Statistics · Statistical Distributions · Year 1

Cumulative Probabilities

A single binomial term is rarely what a question actually wants — "at least", "no more than", "fewer than" all describe a whole range of outcomes, and cumulative probabilities are how you add them up in one step.

10 min read 4 worked examples

The cumulative distribution function

For a random variable $X$, the cumulative probability function gives the total probability of $X$ taking a value up to and including $x$:

$F(x) = P(X\le x) = \displaystyle\sum_{k=0}^{x} P(X=k).$

For $X\sim B(n,p)$, tables of $P(X\le x)$ exist for common values of $n$ and $p$, and every scientific or graphical calculator has a binomial cumulative distribution function built in — so in practice you almost never sum individual terms by hand once $x$ is more than 1 or 2.

A short extract

Here is $P(X\le x)$ for $X\sim B(6,0.4)$, computed term by term:

x0123456
P(X≤x)0.04670.23330.54430.82080.95900.99591.0000

Reading this table tells you, for instance, that $P(X\le3)=0.8208$ directly — no addition required once the table (or calculator) has done the work.

Why P(X≤n) is always exactly 1

The final entry in any cumulative binomial table is always $1$: $X$ can take no value greater than $n$, so the probability of "$X\le n$" is certain. This is a useful check that a table or calculator output is sensible.

Translating the wording

Exam wording rarely says "$P(X\le x)$" directly — it uses everyday phrases that need converting first. Since a binomial random variable only takes whole-number values, strict inequalities can always be rewritten using $\le$ or $\ge$:

PhraseMeansCalculate as
more than 5$X>5$$1-P(X\le5)$
no more than 5$X\le5$$P(X\le5)$
at least 5$X\ge5$$1-P(X\le4)$
fewer than 5$X<5$$P(X\le4)$
at most 5$X\le5$$P(X\le5)$
Strategy — always rewrite in terms of ≤ first

Whatever phrase a question uses, translate it into an expression built entirely from $P(X\le k)$ for some whole number $k$ — using $P(X\ge r)=1-P(X\le r-1)$ and $P(X>r)=1-P(X\le r)$ where needed. Every table and every calculator function is built around $P(X\le x)$, so this is the one form everything else should collapse into.

Worked example 1

No more than / at least, from a cumulative table
A spinner is designed so that $P(\text{red})=0.25$. Priya spins it 15 times. Find the probability that she gets: (a) no more than 3 reds, (b) at least 6 reds.
MethodIdentify the distribution, then translate each phrase
1
Let $X$ be the number of reds in 15 spins. $X\sim B(15, 0.25)$.
Fixed trials, two outcomes, constant p, independent spins.
2
(a) "No more than 3" means $X\le3$: $P(X\le3) = 0.4613$.
Read directly from the cumulative table or calculator.
3
(b) "At least 6" means $X\ge6$: $P(X\ge6) = 1-P(X\le5) = 1-0.8516 = 0.1484$.
Complement of the ≤5 cumulative probability.
▪ solved
Examiner's eye

Write the translation ("no more than 3 means $X\le3$") explicitly before calculating — it is usually worth its own mark, and prevents the easy mistake of computing $P(X\le6)$ instead of $1-P(X\le5)$ for an "at least" statement.

Worked example 2

A probability strictly between two values
Using the same spinner as above ($X\sim B(15,0.25)$), find $P(3<X\le8)$.
MethodSubtract one cumulative probability from another
1
$P(3<X\le8) = P(X\le8) - P(X\le3)$.
The region between two cumulative probabilities is their difference.
2
From tables or a calculator: $P(X\le8)\approx0.9958$, $P(X\le3)=0.4613$.
Look up (or compute) each cumulative value.
3
$P(3<X\le8) \approx 0.9958-0.4613 = 0.5345$.
Subtract.
▪ solved
The key insight

Any "strictly between" probability, however it's phrased, can always be built from two cumulative probabilities subtracted from each other — the only care needed is deciding whether each endpoint is included, which fixes exactly which two values of $k$ to use.

Worked example 3

Working backwards to find a threshold · Challenge
In a game, a contestant has 10 independent attempts, each succeeding with probability $0.2$. The organiser wants the probability of a contestant getting at least $r$ successes to be less than $0.1$. Find the smallest such $r$.
MethodTest successive values of r against the threshold
1
Let $X$ be the number of successes. $X\sim B(10, 0.2)$. Require $P(X\ge r)<0.1$.
Set up the distribution and the condition.
2
$P(X\ge4)=0.1209$, $P(X\ge5)=0.0328$.
Using $P(X\ge r)=1-P(X\le r-1)$ from a calculator.
3
$P(X\ge4)=0.1209 > 0.1$, but $P(X\ge5)=0.0328 < 0.1$.
Compare each against the 0.1 threshold.
4
So the smallest $r$ satisfying the condition is $r=5$.
The first value of r that pushes the probability below 0.1.
▪ found
Strategy — work outward from a sensible starting guess

Rather than testing every value of $r$ from $0$, start near the distribution's mean ($np=2$ here) and move outward — this "threshold-hunting" technique reappears, almost unchanged, when finding critical regions in hypothesis testing.

Common mistake

$P(X<r)$ and $P(X\le r)$ are not the same for a discrete variable — $P(X<r)=P(X\le r-1)$, one step lower. Mixing these up is the single most common error in this topic, and it's always worth writing the "means" column from the translation table explicitly before calculating anything.

Check yourself

$X\sim B(12, 0.3)$. Given $P(X\le2)=0.2528$ and $P(X\le6)=0.9610$, find $P(X\ge3)$ and $P(2<X\le6)$.
(Hint: "at least 3" is the complement of "≤2".)

Method
1
$P(X\ge3) = 1-P(X\le2) = 1-0.2528 = 0.7472$.
2
$P(2<X\le6) = P(X\le6)-P(X\le2) = 0.9610-0.2528 = 0.7082$.
▪ solved