Binomials can be a tricky topic. They are fundamental in many mathematics topics. This page aims to help you with your understanding of binomials, including binomial expansion and binomial distributions. Before we dive in, let's look at some terms and notation.

What is a Binomial?

A binomial is an algebraic expression that contains two different terms connected by addition or subtraction.

Which of the following is a binomial expression?

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Factorial!

What is factorial? Factorial (denoted "!") is the product of an integer and all the integers below it. For example,

\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120. \]

Note: by definition, \(0! = 1 \).

What is the value of 4!?

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Due to the nature of factorials, we can often cancel before dividing. For example,

\[ \begin{aligned} \frac{7!}{4!} &= \frac{7 \times 6 \times 5 \times \cancel{4 \times 3 \times 2 \times 1}}{\cancel{4 \times 3 \times 2 \times 1}} \\ &= 7 \times 6 \times 5 \\ &= 210 \\ \end{aligned} \]

We will see this cancelling in action when we look at combinatorial notation.

Binomial Coefficients

\( \begin{aligned} \text{Let } n &= \text{the total number of events} \\ \text{and } k &= \text{the number of desired outcomes.} \\ \end{aligned} \)

Then we can say the number of ways to get a specific outcome is “n choose k”. We write \[\binom{n}{k}=\frac{n!}{k!\left(n-k\right)!}.\] These values can be calculated using the "\(^nC_r\)" (combinatorial) button on your calulator.

These are the coefficients in a binomial expansion.

How many different ways are there to flip a coin six times and get heads four of those times?

\(k =\)

Answer =

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We have seen how the binomial coefficients can be calculated and what they represent. These coefficients can be represented in Pascal's triangle.

Pascal's Triangle

To construct Pascal's triangle, we begin by placing a 1 in the first row (we call this row 0). To calculate the next row, we use numbers in the subsequent rows. We add the two numbers directly above the number we need to calculate. If there are no numbers to the left or right hand side, we place a zero for that side and proceed with the addition (this means the outer diagonals of the triangle will be 1s).

Can you calculate the 6th row of Pascal's triangle? Change the number of rows to check your answer.

What about the 10th row?

Highlight Multiples of

No. of Rows

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Proof:

\[ \begin{aligned} \binom{n}{k-1}+\binom{n}{k} &= \ \frac{n!}{(k-1)!\left(n-k+1\right)!} + \frac{n!}{k!\left(n-k\right)!} \\ &= \ \frac{n!}{(k-1)!\left(n-k\right)!}\left(\frac{1}{n-k+1} + \frac{1}{k}\right) \\ & = \ \frac{n!}{(k-1)!\left(n-k\right)!} \frac{n+1}{(n-k+1)k} \\ &= \ \frac{(n+1)!}{k!(n+1-k)!} \\ &= \ \binom{n+1}{k} \\ \end{aligned} \]

This identity is actually a mathematical representation of how the rows of Pascal's triangle are constructed. There are also some interesting patterns in Pascal's triangle. To investigate these patterns, click here.

Binomial Expansion

We know that a binomial expression contains two terms, such as \((1+x)\). If we want to square this bracket, we can easily expand it. However, what if we want to calculate \((1+x)^8\)? This would be extremely time consuming. This is where the binomial expansion formula comes in useful.

We can calculate the binomial expansion for positive integer powers using the following formula:

\[ (a+b)^n =a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + ... + \binom{n}{r}a^{n-r}b^r + ... + b^n, \]

where \(n \in \mathbb{N}\) and \( \binom{n}{r} \) is as defined in the binomial coefficient above.

Click the question to reveal the answer.

\[ \begin{aligned} (1 + x)^4 &= \binom{4}{0} 1 + \binom{4}{1} x + \binom{4}{2} x^2 + \binom{4}{3} x^3 + \binom{4}{4} x^4 \\ &= 1 + 4x + 6x^2 + 4x^3 + x^4 \\ \end{aligned} \]

Notice that the coefficients are row 4 from Pascal's triangle, as the binomial is raised to the power of 4.

\[ \begin{aligned} (1 + 2x)^3 &= \binom{3}{0} 1 + \binom{3}{1} (2x) + \binom{3}{2} (2x)^2 + \binom{3}{3} (2x)^3 \\ &= 1 + 3(2x) + 3(2x)^2 + (2x)^3 \\ &= 1 + 6x + 12x^2 + 8x^3 \\ \end{aligned} \]

Here, \(b=2x\). Don't forget that the indices affect every number inside the brackets, including the 2.

\[ \begin{aligned} (2 + x)^5 &= \binom{5}{0} 2^5 + \binom{5}{1} x \times 2^4 + \binom{5}{2} x^2 \times 2^3 + \binom{5}{3} x^3 \times 2^2 + \binom{5}{4} x^4 \times 2 + \binom{5}{5} x^5 \\ &= 32 + 80x + 80x^2 + 40x^3 + 10x^4 + x^5 \\ \end{aligned}\]

Note that this question asks for an answer with increasing powers of \(x\). This means that you should start with the \(x^0\) term, followed by \(x^1\), \(x^2 , \dots , x^5\).

\[ \begin{aligned} (7-x)^4 &= \binom{4}{4} (-x)^4 + \binom{4}{3} 7 \times (-x)^3 + \binom{4}{2} 7^2 \times (-x)^2 + \binom{4}{1} 7^3 \times -x + \binom{4}{0} 7^4 \\ &= x^4 - 28 x^3 + 294 x^2 - 1372 x + 2401 \\ \end{aligned} \]

Take extra care when dealing with terms connected by subtraction. In this example, \(b=-x\), so the terms in our answer have alternating signs (the term is negative when the power is odd).

Binomial Distribution

There is a strong connection between binomial expansion and binomial probabilities, which can be represented using the binomial distribution. The binomial distribution is a discrete probability distribution, which models the number of success in \(n\) independent trials. Given two coins, what are the possible outcomes when we flip both coins?

Flip the coins until you have each of the four possible outcomes.

Outcome 1

Heads
Tails
Heads
Tails

Outcome 2

Heads
Tails
Heads
Tails

Outcome 3

Heads
Tails
Heads
Tails

Outcome 4

Heads
Tails
Heads
Tails
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There are four possible outcomes. For the example above, we are concerned with the order of the outcomes (this is called a permutation). This means getting heads then tails is different from getting tails then heads. However, for a binomial coefficient, we do not care about the order, we only care about the combination of outcomes. For example, if I want the number of ways to get heads once and tails once, my number of desired outcomes is 2 (heads then tails, or tails then heads). For more information on combinations and permutations, click here.

It is easy to list all possible outcomes when we only have two coins. However, what if we have 6 coins? Or 200? We can use a formula to calculate the probabilities of each of these outcomes. The probability of observing \(k\) successes in \(n\) trials is given by \[P(X=k) = \binom{n}{k}p^k(1-p)^{n-k},\]

\( \begin{aligned} \text{where } n &= \ \text{the number of trials,} \\ k &= \ \text{the number of successes}, \\ p &= \ \text{the probability of success on a single trial.} \\ \end{aligned} \)

With 9 coin tosses, what is the probability that 5 are heads? Give you answer as a percentage, to 2 decimal places.

\(k =\)

\(p =\)

Answer =

\(\%\)

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When to use a Binomial Distribution

There are a number of conditions which need to be satisfied in order to use a binomial distribution. We can use the mnemonic BINS to help us remember.

  1. Binary number of outcomes. There are 2 possible outcomes - success and failure.

  2. Independent trials. Knowing the outcome of one trial must not tell us anything about the next.

  3. Number of trials. The number of trials, \(n\), must be fixed in advance.

  4. Same probabilty of success for each trial.

Which of the following are the correct examples for each of the conditions?

Binary Number of Outcomes

Independent Events

Fixed Number of Trials

Same Probability of Success

Submit

The following example is taken from the CCEA GCE Applied Mathematics 2022 paper.

Question: It is known that one in six seeds in a large batch will not germinate when planted.

If \(\frac{1}{6}\) does not germinate, the probability that a seed will germinate is P\((\text{germinate})=\frac{5}{6}\).

Twelve seeds are sown.

We can see that the number of trials is 12. The number of successes is the number of seeds which germinate, which is 9. From part i, the probability of success on a single trial is \(\frac{5}{6}\).
\(\begin{aligned} n &= \ 12 \\ k &= \ 9 \\ p & = \ \frac{5}{6} \\ \end{aligned}\)
Substituting these values into the binomial distrbution formula gives \[P(X=9) = \binom{12}{9} \left(\frac{5}{6} \right)^9 \left(1-\frac{5}{6} \right)^{3} = 0.197.\]

The probability that at most 10 seeds germinate is the same the probability that 11 seeds do not germinate plus the probability that 12 seeds do not germinate. \(\) Therefore, we have

\( \begin{aligned} P(X \leq 10) &= \ 1-\big[P(X=11) + P(X=12)\big] \\ &= \ 1- \left[\binom{12}{11} \left(\frac{5}{6}\right)^{11} \left(1-\frac{5}{6}\right) + \underbrace{\binom{12}{12}}_{=1} \left(\frac{5}{6}\right)^{12} \underbrace{\left(1-\frac{5}{6}\right)^0}_{=1}\right] \\ &= \ 1- \big[0.26918 + 0.11216 \big] \\ &= \ 0.619 \\ \end{aligned} \)

Our expected probability is found by multiplying the number of trials by the probability of success. \[ E(X) = np = 12\left(\frac{5}{6}\right) = 10\] We want to test the probabilities around \(X=10\). We calculated \(X=9\) in part ii, to give a probability of 0.197. We will check \(X=10\) and \(X=11\).

\( \begin{aligned} P(X=10) &= \ \binom{12}{10} \left(\frac{5}{6} \right)^{10} \left(1-\frac{5}{6} \right)^{2} = 0.296 \\ P(X=11) &= \ \binom{12}{11} \left(\frac{5}{6} \right)^{11} \left(1-\frac{5}{6} \right) = 0.269 \\ \end{aligned} \)

where probabilities have been recorded to 3 significant figures. The largest probability is obtained when \(X=10\). This confirms that \(X=10\) is the most likely number of 12 seeds to germinate.

To summarise, let's check that this is a binomial distribution question. There are only 2 outcomes - the seed does germinate or the seed does not germinate. The trials are independent - the outcome of one seed germinating is not affected by whether or not other seeds have germinated. The number of trials is fixed - we are told that twelve seeds are sown. The probability of success remains the same for each trial - we are told that one in six seeds will not germinate and this remains constant throughout the trial. Since all conditions are satisfied, we can use the binomial distribution formula to answer this question.

Binomial Coefficients (Revisited)

Having explored both binomial expansions and binomial distributions, we can now examine the shared feature of these two concepts: the binomial coefficients. What is the reason behind the binomial coefficients being the same in these two formulae?

The binomial coefficients are the same due to their combinatorial nature. They represent the number of possible combinations of choosing \(k\) elements from a set of \(n\) elements. However, the context in which they are used differs.

  • Binomial Expansion: distributing powers of variables \(a\) and \(b\).
  • Binomial Distribution: choosing \(k\) successes from \(n\) independent trials.

In both cases, these coefficients count the ways to distribute or choose elements. Let's investigate this further using two concrete examples.

Example 1 - a binomial expansion problem: The expansion of \((a+b)^5 = (a+b)(a+b)(a+b)(a+b)(a+b)\).
Each term in the expansion represents the number of ways to choose a specific power of \(a\) and \(b\) such that their sum is equal to 5. When we expand this expression, we get the following: \[ a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 +b^5. \] The coefficients of these terms represent the number of ways to obtain that specific combination. That is, there is one way to take 5 of the letter \(a\) (take one from each bracket), there are 5 ways to take 4 of the letter \(a\) and 1 \(b\), and so on.


Example 2 - a binomial distribution problem: Flipping a coin 5 times.
We are considering the different outcomes of flipping a coin 5 times. Each flip can result in either heads (H) or tails (T), and we are interested in counting the different ways of obtaining certain numbers of heads and tails.

How are these examples related? Let's replace \(a\) with heads \((H)\) and \(b\) with tails \((T)\) in Example 1. Therefore, we are looking at the expansion of \((H+T)^5\). Expanding this gives \[ H^5 + 5 H^4T + 10H^3T^2 + 10H^2T^3 + 5HT^4 +T^5,\] where the coefficients of these terms represent the number of ways to obtain a specific combination of heads and tails when flipping a coin 5 times. Thus, the binomial coefficients in the binomial expansion formula (Example 1) are the same as the binomial coefficients in the binomial distribution formula (Example 2), because they both represent the number of ways to obtain specific combinations of the two terms. In both examples, the coefficients represent the number of ways to choose a specific power of a and b (or H and T), such that their sum is equal to 5.

The binomial expansion and binomial distribution share the same binomial coefficients because they are both fundamentally about choosing from two possibilities and counting the different ways those possibilities can be combined.




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