IB Maths AI HLPoisson DistributionPaper 1 & 2~7 min read
Calculating Poisson Probabilities
Once a situation is modelled as X ∼ Po(m), the probabilities come from your GDC — not by hand. Use the Poisson PD for a single value P(X = x) and the Poisson CD for a range P(a ≤ X ≤ b). Just like the binomial, the only real skill is turning the inequality in the question (strict, one-sided, or open) into a clean integer range. The mean m can be any positive real, but X only takes whole numbers.
📘 What you need to know
X is integer-valued: P(non-integer) = P(negative) = 0. Values are 0, 1, 2, … with no upper bound.
Single value: use Poisson PD (a.k.a. PPD / Poisson Pdf) for P(X = x).
Range: use Poisson CD (a.k.a. PCD / Poisson Cdf) for P(a ≤ X ≤ b).
Inputs: PD needs x and m; CD needs lower, upper, and m (many GDCs call m the λ value).
If your GDC only does P(X ≤ b): build everything from it with 1 − … and subtraction (identities below).
Open ranges: no lower limit → use 0; no upper limit → use a large number (9999… or 1099).
Single value: P(X = x)
For one exact value, use the Poisson probability distribution function on your GDC. There’s a formula too, but the exam expects the calculator.
Probability of a single value
P(X = x) = e−mmxx!use the Poisson PD on your GDC
🧭 Recipe — P(X = x) on the GDC
Identifym from the distribution Po(m).
Open the Poisson PD function (PPD / Poisson Pdf).
Enter the x value, then m (often labelled λ).
Read off the probability and round (usually 3 sf).
Calculator notation: many GDCs label the mean λ rather than m — they mean the same thing. Enter the average rate there.
WE 1
A single value
X ∼ Po(6.25). Find P(X = 5).
identify mm = 6.25 (the λ value)use Poisson PD on GDCP(X = 5) = 0.15341…P(X = 5) ≈ 0.153 (3sf)enter x = 5, λ = 6.25 into the PPD function.
Cumulative: P(a ≤ X ≤ b)
For a range of values, use the Poisson cumulative distribution function. Most GDCs take a lower and upper limit directly.
Open-ended ranges: if there’s no lower limit use 0; if there’s no upper limit use a large number (9999… or 1099) — the Poisson has no upper bound. So P(X ≤ b) = P(0 ≤ X ≤ b) and P(X ≥ a) = P(a ≤ X ≤ 9999…).
🤔 Why does P(X < 5) become P(X ≤ 4)?
Because X only takes whole numbers, “less than 5” means the integers 0, 1, 2, 3, 4 — exactly “≤ 4”. This integer trick works for any discrete variable, including the Poisson; it would fail for a continuous one. The same logic gives P(X > 5) = P(X ≥ 6).
🧭 Recipe — pin down the integer range
Find the smallest integerX can take in the range (the lower limit a).
Find the largest integer in the range (the upper limit b); if none, use a large number.
Entera, b, m into the Poisson CD function.
Write the rewritten inequality next to your answer for method marks.
WE 2
A “≤” cumulative probability
Y ∼ Po(4). Find P(Y ≤ 5).
rewrite with a lower limitP(Y ≤ 5) = P(0 ≤ Y ≤ 5)use Poisson CD on GDCP(Y ≤ 5) = 0.78513…P(Y ≤ 5) ≈ 0.785 (3sf)lower = 0, upper = 5, λ = 4.
Inequality identities
If your GDC only gives P(X ≤ x), every other probability can be built from it. These hold because X is a Poisson (integer) variable.
You want
Use P(X ≤ …)
Example
P(X < x)
P(X ≤ x − 1)
P(X < 5) = P(X ≤ 4)
P(X > x)
1 − P(X ≤ x)
P(X > 5) = 1 − P(X ≤ 5)
P(X ≥ x)
1 − P(X ≤ x − 1)
P(X ≥ 5) = 1 − P(X ≤ 4)
P(a ≤ X ≤ b)
P(X ≤ b) − P(X ≤ a − 1)
P(5 ≤ X ≤ 9) = P(X ≤ 9) − P(X ≤ 4)
Turning any inequality into an integer range
List the integers actually included, then read off the smallest and largest.
WE 3
Combining Poissons, then “greater than”
X ∼ Po(6.25) and Y ∼ Po(4) are independent. Find P(X + Y > 7).
form the combined distributionX + Y ∼ Po(6.25 + 4) = Po(10.25)rewrite “> 7” (no upper bound)P(X+Y > 7) = 1 − P(X+Y ≤ 7)Poisson CD, λ = 10.25= 1 − 0.198538… = 0.80146…P(X + Y > 7) ≈ 0.801 (3sf)add the means first, then subtract P(≤ 7) from 1.
WE 4
Using “at least” (≥)
X ∼ Po(6.25). Find P(X ≥ 8).
rewrite using the ≥ identityP(X ≥ 8) = 1 − P(X ≤ 7)Poisson CD: P(X ≤ 7)= 0.70890… → 1 − 0.70890…P(X ≥ 8) ≈ 0.291 (3sf)“at least 8” includes 8, so subtract P(X ≤ 7), not P(X ≤ 8).
WE 5
A strict double inequality
X ∼ Po(6.25). Find P(3 < X < 8).
identify the integers includedP(3 < X < 8) = P(4 ≤ X ≤ 7)use Poisson CD on GDClower = 4, upper = 7, λ = 6.25P(3 < X < 8) ≈ 0.579 (3sf)strict both ends: bump the lower up by 1 and the upper down by 1.
💡 Top tips
PD for one value, CD for a range — pick the right GDC function first.
List the integers in the range, then take the smallest and largest as a and b.
Strict < / > shift the limit by 1: P(X < x) = P(X ≤ x − 1), P(X > x) = P(X ≥ x + 1).
No lower/upper limit? use 0 or a large number (9999…).
Combine Poissons first (add the means) before finding P(X + Y …).
Watch the λ label — your GDC may call the mean λ instead of m.
⚠ Common mistakes
Confusing < with ≤ — forgetting to shift the limit by 1 for strict inequalities.
Using PD for a range (or CD for a single value) — wrong function entirely.
Off-by-one on P(X ≥ x) — it’s 1 − P(X ≤ x − 1), not 1 − P(X ≤ x).
Adding the SDs (or √m) when combining Poissons — add the means instead.
Rounding too early, then losing accuracy in a follow-up calculation.
Not stating the rewritten inequality, so a mistyped GDC entry earns nothing.
That completes the Poisson Distribution unit! You can now check the conditions, set up X ∼ Po(m), scale rates, combine independent Poissons, and calculate exact and cumulative probabilities on your GDC. Next this feeds into hypothesis testing, where you’ll test claims about a Poisson mean using the very same PD / CD machinery.
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