Binomial Distribution

MATBinomial Distribution

A binomial distribution models the number of successes $X$ in $n$ independent Bernoulli trials, each with the same probability of success $p$ (and failure $q = 1-p$ ). It is examined heavily because almost any "repeated identical trial" set-up — dice thrown several times, defective items in a sample, guessing MCQs, repeated shots at a target — reduces to computing $P(X=r)$ , or the mean $np$ and variance $npq$ .

Binomial probability (general term)

P(X = r) = \binom{n}{r} p^{r} q^{\,n-r}, \quad r = 0,1,2,\dots,n

n

= number of trials,

p

= probability of success in one trial,

q = 1-p

r

= number of successes;

\binom{n}{r} = \dfrac{n!}{r!\,(n-r)!}

At least one success

P(X \ge 1) = 1 - P(X = 0) = 1 - q^{\,n}

Use the complement instead of summing

r=1

n

. Here

q^{n} = P(\text{no success at all})

Mean of a binomial distribution

\mu = E(X) = np

\mu

is the expected number of successes in

n

trials.

Variance and standard deviation

\operatorname{Var}(X) = npq, \qquad \sigma = \sqrt{npq}

\operatorname{Var}(X) = npq

with

q = 1-p

; standard deviation

\sigma = \sqrt{npq}

. Since

q < 1

, always

\operatorname{Var}(X) < \mu

Recovering parameters from mean and variance

q = \dfrac{\operatorname{Var}(X)}{\mu}, \quad p = 1 - q, \quad n = \dfrac{\mu}{p}

Given

\mu = np

and

\operatorname{Var}(X) = npq

, divide to get

q

, then

p

and

n

. Applies to the "mean and variance given, find

n,p

" type problem.

Four conditions for a binomial setting: (i) a fixed number $n$ of trials, (ii) only two outcomes per trial (success/failure), (iii) trials are independent, (iv) $p$ is constant across trials.
Always identify $n$ , $p$ and $q$ first, and write the general term $P(X=r) = \binom{n}{r} p^{r} q^{n-r}$ before substituting.
"At least $k$ " means sum from $r=k$ to $n$ ; "at most $k$ " means sum from $r=0$ to $k$ . Use the complement $1 - P(X=0)$ for "at least one".
For a fair die, success "a six" has $p = \dfrac{1}{6}$ , $q = \dfrac{5}{6}$ ; for a fair coin $p = q = \dfrac{1}{2}$ ; for guessing a 4-option MCQ $p = \dfrac{1}{4}$ .
Mean $= np$ and variance $= npq$ , so the variance is always less than the mean; if a question gives variance $>$ mean the data cannot be binomial.
All $n+1$ probabilities $P(X=0), P(X=1), \dots, P(X=n)$ sum to $1$ , since $\sum \binom{n}{r} p^{r} q^{n-r} = (q+p)^{n} = 1$ .
Sampling "with replacement" or independent repetitions keeps $p$ constant (binomial); drawing "without replacement" changes $p$ each draw, so it is NOT binomial.

Condition	What it means	What breaks it
Fixed $n$	The number of trials is decided in advance	'Keep going until the first success'
Two outcomes	Each trial is a success or a failure	A die scored $1$ – $6$ rather than 'six or not six'
Constant $p$	The success probability is the same every trial	Drawing without replacement
Independence	One trial's result does not affect another's	Drawing without replacement

All four must hold before you may write

P(X = r) = {}^{n}C_{r}\,p^{r}q^{\,n-r}

. Sampling *without* replacement breaks the third and fourth, and the distribution becomes hypergeometric, not binomial.

Worked example · 3 marks

A fair die is thrown

5

times. Find the probability of getting exactly two sixes.

Each throw is a success (a six) or not, $n = 5$ is fixed, $p = \dfrac{1}{6}$ is constant and throws are independent — so $X \sim B\!\left(5, \dfrac{1}{6}\right)$ .
$P(X = r) = {}^{n}C_{r}\,p^{r}q^{\,n-r}$ with $q = \dfrac{5}{6}$ .
$P(X = 2) = {}^{5}C_{2}\left(\dfrac{1}{6}\right)^{2}\left(\dfrac{5}{6}\right)^{3} = 10\times\dfrac{1}{36}\times\dfrac{125}{216}$ .
$= \dfrac{1250}{7776}$ .

Answer

P(X = 2) = \dfrac{625}{3888} \approx 0.161

Where the marks go

Mixing up success and failure exponents: in $\binom{n}{r} p^{r} q^{n-r}$ the power of $p$ equals the number of successes $r$ , not $n-r$ .
Forgetting the combinatorial coefficient $\binom{n}{r}$ and writing only $p^{r} q^{n-r}$ — this undercounts the arrangements.
Using $\operatorname{Var}(X) = np$ or $\sqrt{np}$ instead of $npq$ ; the variance carries the extra factor $q$ .
Computing "at least one" by summing many terms and making arithmetic slips, instead of the clean complement $1 - q^{n}$ .

How the board asks it

Numericalgeneral term of a binomial distribution
A die is thrown $6$ times. If getting an odd number is a success, find the probability of (i) exactly $5$ successes, (ii) at most $2$ successes.
Numericalat least one success using the complement
The probability that a bulb produced by a factory is defective is $0.05$ . If a sample of $10$ bulbs is drawn, find the probability that the sample contains at least one defective bulb.
Numericalmean and variance of a binomial distribution
A pair of dice is thrown $4$ times. Getting a doublet is considered a success. Find the mean and variance of the number of successes.
Numericalrecovering parameters from given mean and variance
The mean and variance of a binomial distribution are $4$ and $\dfrac{4}{3}$ respectively. Find $P(X \geq 1)$ .
Give reasonsbinomial conditions: sampling with vs without replacement
Five cards are drawn one by one, without replacement, from a well-shuffled pack of $52$ cards. State, with reasons, whether the number of kings drawn follows a binomial distribution.

Practise this topic

PreviousRandom Variables & Probability Distributions (Mean, Variance)

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.