviation $\sigma=\sqrt{np(1-p)}=\sqrt{25\times0.04\times0.96}=\sqrt{0.96}=0.980$ (3 s.f.). (A1)
- (c) Lifetime is continuous, so use the normal: $P(L>1000)=$ normalcdf$(1000,\ 1\text{E}99,\ 1200,\ 150)=0.909$ (3 s.f.). (M1)(A1)
- (d) The shortest-lived $5$% is the left tail, so the area to the left is $0.05$: $L=$ invNorm$(0.05,\ 1200,\ 150)=953$ hours (3 s.f.). (M1)(A1)
Examiner's note: the first mark in each part is choosing the model from the wording — a fixed $25$ trials with constant $4$% is binomial; a continuous lifetime is normal. Keep the units (hours) and round to 3 s.f. unless told otherwise. Two easy marks routinely vanish: using binompdf where "at least" needs $1-$ binomcdf$(n,p,k-1)$, and giving invNorm the right-tail $0.95$ when the "shortest-lived $5$%" sits in the left tail at $0.05$.
Common Student Questions
What's the difference between binompdf and binomcdf — and how do I do "at least"?
binompdf$(n,p,k)$ gives the probability of exactly $k$ successes; binomcdf$(n,p,k)$ gives $P(X\le k)$, the probability of at most $k$. For "at least $k$", use $P(X\ge k)=1-$ binomcdf$(n,p,k-1)$ — note the $k-1$, because "at least $2$" excludes only the $0$ and $1$ outcomes. Match the wording to the command before you type: "exactly" is pdf, any range is cdf.
On the inverse normal, do I enter the area to the left or the right?
Always the area to the left of the value. invNorm answers the question "what value has this much probability below it?" So for the "top $5$%" — a right-tail cut-off — enter $0.95$; for the "bottom $5$%", enter $0.05$. Sketch the curve and shade the tail you were given, then read off which side the area sits on. Getting the wrong tail is the most common inverse-normal mistake.
How do I know whether to use the binomial, the Poisson or the normal?
Ask what is being counted or measured. A fixed number $n$ of independent yes/no trials with constant probability $p$ $\to$ binomial. A count of events happening at a steady average rate over an interval, with no fixed maximum $\to$ Poisson. A continuous measurement clustered symmetrically around a mean $\to$ normal. In AI the context words point straight at the model: "in a sample of $20$" (binomial), "on average $3$ per hour" (Poisson), "the mass is normally distributed" (normal).
The question gives me the variance — what do I put into normalcdf?
The standard deviation $\sigma$, which is $\sqrt{\text{variance}}$. IB writes a normal as $N(\mu,\sigma^{2})$, so a stated "variance of $16$" means $\sigma=4$ — enter $4$, never $16$. The binomial is the mirror trap: its formula gives you the variance $np(1-p)$, so you must take the square root to get the standard deviation. Whenever a distribution hands you a variance, root it before it goes near a probability.
How do I decide whether a game is "fair"?
Work out the expected net gain: $E(\text{gain})=E(\text{payout})-\text{stake}$, where $E(\text{payout})=\sum x\,P(X=x)$. The game is fair when this equals $0$, favourable to the player when positive, and favourable to the house when negative. The classic error is stopping at $E(\text{payout})$ and forgetting to subtract what it cost to play: if the expected payout equals the stake the game is fair, and if it falls short of the stake the house comes out ahead.
What's NOT in this cheatsheet
This page gives you the formulas and traps for free. The full Photon Academy IB Math AI HL library (enrolled students, or the lifetime resource library) adds:
- Notes PDF — every distribution, formula and GDC keystroke sequence, with fully worked binomial, normal and Poisson examples.
- Tutorial booklet — 25+ AI HL-style questions from distribution-table fluency to full modelling applications (defect rates, lifetimes, event counts).
- Tutorial Solutions — mark-scheme-style solutions with M1/A1 annotations and the exact GDC screens to reproduce.
- Predicted-paper questions — the probability-distribution question types most likely to appear in the next AI HL session.
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