Learn9709Paper 5 · Probability & Statistics 1Discrete random variablesThe binomial distribution

Discrete random variables · Subtopic 3 of 4

The Binomial Distribution

Count the successes in $n$ independent trials, each with the same chance $p$ . The probability of exactly $r$ successes is $\binom nr p^r(1-p)^{n-r}$ : choose which trials succeed, then multiply.

Cambridge 9709 Paper 5Edexcel 9MA0 Paper 3Edexcel 9FM0 OptionAQA 7357 Paper 3≈ 40 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

State the conditions for a binomial model, $X\sim B(n,p)$ .
Calculate $P(X=r)=\binom nr p^r q^{n-r}$ , where $q=1-p$ .
Find cumulative probabilities such as $P(X\le2)$ and $P(X\ge1)$ .
Use $E(X)=np$ and $\text{Var}(X)=npq$ .

Pause and think

A student guesses all $4$ answers on a multiple-choice quiz, each with $4$ options. Why is the probability of getting exactly $1$ right not just $\tfrac14\times\left(\tfrac34\right)^3$ ?

Section 2 of 14

Core idea

Start hereTen free throws

A basketball player makes 30% of free throws. In 10 attempts, what's the chance of exactly 3 baskets? Each throw is a yes/no trial with the same chance, independent of the others. That pattern is so common it has its own distribution: the binomial.

The big idea

$X\sim B(n,p)$ counts successes in $n$ independent trials with constant success probability $p$ . $P(X=r)=\binom nrp^r(1-p)^{n-r}$ ; $E(X)=np$ ; $\operatorname{Var}(X)=np(1-p)$ .

1. Conditions

Fixed number of trials $n$ .
Each trial: success or failure.
Constant probability $p$ .
Independent trials.

2. The formula

P(X=r)

$\text{one order (e.g. }SSS\,FFFFFFF):\ p^r(1-p)^{n-r}$
Independent: multiply.
$\text{number of orders}=\binom nr$
Choose which $r$ trials succeed.
$P(X=r)=\binom nrp^r(1-p)^{n-r}$
Each order equally likely; add them.

B(10, p): the shape as p changes

p = 0.10

Small

p

: the bars pile up near

0

p=0.5

: symmetric. Large

p

: piled up near

10

. The peak is always near the mean

np

Worked example $X\sim B(10,0.3)$

Find $P(X=3)$ , $P(X\le2)$ , $P(X\ge4)$ , mean and variance.

Goal: Values.
Strategy: Formula; complement for "at least".

$P(X=3)=120(0.027)(0.08235)=0.267$
$\binom{10}3=120$ .
$P(X\le2)=0.0282+0.1211+0.2335=0.383$
$r=0,1,2$ .
$P(X\ge4)=1-(0.383+0.267)=0.350$
Complement of $X\le3$ .
$E=3,\ \operatorname{Var}=2.1$
$np$ , $np(1-p)$ .

Answer

0.267

;

0.383

;

0.350

;

3

;

2.1

✓ CheckThe most likely value is

3=E(X)

✓.

Worked exampleFinding n

How many rolls of a die so that P(at least one six) > 0.9?

Goal: Least $n$ .
Strategy: Complement; logs.

$1-(\tfrac56)^n>0.9\Rightarrow(\tfrac56)^n<0.1$
Rearrange.
$n>\frac{\ln0.1}{\ln(5/6)}=12.6$
$\ln\frac56<0$ : flip.

Answer

13

rolls.

✓ Check

(\frac56)^{13}=0.093<0.1

✓;

(\frac56)^{12}=0.112

✗.

Common misconceptionWithout replacement Drawing cards without replacement changes

p

: not binomial.

Pause and think

Why isn't "number of aces in 5 cards from a pack" binomial?

Summary

Four conditions.
$\binom nrp^rq^{n-r}$ .
$np$ , $npq$ .

Trial: One repetition.
Binomial distribution: Number of successes in n independent trials.

X ~ B(10, 0.3)

Each bar is

P(X=r)=\binom{10}{r}(0.3)^r(0.7)^{10-r}

. The distribution peaks near the mean

np=3

and is skewed to the right because

p<0.5

HistoryBernoulli's Art of Conjecturing. Jacob Bernoulli worked for twenty years on Ars Conjectandi ("The Art of Conjecturing"), published in 1713, eight years after his death. It analysed repeated success-or-failure trials (now called Bernoulli trials) and proved the first law of large numbers: over many trials, the proportion of successes settles down to

p

Fun factThe bean machine. Francis Galton built a board in which balls bounce off rows of pins, going left or right with equal probability at each pin. The slot a ball lands in follows a binomial distribution, and thousands of balls pile up into the familiar hump shape. Desktop versions are sold as toys.

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