IB Mathematics Analysis and Approaches Topic SL…

Binomial Distribution: Counting Successes In Repeated Trials

Flip a coin 10 times and count the heads.
Test 50 components and count how many come out faulty.
Guess on 20 multiple-choice questions and count how many you get right.
These look like three different problems, but they share one skeleton: the same yes-or-no trial repeated a fixed number of times, with the successes counted.
The binomial distribution is the single model that answers all three.
Until now you could only find the probability of one specific run of outcomes, like heads then tails then heads.
The binomial distribution bundles together every run that has the same number of successes.
So you can jump straight to a question like the chance of exactly 3 heads, without listing every way it could happen.

When A Situation Is Binomial: The Four Conditions

Check all four conditions before using the model, because if even one fails the binomial distribution does not apply.

A fixed number of trials, $n$: You decide in advance how many times the trial runs, such as 20 questions or 50 components.
Only two outcomes each trial: Each trial ends in a success or a failure, and you choose which outcome to call a success.
The trials are independent: The result of one trial does not change the probabilities for any other trial.
A constant probability of success, $p$: The same probability $p$ applies on every single trial.

Note

Independent plus constant $p$ is the pair students most often lose.
Drawing objects without replacement breaks both, because each draw changes what is left.
If the number of trials is not fixed in advance, the model does not fit either.

Writing It Down: The Notation $X \sim B(n, p)$

Let $X$ be the number of successes across the $n$ trials, which makes $X$ a discrete random variable.
You summarise which binomial model it follows in a single line of notation. $$X \sim B(n,\, p)$$
1. $n$ is the number of trials.
2. $p$ is the probability of success on a single trial.
3. $X$ can take any whole-number value from $0$ up to $n$.

Building The Probability Formula From A Single Sequence

Build it yourself with $n = 5$ trials and exactly $k = 2$ successes.
Start with one particular order, say success, success, failure, failure, failure ($SSFFF$).
Because the trials are independent, the probability of that exact order multiplies out: $$P(SSFFF) = p \times p \times (1-p) \times (1-p) \times (1-p) = p^{2}(1-p)^{3}$$
1. Every other order with 2 successes and 3 failures has the same probability $p^{2}(1-p)^{3}$, because multiplication ignores order.
So the only thing left is to count how many such orders there are.
1. Choosing which 2 of the 5 trials are the successes is a combinations count, $\binom{5}{2} = 10$.
2. Adding those 10 equally likely orders gives $P(X = 2) = \binom{5}{2}\, p^{2}(1-p)^{3}$.
Replacing the specific numbers with $n$, $k$ and $n-k$ gives the general result. $$P(X = k) = \binom{n}{k}\, p^{k}(1-p)^{\,n-k}, \qquad k = 0, 1, 2, \ldots, n$$
1. $\binom{n}{k}$ counts the orders, that is which of the $n$ trials are the successes.
2. $p^{k}$ is the probability of the $k$ successes.
3. $(1-p)^{\,n-k}$ is the probability of the $n-k$ failures.

Note

The count $\binom{n}{k}$ is the same $n\mathrm{C}r$ from the binomial theorem in Topic 1.9.
Find it with the formula $\binom{n}{k} = \dfrac{n!}{k!(n-k)!}$ or straight from your GDC.

Exam technique

The formula booklet, under Topic 4, lists the binomial distribution with its mean $np$ and variance $np(1-p)$.
In the exam you are expected to find binomial probabilities with your GDC, not by hand.
So the probability formula above explains where the GDC's number comes from, and the syllabus never asks you to prove the mean or variance.

Finding Binomial Probabilities On The GDC

Two calculator commands do all the work, and everything else is rewriting the question to fit one of them.

GDC command	What it finds	Matches the words
binompdf(n, p, k)	P(X = k)	exactly k
binomcdf(n, p, k)	P(X <= k)	at most k, k or fewer

For "at least", "more than" and "between", rewrite the event using these two commands:

In words	As a probability	On the GDC
at most k	P(X <= k)	binomcdf(n, p, k)
fewer than k	P(X <= k-1)	binomcdf(n, p, k-1)
at least k	1 - P(X <= k-1)	1 - binomcdf(n, p, k-1)
more than k	1 - P(X <= k)	1 - binomcdf(n, p, k)
between a and b (inclusive)	P(X <= b) - P(X <= a-1)	binomcdf(n,p,b) - binomcdf(n,p,a-1)

Common Mistake

"At least $k$" uses $k-1$, not $k$, inside the cumulative command.
$P(X \ge k) = 1 - P(X \le k-1)$, because $X = k$ itself must stay on the "at least" side.
Getting this boundary wrong by one is the most common binomial slip in exams.

Example question

A test has 12 multiple-choice questions, each with 4 options and exactly one correct answer. A student guesses every question at random. Find the probability that the student gets (a) exactly 4 correct, and (b) at least 4 correct.

Solution

Model the number correct as $X \sim B(12,\, 0.25)$, since $n = 12$ and $p = \tfrac{1}{4}$.
(a): $P(X = 4) = \texttt{binompdf}(12, 0.25, 4) = 0.194$ (3 s.f.).
(b): rewrite "at least 4" as $P(X \ge 4) = 1 - P(X \le 3)$.
$= 1 - \texttt{binomcdf}(12, 0.25, 3) = 1 - 0.649 = 0.351$ (3 s.f.).

Mean And Variance: The Centre And The Spread

You do not need the whole distribution to describe a binomial, because two summary numbers place its centre and its spread.
The mean is easy to believe: with probability $p$ of success on each trial, across $n$ trials you expect a fraction $p$ of them to succeed, which is $np$.

$$E(X) = np \qquad\qquad \operatorname{Var}(X) = np(1-p)$$

Note

Both formulas are given in the formula booklet, so you select and apply them rather than recall them.
The syllabus states you will not be asked to prove either result, so use them directly.
The spread $np(1-p)$ is largest when $p = 0.5$ and shrinks as $p$ moves toward 0 or 1.

Standard Deviation And A Worked Example

The standard deviation is the square root of the variance, so $\sigma = \sqrt{np(1-p)}$, in the same units as the count itself.

Example question

A biased coin lands heads with probability $0.3$. It is tossed 40 times. Find the mean, variance and standard deviation of the number of heads.

Solution

Identify $X \sim B(40,\, 0.3)$, so $n = 40$ and $p = 0.3$.
Mean: $E(X) = np = 40 \times 0.3 = 12$ heads.
Variance: $\operatorname{Var}(X) = np(1-p) = 40 \times 0.3 \times 0.7 = 8.4$.
Standard deviation: $\sigma = \sqrt{8.4} = 2.90$ (3 s.f.).

Spotting Binomial And Non-Binomial Situations

The exam often hides whether a situation is binomial, so test each scenario against the four conditions before modelling.

Scenario	Binomial?	Reason
Flip a fair coin 10 times, count heads	Yes	Fixed n, constant p = 0.5, independent, two outcomes
Roll a die 20 times, count sixes	Yes	Six versus not-six, p = 1/6 stays constant
Draw 5 cards without replacement, count red	No	p changes each draw, so trials are not independent
Ask people until the first says yes	No	The number of trials is not fixed in advance

Common Mistake

The classic trap is sampling without replacement, where $p$ shifts on every draw.
If the population is very large, sampling without replacement is close enough to binomial to be a reasonable model.
State that modelling assumption when a question asks you to justify using a binomial distribution.

See How The Shape Depends On $n$ And $p$

The binomial is not always a tidy bell, because its shape swings with $p$ and only smooths out as $n$ grows.
Watching those two dials move is the fastest way to build intuition for it.

Tip

Two facts stick: the distribution is symmetric only when $p = 0.5$, and as $n$ grows a binomial starts to resemble the normal curve of the next subtopic.

Active recall

State the four conditions a situation must meet to be modelled by a binomial distribution.
In the formula $P(X = k) = \binom{n}{k} p^{k}(1-p)^{n-k}$, what does the factor $\binom{n}{k}$ count?
Which two binomial results are given in the formula booklet, and which one does the syllabus say you will never be asked to prove?
How do you rewrite $P(X \ge k)$ so it can be found with a single cumulative GDC command?
For $X \sim B(n, p)$, write down the mean and the variance, and state the value of $p$ that makes the spread largest.

Binomial Distribution: Counting Successes In Repeated Trials

Flip a coin 10 times and count the heads.
Test 50 components and count how many come out faulty.
Guess on 20 multiple-choice questions and count how many you get right.
These look like three different problems, but they share one skeleton: the same yes-or-no trial repeated a fixed number of times, with the successes counted.
The binomial distribution is the single model that answers all three.
Until now you could only find the probability of one specific run of outcomes, like heads then tails then heads.
The binomial distribution bundles together every run that has the same number of successes.
So you can jump straight to a question like the chance of exactly 3 heads, without listing every way it could happen.

When A Situation Is Binomial: The Four Conditions

Check all four conditions before using the model, because if even one fails the binomial distribution does not apply.

A fixed number of trials, $n$: You decide in advance how many times the trial runs, such as 20 questions or 50 components.
Only two outcomes each trial: Each trial ends in a success or a failure, and you choose which outcome to call a success.
The trials are independent: The result of one trial does not change the probabilities for any other trial.
A constant probability of success, $p$: The same probability $p$ applies on every single trial.

Note

Independent plus constant $p$ is the pair students most often lose.
Drawing objects without replacement breaks both, because each draw changes what is left.
If the number of trials is not fixed in advance, the model does not fit either.

Writing It Down: The Notation $X \sim B(n, p)$

Let $X$ be the number of successes across the $n$ trials, which makes $X$ a discrete random variable.
You summarise which binomial model it follows in a single line of notation. $$X \sim B(n,\, p)$$
1. $n$ is the number of trials.
2. $p$ is the probability of success on a single trial.
3. $X$ can take any whole-number value from $0$ up to $n$.

Building The Probability Formula From A Single Sequence

Build it yourself with $n = 5$ trials and exactly $k = 2$ successes.
Start with one particular order, say success, success, failure, failure, failure ($SSFFF$).
Because the trials are independent, the probability of that exact order multiplies out: $$P(SSFFF) = p \times p \times (1-p) \times (1-p) \times (1-p) = p^{2}(1-p)^{3}$$
1. Every other order with 2 successes and 3 failures has the same probability $p^{2}(1-p)^{3}$, because multiplication ignores order.
So the only thing left is to count how many such orders there are.
1. Choosing which 2 of the 5 trials are the successes is a combinations count, $\binom{5}{2} = 10$.
2. Adding those 10 equally likely orders gives $P(X = 2) = \binom{5}{2}\, p^{2}(1-p)^{3}$.
Replacing the specific numbers with $n$, $k$ and $n-k$ gives the general result. $$P(X = k) = \binom{n}{k}\, p^{k}(1-p)^{\,n-k}, \qquad k = 0, 1, 2, \ldots, n$$
1. $\binom{n}{k}$ counts the orders, that is which of the $n$ trials are the successes.
2. $p^{k}$ is the probability of the $k$ successes.
3. $(1-p)^{\,n-k}$ is the probability of the $n-k$ failures.

Note

The count $\binom{n}{k}$ is the same $n\mathrm{C}r$ from the binomial theorem in Topic 1.9.
Find it with the formula $\binom{n}{k} = \dfrac{n!}{k!(n-k)!}$ or straight from your GDC.

Exam technique

The formula booklet, under Topic 4, lists the binomial distribution with its mean $np$ and variance $np(1-p)$.
In the exam you are expected to find binomial probabilities with your GDC, not by hand.
So the probability formula above explains where the GDC's number comes from, and the syllabus never asks you to prove the mean or variance.

Finding Binomial Probabilities On The GDC

Two calculator commands do all the work, and everything else is rewriting the question to fit one of them.

GDC command	What it finds	Matches the words
binompdf(n, p, k)	P(X = k)	exactly k
binomcdf(n, p, k)	P(X <= k)	at most k, k or fewer

For "at least", "more than" and "between", rewrite the event using these two commands:

In words	As a probability	On the GDC
at most k	P(X <= k)	binomcdf(n, p, k)
fewer than k	P(X <= k-1)	binomcdf(n, p, k-1)
at least k	1 - P(X <= k-1)	1 - binomcdf(n, p, k-1)
more than k	1 - P(X <= k)	1 - binomcdf(n, p, k)
between a and b (inclusive)	P(X <= b) - P(X <= a-1)	binomcdf(n,p,b) - binomcdf(n,p,a-1)

Common Mistake

"At least $k$" uses $k-1$, not $k$, inside the cumulative command.
$P(X \ge k) = 1 - P(X \le k-1)$, because $X = k$ itself must stay on the "at least" side.
Getting this boundary wrong by one is the most common binomial slip in exams.

Example question

Solution

Model the number correct as $X \sim B(12,\, 0.25)$, since $n = 12$ and $p = \tfrac{1}{4}$.
(a): $P(X = 4) = \texttt{binompdf}(12, 0.25, 4) = 0.194$ (3 s.f.).
(b): rewrite "at least 4" as $P(X \ge 4) = 1 - P(X \le 3)$.
$= 1 - \texttt{binomcdf}(12, 0.25, 3) = 1 - 0.649 = 0.351$ (3 s.f.).

Mean And Variance: The Centre And The Spread

You do not need the whole distribution to describe a binomial, because two summary numbers place its centre and its spread.
The mean is easy to believe: with probability $p$ of success on each trial, across $n$ trials you expect a fraction $p$ of them to succeed, which is $np$.

$$E(X) = np \qquad\qquad \operatorname{Var}(X) = np(1-p)$$

Note

Both formulas are given in the formula booklet, so you select and apply them rather than recall them.
The syllabus states you will not be asked to prove either result, so use them directly.
The spread $np(1-p)$ is largest when $p = 0.5$ and shrinks as $p$ moves toward 0 or 1.

Standard Deviation And A Worked Example

The standard deviation is the square root of the variance, so $\sigma = \sqrt{np(1-p)}$, in the same units as the count itself.

Example question

A biased coin lands heads with probability $0.3$. It is tossed 40 times. Find the mean, variance and standard deviation of the number of heads.

Solution

Identify $X \sim B(40,\, 0.3)$, so $n = 40$ and $p = 0.3$.
Mean: $E(X) = np = 40 \times 0.3 = 12$ heads.
Variance: $\operatorname{Var}(X) = np(1-p) = 40 \times 0.3 \times 0.7 = 8.4$.
Standard deviation: $\sigma = \sqrt{8.4} = 2.90$ (3 s.f.).

Spotting Binomial And Non-Binomial Situations

The exam often hides whether a situation is binomial, so test each scenario against the four conditions before modelling.

Scenario	Binomial?	Reason
Flip a fair coin 10 times, count heads	Yes	Fixed n, constant p = 0.5, independent, two outcomes
Roll a die 20 times, count sixes	Yes	Six versus not-six, p = 1/6 stays constant
Draw 5 cards without replacement, count red	No	p changes each draw, so trials are not independent
Ask people until the first says yes	No	The number of trials is not fixed in advance

Common Mistake

The classic trap is sampling without replacement, where $p$ shifts on every draw.
If the population is very large, sampling without replacement is close enough to binomial to be a reasonable model.
State that modelling assumption when a question asks you to justify using a binomial distribution.

See How The Shape Depends On $n$ And $p$

The binomial is not always a tidy bell, because its shape swings with $p$ and only smooths out as $n$ grows.
Watching those two dials move is the fastest way to build intuition for it.

Tip

Two facts stick: the distribution is symmetric only when $p = 0.5$, and as $n$ grows a binomial starts to resemble the normal curve of the next subtopic.

Active recall

State the four conditions a situation must meet to be modelled by a binomial distribution.
In the formula $P(X = k) = \binom{n}{k} p^{k}(1-p)^{n-k}$, what does the factor $\binom{n}{k}$ count?
Which two binomial results are given in the formula booklet, and which one does the syllabus say you will never be asked to prove?
How do you rewrite $P(X \ge k)$ so it can be found with a single cumulative GDC command?
For $X \sim B(n, p)$, write down the mean and the variance, and state the value of $p$ that makes the spread largest.

IB Mathematics Analysis and Approaches SL 4.8—binomial Distribution Notes

IB Mathematics Analysis and Approaches SL 4.8—binomial Distribution Notes

Binomial Distribution: Counting Successes In Repeated Trials

When A Situation Is Binomial: The Four Conditions

Writing It Down: The Notation $X \sim B(n, p)$

Building The Probability Formula From A Single Sequence

Finding Binomial Probabilities On The GDC

Mean And Variance: The Centre And The Spread

Standard Deviation And A Worked Example

Spotting Binomial And Non-Binomial Situations

See How The Shape Depends On $n$ And $p$

Binomial Distribution: Counting Successes In Repeated Trials

When A Situation Is Binomial: The Four Conditions

Writing It Down: The Notation $X \sim B(n, p)$

Building The Probability Formula From A Single Sequence

Finding Binomial Probabilities On The GDC

Mean And Variance: The Centre And The Spread

Standard Deviation And A Worked Example

Spotting Binomial And Non-Binomial Situations

See How The Shape Depends On $n$ And $p$

Number and Algebra

Functions

Geometry & Trigonometry

Statistics & Probability

Calculus

Number and Algebra

Functions

Geometry & Trigonometry

Statistics & Probability

Calculus

IB Mathematics Analysis and Approaches SL 4.8—binomial Distribution Notes

IB Mathematics Analysis and Approaches SL 4.8—binomial Distribution Notes

Binomial Distribution: Counting Successes In Repeated Trials

When A Situation Is Binomial: The Four Conditions

Writing It Down: The Notation $X \sim B(n, p)$

Building The Probability Formula From A Single Sequence

Finding Binomial Probabilities On The GDC

Mean And Variance: The Centre And The Spread

Standard Deviation And A Worked Example

Spotting Binomial And Non-Binomial Situations

See How The Shape Depends On $n$ And $p$

Binomial Distribution: Counting Successes In Repeated Trials

When A Situation Is Binomial: The Four Conditions

Writing It Down: The Notation $X \sim B(n, p)$

Building The Probability Formula From A Single Sequence

Finding Binomial Probabilities On The GDC

Mean And Variance: The Centre And The Spread

Standard Deviation And A Worked Example

Spotting Binomial And Non-Binomial Situations

See How The Shape Depends On $n$ And $p$