1.4.2 Link to binomial probabilities

What you'll learn

How the binomial expansion is connected to repeated trials with two possible outcomes.
Why $(nr)pr(1−p)n−r\binom{n}{r}p^r(1-p)^{n-r}$ gives the probability of exactly $r$ successes.
How to use binomial coefficients to calculate individual and combined probabilities.
How the identity $p+(1-p))^n=1$ confirms that all possible probabilities sum to 1.

Prerequisite: the binomial expansion

For a positive integer $n$ , the binomial expansion of $a+b)^n$ is

(a+b)n=∑r=0n(nr)arbn−r.(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^r b^{n-r}.

The symbol $(nr)\binom{n}{r}$ is called a binomial coefficient and is calculated using

(nr)=n!r!(n−r)!.\binom{n}{r}=\frac{n!}{r!(n-r)!}.

Here, the exclamation mark denotes a factorial:

n!=n(n−1)(n−2)⋯2⋅1.n!=n(n-1)(n-2)\cdots 2\cdot 1.

For example,

(52)=5!2!3!=10.\binom{5}{2}=\frac{5!}{2!3!}=10.

Definition

Combinations

The binomial coefficient $(nr)\binom{n}{r}$ is the number of different ways to choose $r$ objects from $n$ objects when the order of selection does not matter.

The general term in the expansion of $a+b)^n$ is

(nr)arbn−r,\binom{n}{r}a^r b^{n-r},

where $r$ takes the integer values from 0 to $n$ .

Example

Expanding a binomial

Expand $p+q)^4$ .

Apply the binomial theorem with $n = 4$ :
$(p+q)4=∑r=04(4r)prq4−r.(p+q)^4=\sum_{r=0}^{4}\binom{4}{r}p^r q^{4-r}.$
Use the binomial coefficients 1, 4, 6, 4 and 1:
$p+q)^4=q^4+4pq^3+6p^2q^2+4p^3q+p^4.$
Read each term as representing a different possible number of factors contributing a $p$ : from no factors contributing $p$ , in the term $q^4$ , to all four contributing $p$ , in the term $p^4$ .

The binomial probability model

A Bernoulli trial is a trial with exactly two possible outcomes. These outcomes are usually called success and failure, although a “success” can mean any outcome you choose to count.

Suppose that:

there are a fixed number $n$ of trials;
each trial has two possible outcomes;
the trials are independent;
the probability of success is the same on every trial.

Let the probability of success be $p$ . The probability of failure is then

q = 1 - p .

If the random variable $X$ counts the number of successes, then $X$ has a binomial distribution.

Definition

Binomial distribution

The notation

X∼B⁡(n,p)X\sim\operatorname{B}(n,p)

means that $X$ counts the number of successes in $n$ independent trials, each having constant probability of success $p$ .

The possible values of $X$ are the integers from 0 to $n$ .

Common Mistake

When the binomial model does not apply

The binomial model is not valid if the trials are dependent, the success probability changes between trials, there are more than two relevant outcomes, or the number of trials is not fixed.

Probability of one particular sequence

Suppose there are $n$ independent trials and you consider one particular sequence containing:

$r$ successes;
$n - r$ failures.

Because the trials are independent, you multiply their probabilities. The probability of this one particular sequence is therefore

p^r(1-p)^{n-r}.

For example, in four trials, the probability of the particular sequence success, failure, success, failure is

p(1-p)p(1-p)=p^2(1-p)^2.

However, this is only one of several possible arrangements containing two successes.

Counting all possible arrangements

To find the probability of exactly $r$ successes, you must count how many different arrangements contain $r$ successes among the $n$ trials.

That number is

(nr).\binom{n}{r}.

Every such arrangement has probability $p^r(1-p)^{n-r}$ . Therefore,

P⁡(X=r)=(nr)pr(1−p)n−r.\operatorname{P}(X=r)=\binom{n}{r}p^r(1-p)^{n-r}.

Key Idea

The link to the binomial expansion

In the expansion of

p+(1-p))^n,

the term $(nr)pr(1−p)n−r\binom{n}{r}p^r(1-p)^{n-r}$ is exactly the probability of obtaining $r$ successes in $n$ independent trials.

The three parts of the probability formula have distinct meanings:

$(nr)\binom{n}{r}$ counts the possible arrangements of the successes.
$p^r$ gives the probability contribution from the $r$ successes.
$1-p)^{n-r}$ gives the probability contribution from the $n - r$ failures.

Example

Finding the probability of exactly three successes

A biased coin lands on heads with probability $0.4$ . It is tossed five times. Find the probability of exactly three heads.

Let $X$ be the number of heads. The trials are independent, the number of trials is fixed and the probability of heads is constant, so
$X∼B⁡(5,0.4).X\sim\operatorname{B}(5,0.4).$
Substitute $n = 5$ , $r = 3$ , $p = 0.4$ and $1 - p = 0.6$ :
$P⁡(X=3)=(53)(0.4)3(0.6)2.\operatorname{P}(X=3)=\binom{5}{3}(0.4)^3(0.6)^2.$
Evaluate the binomial coefficient and the powers:
$P⁡(X=3)=10(0.4)3(0.6)2=0.2304.\begin{aligned} \operatorname{P}(X=3) &=10(0.4)^3(0.6)^2\\ &=0.2304. \end{aligned}$

Common Mistake

Forgetting the binomial coefficient

The expression $p^r(1-p)^{n-r}$ gives the probability of one particular ordering. For exactly $r$ successes in any order, you must multiply by $(nr)\binom{n}{r}$ .

Seeing the terms directly in an expansion

Consider four independent trials with success probability $p$ and failure probability $q = 1 - p$ :

q+p)^4=q^4+4pq^3+6p^2q^2+4p^3q+p^4.

Each term represents one possible number of successes:

$q^4$ represents no successes.
$4pq^3$ represents exactly one success.
$6p^2q^2$ represents exactly two successes.
$4p^3q$ represents exactly three successes.
$p^4$ represents exactly four successes.

The coefficient of each term counts the number of possible arrangements. For instance, the coefficient 6 in $6p^2q^2$ tells you that there are six ways to place two successes among four trials.

Example

Using an expansion to identify a probability

A trial has success probability $p = 0.3$ and is repeated independently four times. Find the probability of exactly two successes.

In the expansion of $q+p)^4$ , the term representing two successes is
$6p^2q^2.$
Since $q = 1 - p = 0.7$ , substitute the probabilities:
$P⁡(X=2)=6(0.3)2(0.7)2.\operatorname{P}(X=2)=6(0.3)^2(0.7)^2.$
Evaluate the expression:
$P⁡(X=2)=0.2646.\operatorname{P}(X=2)=0.2646.$

Why all the probabilities sum to 1

The number of successes must be exactly one of the integers from 0 to $n$ . These possibilities are mutually exclusive, meaning that no two can occur at the same time, and exhaustive, meaning that together they include every possible outcome.

Therefore,

∑r=0nP⁡(X=r)=1.\sum_{r=0}^{n}\operatorname{P}(X=r)=1.

Using the binomial probability formula gives

∑r=0n(nr)pr(1−p)n−r.\sum_{r=0}^{n}\binom{n}{r}p^r(1-p)^{n-r}.

By the binomial theorem, this is

p+(1-p))^n=1^n=1.

This is an important reason that the binomial expansion and the binomial distribution fit together perfectly.

Example

Finding a missing probability

Let $X∼B⁡(3,p)X\sim\operatorname{B}(3,p)$ . Use the binomial expansion to show that $P⁡(X≥1)=1−(1−p)3\operatorname{P}(X\geq 1)=1-(1-p)^3$ .

The complementary event to $X≥1X\geq 1$ is $X = 0$ , because either there is at least one success or there are no successes.
The probability of no successes is the first term in the expansion of $1-p)+p)^3$ :
$P⁡(X=0)=(30)p0(1−p)3=(1−p)3.\operatorname{P}(X=0)=\binom{3}{0}p^0(1-p)^3=(1-p)^3.$
Since the probabilities of all possible values sum to 1,
$P⁡(X≥1)=1−P⁡(X=0)=1−(1−p)3.\operatorname{P}(X\geq 1)=1-\operatorname{P}(X=0)=1-(1-p)^3.$

Tip

At least one success

“At least one” is often quickest to calculate using the complement:

P⁡(X≥1)=1−P⁡(X=0).\operatorname{P}(X\geq 1)=1-\operatorname{P}(X=0).

This avoids adding the probabilities for one, two, three, and so on.

Probabilities involving several values

If a question asks for a range of values, add the relevant terms of the binomial expansion.

For example,

P⁡(X≤2)=P⁡(X=0)+P⁡(X=1)+P⁡(X=2).\operatorname{P}(X\leq 2) = \operatorname{P}(X=0)+\operatorname{P}(X=1)+\operatorname{P}(X=2).

Be careful when translating inequalities:

“fewer than 3” means $X < 3$ , so $X$ can be 0, 1 or 2;
“at most 3” means $X≤3X\leq 3$ ;
“more than 3” means $X > 3$ ;
“at least 3” means $X≥3X\geq 3$ .

Common Mistake

Misreading at least and at most

“At least $r$ ” includes $r$ , whereas “more than $r$ ” does not. Similarly, “at most $r$ ” includes $r$ , whereas “fewer than $r$ ” does not.

Exam technique

In the exam

Check that the situation has a fixed number of independent trials, two outcomes per trial and a constant success probability.
Define what counts as a success, then identify $n$ , $r$ , $p$ and the failure probability $1 - p$ .
For exactly $r$ successes, use $(nr)pr(1−p)n−r\binom{n}{r}p^r(1-p)^{n-r}$ and make sure the exponents add to $n$ .
For “at least one”, consider using the complement $1−P⁡(X=0)1-\operatorname{P}(X=0)$ .
Check that your answer lies between 0 and 1 and that any requested rounding is applied only at the end.

Self review

Check yourself

Why is $(nr)\binom{n}{r}$ needed when calculating the probability of exactly $r$ successes?
Which term in the expansion of $q+p)^7$ represents exactly five successes?
How would you calculate the probability of at least one success in six independent trials?

What you'll learn

How the binomial expansion is connected to repeated trials with two possible outcomes.
Why $(nr)pr(1−p)n−r\binom{n}{r}p^r(1-p)^{n-r}$ gives the probability of exactly $r$ successes.
How to use binomial coefficients to calculate individual and combined probabilities.
How the identity $p+(1-p))^n=1$ confirms that all possible probabilities sum to 1.

Prerequisite: the binomial expansion

For a positive integer $n$ , the binomial expansion of $a+b)^n$ is

(a+b)n=∑r=0n(nr)arbn−r.(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^r b^{n-r}.

The symbol $(nr)\binom{n}{r}$ is called a binomial coefficient and is calculated using

(nr)=n!r!(n−r)!.\binom{n}{r}=\frac{n!}{r!(n-r)!}.

Here, the exclamation mark denotes a factorial:

n!=n(n−1)(n−2)⋯2⋅1.n!=n(n-1)(n-2)\cdots 2\cdot 1.

For example,

(52)=5!2!3!=10.\binom{5}{2}=\frac{5!}{2!3!}=10.

Definition

Combinations

The binomial coefficient $(nr)\binom{n}{r}$ is the number of different ways to choose $r$ objects from $n$ objects when the order of selection does not matter.

The general term in the expansion of $a+b)^n$ is

(nr)arbn−r,\binom{n}{r}a^r b^{n-r},

where $r$ takes the integer values from 0 to $n$ .

Example

Expanding a binomial

Expand $p+q)^4$ .

Apply the binomial theorem with $n = 4$ :
$(p+q)4=∑r=04(4r)prq4−r.(p+q)^4=\sum_{r=0}^{4}\binom{4}{r}p^r q^{4-r}.$
Use the binomial coefficients 1, 4, 6, 4 and 1:
$p+q)^4=q^4+4pq^3+6p^2q^2+4p^3q+p^4.$
Read each term as representing a different possible number of factors contributing a $p$ : from no factors contributing $p$ , in the term $q^4$ , to all four contributing $p$ , in the term $p^4$ .

The binomial probability model

A Bernoulli trial is a trial with exactly two possible outcomes. These outcomes are usually called success and failure, although a “success” can mean any outcome you choose to count.

Suppose that:

there are a fixed number $n$ of trials;
each trial has two possible outcomes;
the trials are independent;
the probability of success is the same on every trial.

Let the probability of success be $p$ . The probability of failure is then

q = 1 - p .

If the random variable $X$ counts the number of successes, then $X$ has a binomial distribution.

Definition

Binomial distribution

The notation

X∼B⁡(n,p)X\sim\operatorname{B}(n,p)

means that $X$ counts the number of successes in $n$ independent trials, each having constant probability of success $p$ .

The possible values of $X$ are the integers from 0 to $n$ .

Common Mistake

When the binomial model does not apply

The binomial model is not valid if the trials are dependent, the success probability changes between trials, there are more than two relevant outcomes, or the number of trials is not fixed.

Probability of one particular sequence

Suppose there are $n$ independent trials and you consider one particular sequence containing:

$r$ successes;
$n - r$ failures.

Because the trials are independent, you multiply their probabilities. The probability of this one particular sequence is therefore

p^r(1-p)^{n-r}.

For example, in four trials, the probability of the particular sequence success, failure, success, failure is

p(1-p)p(1-p)=p^2(1-p)^2.

However, this is only one of several possible arrangements containing two successes.

Counting all possible arrangements

To find the probability of exactly $r$ successes, you must count how many different arrangements contain $r$ successes among the $n$ trials.

That number is

(nr).\binom{n}{r}.

Every such arrangement has probability $p^r(1-p)^{n-r}$ . Therefore,

P⁡(X=r)=(nr)pr(1−p)n−r.\operatorname{P}(X=r)=\binom{n}{r}p^r(1-p)^{n-r}.

Key Idea

The link to the binomial expansion

In the expansion of

p+(1-p))^n,

the term $(nr)pr(1−p)n−r\binom{n}{r}p^r(1-p)^{n-r}$ is exactly the probability of obtaining $r$ successes in $n$ independent trials.

The three parts of the probability formula have distinct meanings:

$(nr)\binom{n}{r}$ counts the possible arrangements of the successes.
$p^r$ gives the probability contribution from the $r$ successes.
$1-p)^{n-r}$ gives the probability contribution from the $n - r$ failures.

Example

Finding the probability of exactly three successes

A biased coin lands on heads with probability $0.4$ . It is tossed five times. Find the probability of exactly three heads.

Let $X$ be the number of heads. The trials are independent, the number of trials is fixed and the probability of heads is constant, so
$X∼B⁡(5,0.4).X\sim\operatorname{B}(5,0.4).$
Substitute $n = 5$ , $r = 3$ , $p = 0.4$ and $1 - p = 0.6$ :
$P⁡(X=3)=(53)(0.4)3(0.6)2.\operatorname{P}(X=3)=\binom{5}{3}(0.4)^3(0.6)^2.$
Evaluate the binomial coefficient and the powers:
$P⁡(X=3)=10(0.4)3(0.6)2=0.2304.\begin{aligned} \operatorname{P}(X=3) &=10(0.4)^3(0.6)^2\\ &=0.2304. \end{aligned}$

Common Mistake

Forgetting the binomial coefficient

The expression $p^r(1-p)^{n-r}$ gives the probability of one particular ordering. For exactly $r$ successes in any order, you must multiply by $(nr)\binom{n}{r}$ .

Seeing the terms directly in an expansion

Consider four independent trials with success probability $p$ and failure probability $q = 1 - p$ :

q+p)^4=q^4+4pq^3+6p^2q^2+4p^3q+p^4.

Each term represents one possible number of successes:

$q^4$ represents no successes.
$4pq^3$ represents exactly one success.
$6p^2q^2$ represents exactly two successes.
$4p^3q$ represents exactly three successes.
$p^4$ represents exactly four successes.

The coefficient of each term counts the number of possible arrangements. For instance, the coefficient 6 in $6p^2q^2$ tells you that there are six ways to place two successes among four trials.

Example

Using an expansion to identify a probability

A trial has success probability $p = 0.3$ and is repeated independently four times. Find the probability of exactly two successes.

In the expansion of $q+p)^4$ , the term representing two successes is
$6p^2q^2.$
Since $q = 1 - p = 0.7$ , substitute the probabilities:
$P⁡(X=2)=6(0.3)2(0.7)2.\operatorname{P}(X=2)=6(0.3)^2(0.7)^2.$
Evaluate the expression:
$P⁡(X=2)=0.2646.\operatorname{P}(X=2)=0.2646.$

Why all the probabilities sum to 1

Therefore,

∑r=0nP⁡(X=r)=1.\sum_{r=0}^{n}\operatorname{P}(X=r)=1.

Using the binomial probability formula gives

∑r=0n(nr)pr(1−p)n−r.\sum_{r=0}^{n}\binom{n}{r}p^r(1-p)^{n-r}.

By the binomial theorem, this is

p+(1-p))^n=1^n=1.

This is an important reason that the binomial expansion and the binomial distribution fit together perfectly.

Example

Finding a missing probability

Let $X∼B⁡(3,p)X\sim\operatorname{B}(3,p)$ . Use the binomial expansion to show that $P⁡(X≥1)=1−(1−p)3\operatorname{P}(X\geq 1)=1-(1-p)^3$ .

The complementary event to $X≥1X\geq 1$ is $X = 0$ , because either there is at least one success or there are no successes.
The probability of no successes is the first term in the expansion of $1-p)+p)^3$ :
$P⁡(X=0)=(30)p0(1−p)3=(1−p)3.\operatorname{P}(X=0)=\binom{3}{0}p^0(1-p)^3=(1-p)^3.$
Since the probabilities of all possible values sum to 1,
$P⁡(X≥1)=1−P⁡(X=0)=1−(1−p)3.\operatorname{P}(X\geq 1)=1-\operatorname{P}(X=0)=1-(1-p)^3.$

Tip

At least one success

“At least one” is often quickest to calculate using the complement:

P⁡(X≥1)=1−P⁡(X=0).\operatorname{P}(X\geq 1)=1-\operatorname{P}(X=0).

This avoids adding the probabilities for one, two, three, and so on.

Probabilities involving several values

If a question asks for a range of values, add the relevant terms of the binomial expansion.

For example,

P⁡(X≤2)=P⁡(X=0)+P⁡(X=1)+P⁡(X=2).\operatorname{P}(X\leq 2) = \operatorname{P}(X=0)+\operatorname{P}(X=1)+\operatorname{P}(X=2).

Be careful when translating inequalities:

“fewer than 3” means $X < 3$ , so $X$ can be 0, 1 or 2;
“at most 3” means $X≤3X\leq 3$ ;
“more than 3” means $X > 3$ ;
“at least 3” means $X≥3X\geq 3$ .

Common Mistake

Misreading at least and at most

“At least $r$ ” includes $r$ , whereas “more than $r$ ” does not. Similarly, “at most $r$ ” includes $r$ , whereas “fewer than $r$ ” does not.

Exam technique

In the exam

Check that the situation has a fixed number of independent trials, two outcomes per trial and a constant success probability.
Define what counts as a success, then identify $n$ , $r$ , $p$ and the failure probability $1 - p$ .
For exactly $r$ successes, use $(nr)pr(1−p)n−r\binom{n}{r}p^r(1-p)^{n-r}$ and make sure the exponents add to $n$ .
For “at least one”, consider using the complement $1−P⁡(X=0)1-\operatorname{P}(X=0)$ .
Check that your answer lies between 0 and 1 and that any requested rounding is applied only at the end.

Self review

Check yourself

Why is $(nr)\binom{n}{r}$ needed when calculating the probability of exactly $r$ successes?
Which term in the expansion of $q+p)^7$ represents exactly five successes?
How would you calculate the probability of at least one success in six independent trials?

What you'll learn

Prerequisite: the binomial expansion

The binomial probability model

Probability of one particular sequence

Counting all possible arrangements

Seeing the terms directly in an expansion

Why all the probabilities sum to 1

Probabilities involving several values

Review 1.4.2 Link to binomial probabilities by teaching Genie

1.4.2 Link to binomial probabilities

What you'll learn

Prerequisite: the binomial expansion

The binomial probability model

Probability of one particular sequence

Counting all possible arrangements

Seeing the terms directly in an expansion

Why all the probabilities sum to 1

Probabilities involving several values

Review 1.4.2 Link to binomial probabilities by teaching Genie

1.4.2 Link to binomial probabilities

What you'll learn

Prerequisite: the binomial expansion

The binomial probability model

Probability of one particular sequence

Counting all possible arrangements

Seeing the terms directly in an expansion

Why all the probabilities sum to 1

Probabilities involving several values

Review 1.4.2 Link to binomial probabilities by teaching Genie

1.4.2 Link to binomial probabilities Revision Guide

1.4.2 Link to binomial probabilities

What you'll learn

Prerequisite: the binomial expansion

The binomial probability model

Probability of one particular sequence

Counting all possible arrangements

Seeing the terms directly in an expansion

Why all the probabilities sum to 1

Probabilities involving several values

Review 1.4.2 Link to binomial probabilities by teaching Genie

1.4.2 Link to binomial probabilities Revision Guide