IB Mathematics Applications & Interpretation Topic SL…

Binomial Distribution: Counting Successes In Repeated Trials

Flip a coin 10 times and count the heads.
Test 50 phone screens off a production line and count how many are 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.

Note

Until now you could only find the probability of one specific run, such as 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 order it could happen in.

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 screens.
Only two outcomes each trial: every 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 one 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

$\binom{n}{k}$, read "n choose k", is the number of ways to place the $k$ successes among the $n$ trials.
Find it with the $n\mathrm{C}r$ key on your GDC, or with $\binom{n}{k} = \dfrac{n!}{k!(n-k)!}$.

What The Formula Booklet Gives You

Under Topic 4, the formula booklet lists the binomial distribution with its mean $np$ and variance $np(1-p)$.
The probability formula above is there to explain where the calculator's number comes from, not to be evaluated by hand.
Both SL papers are calculator papers, so in the exam you always find binomial probabilities on the GDC.

Exam technique

Because the mean and variance are given in the booklet, the marks are for selecting and applying them, not for recalling them.
The syllabus states you will never be asked to prove the mean or the variance, so use them directly.
Quote the formula you use, substitute the numbers, then write the value: the substitution is the method mark.

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.)

Example question

A machine produces microchips, and on average 8% are defective. A quality inspector tests a random sample of 25 chips. Find the probability that the sample contains (a) exactly 2 defective chips, (b) more than 3 defective chips, and (c) between 1 and 4 defective chips inclusive.

Solution

Model the number of defective chips as $X \sim B(25,\, 0.08)$.
(a) $P(X = 2) = \texttt{binompdf}(25, 0.08, 2) = 0.282$ (3 s.f.)
(b) rewrite "more than 3" as $P(X > 3) = 1 - P(X \le 3)$.
- $= 1 - \texttt{binomcdf}(25, 0.08, 3) = 0.135$ (3 s.f.)
(c) rewrite "between 1 and 4 inclusive" as $P(1 \le X \le 4) = P(X \le 4) - P(X \le 0)$.
- $= \texttt{binomcdf}(25, 0.08, 4) - \texttt{binomcdf}(25, 0.08, 0) = 0.830$ (3 s.f.).

Example question

A biased coin lands heads with probability $0.4$. It is tossed $n$ times. Find the smallest value of $n$ for which the probability of getting at least one head is greater than $0.99$.

Solution

"At least one head" is the complement of no heads, so $P(X \ge 1) = 1 - P(X = 0) = 1 - 0.6^{\,n}$.
Require $1 - 0.6^{\,n} > 0.99$, which rearranges to $0.6^{\,n} < 0.01$.
Take logs: $n > \dfrac{\ln 0.01}{\ln 0.6} = 9.02$, so $n$ must be a whole number above $9.02$.
The smallest such value is $n = 10$; a GDC check gives $1 - 0.6^{10} = 0.994 > 0.99$.

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 of $n$ 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 spread $np(1-p)$ is largest when $p = 0.5$ and shrinks as $p$ moves toward 0 or 1.
The variance is $np(1-p)$; the standard deviation is its square root, $\sqrt{np(1-p)}$.

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.

Example question

In a large city, 30% of adults cycle to work. A researcher stops 15 adults at random and counts how many cycle to work. (a) State the two conditions, beyond a fixed number of trials and two outcomes, that must hold for a binomial model. (b) Assuming a binomial model, find the probability that at least 5 of the 15 adults cycle to work.

Solution

(a) The trials must be independent, and the probability of success must stay constant at $p = 0.3$ for each adult.
- Because the city is large, sampling without replacement barely changes $p$, so both conditions are reasonable.
(b) Model the number who cycle as $X \sim B(15,\, 0.3)$.
- $P(X \ge 5) = 1 - P(X \le 4) = 1 - \texttt{binomcdf}(15, 0.3, 4) = 0.485$ (3 s.f.)

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.
It is symmetric only when $p = 0.5$; for $p$ below $0.5$ it leans right, and for $p$ above $0.5$ it leans left.
As $n$ grows the histogram smooths toward the bell of the normal distribution in the next subtopic, which is why the two are linked.

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 phone screens off a production line and count how many are 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.

Note

Until now you could only find the probability of one specific run, such as 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 order it could happen in.

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 screens.
Only two outcomes each trial: every 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 one 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

$\binom{n}{k}$, read "n choose k", is the number of ways to place the $k$ successes among the $n$ trials.
Find it with the $n\mathrm{C}r$ key on your GDC, or with $\binom{n}{k} = \dfrac{n!}{k!(n-k)!}$.

What The Formula Booklet Gives You

Under Topic 4, the formula booklet lists the binomial distribution with its mean $np$ and variance $np(1-p)$.
The probability formula above is there to explain where the calculator's number comes from, not to be evaluated by hand.
Both SL papers are calculator papers, so in the exam you always find binomial probabilities on the GDC.

Exam technique

Because the mean and variance are given in the booklet, the marks are for selecting and applying them, not for recalling them.
The syllabus states you will never be asked to prove the mean or the variance, so use them directly.
Quote the formula you use, substitute the numbers, then write the value: the substitution is the method mark.

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.)

Example question

Solution

Model the number of defective chips as $X \sim B(25,\, 0.08)$.
(a) $P(X = 2) = \texttt{binompdf}(25, 0.08, 2) = 0.282$ (3 s.f.)
(b) rewrite "more than 3" as $P(X > 3) = 1 - P(X \le 3)$.
- $= 1 - \texttt{binomcdf}(25, 0.08, 3) = 0.135$ (3 s.f.)
(c) rewrite "between 1 and 4 inclusive" as $P(1 \le X \le 4) = P(X \le 4) - P(X \le 0)$.
- $= \texttt{binomcdf}(25, 0.08, 4) - \texttt{binomcdf}(25, 0.08, 0) = 0.830$ (3 s.f.).

Example question

A biased coin lands heads with probability $0.4$. It is tossed $n$ times. Find the smallest value of $n$ for which the probability of getting at least one head is greater than $0.99$.

Solution

"At least one head" is the complement of no heads, so $P(X \ge 1) = 1 - P(X = 0) = 1 - 0.6^{\,n}$.
Require $1 - 0.6^{\,n} > 0.99$, which rearranges to $0.6^{\,n} < 0.01$.
Take logs: $n > \dfrac{\ln 0.01}{\ln 0.6} = 9.02$, so $n$ must be a whole number above $9.02$.
The smallest such value is $n = 10$; a GDC check gives $1 - 0.6^{10} = 0.994 > 0.99$.

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 of $n$ 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 spread $np(1-p)$ is largest when $p = 0.5$ and shrinks as $p$ moves toward 0 or 1.
The variance is $np(1-p)$; the standard deviation is its square root, $\sqrt{np(1-p)}$.

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.

Example question

Solution

(a) The trials must be independent, and the probability of success must stay constant at $p = 0.3$ for each adult.
- Because the city is large, sampling without replacement barely changes $p$, so both conditions are reasonable.
(b) Model the number who cycle as $X \sim B(15,\, 0.3)$.
- $P(X \ge 5) = 1 - P(X \le 4) = 1 - \texttt{binomcdf}(15, 0.3, 4) = 0.485$ (3 s.f.)

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.
It is symmetric only when $p = 0.5$; for $p$ below $0.5$ it leans right, and for $p$ above $0.5$ it leans left.
As $n$ grows the histogram smooths toward the bell of the normal distribution in the next subtopic, which is why the two are linked.

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 Applications & Interpretation SL 4.8—binomial Distribution Notes

IB Mathematics Applications & Interpretation 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

What The Formula Booklet Gives You

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

What The Formula Booklet Gives You

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 and Trigonometry

Statistics and Probability

Calculus

Number and Algebra

Functions

Geometry and Trigonometry

Statistics and Probability

Calculus

IB Mathematics Applications & Interpretation SL 4.8—binomial Distribution Notes

IB Mathematics Applications & Interpretation 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

What The Formula Booklet Gives You

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

What The Formula Booklet Gives You

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 $p$