Don't leave your Extension 1 maths to chance, read this guide to make an E4 an odds on favourite!
Feel like the odds are stacked against for Year 11 Extension 1 Probability? In this article, we will address all the key syllabus points for Year 11 Maths Ext 1 Probability so you don’t leave your marks to chance!
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Probability and Statistics is extensively used in mathematics, statistics, finance, science, artificial intelligence and many more areas of study.
Surprisingly, the topic that underpins hard probability problems is an area of Mathematics called combinatorics (meaning counting!) Indeed, as probabilities are the ratios of the number of ways an event can happen to the total number of possibilities, the step is to develop a theory of counting.
We are going to look at the following NESA syllabus points.
NESA dictates that students need to learn the following concepts to demonstrate proficiency in Probability and Statistics.
Students only need to be familiar with arithmetic for combinatorics. Students should already be familiar with the basic concepts of probability. They should also understand basic algebraic techniques and expansion to understand the concepts explored in the following guide.
Combinatorics (also known as enumeration) is the study of counting. It sounds deceivingly simple; however, it is independently very difficult and also the foundation behind probability.Before we begin, it will be very useful to understand:
The following notation will be used extensively:
\(n!=n(n1)(n2)…3×2×1\)
For example: \(5!=5×4×3×2×1=120\)
Properties of Factorials:

Combinations is about unordered selections. It counts the number of ways you can choose \(k\) objects out of \(n\) if order does not matter. This could be choosing players to be in a football team, or choosing members to be in a committee, or choosing cards from a pack of cards.
Combinations are also called “unordered selections without replacement.”
Unordered Selections without replacement
The number of ways you can choose \(k\) objects from \(n\) is denoted as \(^nC_k\) or \({n \choose k}\) and pronounced as “n choose k.” It is calculated by the formula: \(^nC_k = {n \choose k} = \frac{n!}{k!(nk!)}\) 
Example:
How many ways can you choose a team of \(10\) players from a group of \(30\)?
Solution:
This is an unordered selection without replacement. There are \({30 \choose 10}\) ways to do this.
Note: You do not need to provide the numerical value of your coefficient, because quite often, they are very large. (In this case, \(30\) choose \(10\) is \(30045015\)). It is acceptable (and recommended) to state \(30\) choose \(10\) as your final answer.
Many counting problems, however, involve multiple steps (or decisions). To find the total number of possibilities in such multistage processes we need something called the multiplication principle.
Simply, it says:
If there are \(m\) ways of making decision \(A\) and \(n\) ways of making decision \(B\); the total number of ways making decision \(A\) and then decision \(B\) is \(m×n\).
Example:
Suppose Adam has \(12\) different pairs of socks and \(14\) different pairs of shoes. How many different combinations of shoes and socks can he wear?
Solution:
Using the multiplication principle – there are \(12\) objects in bag \(A\) (the pairs of socks) and \(14\) objects in bag \(B\) (the pairs of shoes). Hence, the number of ways of choosing one pair of socks and then one pair of shoes is \(12×14=168\) different combinations.
And this can be easily generalised to any number of decisions;
Multiplication Principle: (The fundamental counting principle)
Suppose we have \(k\) decisions to make. If there are: \(n_1\) ways to make the first decision, \(n_2\) ways to make the second decision,\( …, n_k\) ways to make the last decision; The total ways of making the first decision and then the second decision and then the third decision … and then the last decision is: \(n_1×n_2×⋯×n_k\) 
NOTE: In the multiplication principle, order MATTERS.
For example, you are to choose a twodigit number, there are \(9\) options for the first number, and \(10\) options for the second number, so \(9×10=90\) options total. This would treat \(34\) and \(43\) as different numbers, even though they contain the same digits.
This is in contrast to combinations, where order did not matter.
Example:
Suppose James has \(10\) options for what to have for breakfast, \(6\) options for what to have for lunch and \(9\) options for what to have for dinner on a particular day.
How many different possibilities of breakfast, lunch and dinner meal plans can he have?
Solution:
For James:
Hence, by the multiplication principle, there are
\(10×6×9=540\)possibilities he can have on that day.
There are three special cases of the multiplication principle:
We will see each case one at a time.
Ordered Selection with replacement
Suppose a decision is to be made \(r\) times (in order) with the same \(n\) options for each decision. Then, by the multiplication principle, there are \(n^r\) total ways you can make a sequence of \(r\) decisions. \(\underbrace{(n×n×⋯×n)}_{\hbox{(r times )}}=n^r\)possibilities 
Example:
How many \(4\) digit codes can be made from the numbers \(1\) to \(9\) if replacement is allowed? (I.e. you can use a number multiple times – like \(1125\) or \(3333\))
Solution:
Using the multiplication principles, there are \(9\) ways to make the first decision (selecting the first digit of the code), there are \(9\) ways to make the second decision (selecting the second digit of the code) and so on.
\(9×9×9×9=9^4=6561\)
Hence, there are \(6561\) possible codes that can be made.
Permutations
A permutation, in general, is the number of rearrangements for a set of objects. The number of permutations for n objects is \(n!=n×(n1)×⋯×3×2×1\). 
Example 1:
Suppose that three friends, Adam (A), Betty (B) and Charles (C), line up for a bus. In how many different ways can this be done?
Solution:
Method 1 (Manually listing cases)
If we simply list out all the cases, we can see there would be six possibilities:
\(ABC, ACB, BAC, BCA, CAB, CBA.\)
Method 2 (Using the multiplication principle)
Then, by the multiplication principle, there are
\(3×2×1=3!=6\)ways to arrange the three people in a line, as we manually listed above.
Note: You do not need to write out the decisions for the multiplication principle each time. You can immediately write \(3!\) as your final answer.
Example 2:
Suppose at a bank there are three active counters and \(5\) people sent to the first counter, \(6\) people sent to the second counter and \(3\) people to the last. In how many different possibilities are there for the order people are served?
Solution:
For each counter:
Hence, by the multiplication principle, there are in total:
\(5!×6!×3!\)
possibilities.
Example 3: (Circular arrangements)
Suppose 4 people are sitting around a circle table. In how many different ways can they be arranged? (Note: Be careful! Arranging people around a circle is not the same as arranging people in a line)
Solution:
Let the four people be \(A, B, C, D\).
Consider a circle:
The number of arrangements would not simply be \(4!\) like before. Because some permutations actually lead to the same arrangement! For example:
would be counted in \(4!\) as a different arrangement, but actually, it’s the same! It is just a \(90\)degree rotation of the first table. How do we avoid this?
We have to fix one of the people and then change the ordering of the rest of the people. So, for example, in the first image if we fix \(A\) in that position and rearrange the remaining three people, there are \(3!\) Ways of doing this.
Hence, the answer is \(3!=6\) ways. (Try to list them out yourself!)
Ordered Selections without replacement
An ordered selection without replacement is just one step further than a combination. Out of \(n\) selections, choose \(k\) of them, but this time, allow for different arrangements. How many different possibilities are there? There are \({n \choose k}\) number of ways to make the first decision (choosing \(k\) letters), and there are \(k!\) Number of ways of making the second decision (deciding how to arrange them). Hence, by the multiplication principle, the total possibilities are \({n \choose k}×k!\) We call this “n permute k”: \(^nP_k={n \choose k}K!\) 
Example:
How many \(4\) digit codes can be made from the numbers \(1\) to \(9\) if replacement is not allowed?
(E.g. \(8752\) would be allowed, but \(1198\) would not be allowed)
Solution:
This is an ordered selection without replacement.
Hence, by the multiplication principle, there are
\({9 \choose 4}×4!=126×4!=3024\)
possible codes you can make.
Note: You do not need to write out the decisions for the multiplication principle each time. You can immediately write \({9 \choose 4}×4!\) as your final answer.
The pigeonhole principle is another concept which seems quite simple and obvious. However, as you will see in questions, it can be used to prove highly nontrivial and unexpected results.
The Pigeonhole Principle:
If there are \(n+1\) pigeons to be placed into n pigeonholes, then at least one pigeonhole will contain at least \(2\) pigeons. 
Example 1:
A dinner party is held with \(27\) guests. Prove that there must be at least two guests whose first names start with the same letter.
Solution:
There are \(26\) possibilities for what the first letter for each guest could be. As there are \(27\) guests, we can define:
Hence, by the pigeonhole principle, there must be at least one letter (pigeonhole) that two guests (pigeons) share as the first letter in their first name. In other words, there must be at least two guests whose first names start with the same letter.
Example 2: (Harder Example)
At the same dinner party with \(27\) guests, some of the guests shake hands when they meet. Use the pigeonhole principle to show that there is a pair of guests who shake hands with the same number of people.
Solution:
The set of the possible number of handshakes for any particular guest is \(\{{0,1,2,…,26}\}\). Note that \(27\) is not included as we assume a guest cannot shake hands with himself.
There are \(27\) guests (pigeons) and \(27\) possibilities (pigeonholes) for the number of handshakes each guest could have had – the pigeonhole principle is not yet applicable (we need \(2\)6 pigeonholes)!
However, note that if a guest has made \(0\) handshakes, there must be no one in the party that has made \(26\) handshakes (i.e. somebody that has shaken hands with everyone else).
We can form two cases based on whether or not everyone has shaken hands with at least one other guest.
Case 1: If everyone has shaken hands with at least one other guest, then there cannot be a guest with \(0\) handshakes. Hence, the \(26\) possible number of handshakes (pigeonholes) for each guest is \(\{{1,2,…,26}\}\) and there are \(27\) guests (pigeons), so by the pigeonhole principle, there must be at least two guests with the same number of handshakes.
Case 2: If not everyone has shaken hands with at least one other guest, then there cannot be guest with \(26\) handshakes. Hence, the \(26\) possible number of handshakes (pigeonholes) for each guest is \(\{{0,1,2,…,25}\}\) and there are \(27\) guests (pigeons), so by the pigeonhole principle, there must be at least two guests with the same number of handshakes.
Hence, in any case, there must be at least two guests with the same number of handshakes.
There is also a simple extension to the pigeonhole principle called the generalised pigeonhole principle.
The generalised pigeonhole principle:
If there are \(n\) pigeons to be placed into k pigeonholes, then at least one pigeonhole will contain at least \(\frac{n}{k}\) pigeons. 
Example:
On a TV show, there are \(16\) contestants who are split into \(3\) teams. Prove that there is at least one team with at least \(6\) contestants.
Solution:
Let the \(16\) contestants be the pigeons and the \(3\) teams by the pigeonholes. Then immediately, by the generalised pigeonhole principle, there must be at least \(\frac{16}{3}=5.33\) or equivalently (rounding up) \(6\) contestants in at least one team.
The binomial theorem is an efficient method of expanding out all the higher powers of an expression in the form of \((a+b)^n\) for arbitrary values of \(a,b\) and \(n\). Learning how to conduct binomial expansion quickly is essential as it acts as a foundation to subsequent extension \(1\) theory in topics such as binomial identities.
Suppose we wanted to expand out \((a+b)^4\). To do so, we can undertake a very tedious method of expanding to get our answer:
\begin{align*} (a+b)^4&=(a+b) (a+b)^3\\ &=(a+b)(a^3+3a^2 b+3ab^2+b^3)\\ &=a^4+4a^3 b+6a^2 b^2+4ab^3+b^4\\ \end{align*} 
Is there a quicker way? What happens if we wanted to expand \((a+b)^32\). Is there a shortcut?
It turns out that the value of the coefficients of our expansion can be quickly determined by drawing up Pascal’s Triangle. Every number in Pascal’s triangle is the sum of the two numbers above to the right and left. The outer numbers of the Pascal’s triangle are always \(1\).
Write down the powers of a and b for each term, with the powers of a decreasing and power of b increasing. You must ensure that the sum of these powers is always \(n\).If we go back and look at the previous example where we expanded \((a+b)^4\), we can quickly see that the coefficients of our expansion correspond to the 5th line of Pascal’s triangle (i.e. \(1, 4, 6, 4,1\)). This is true in general for \((a+b)^n\) where we can obtain the coefficients of our expansion as the \((n+1)\)th line of Pascal’s Triangle. Hence, our method of expanding is as follows:
Then, use Pascal’s triangle to determine the coefficients of each term. Remember that the order we assign these coefficients matter!
Example: Expand \((2x+y)^3\)
We note here that the coefficients of our expansion should correspond to the fourth line of Pascal’s triangle (\(1, 3, 3, 1\)). Hence, we can write out the expansion as follows:
\begin{align*} (2x+3y)^3&=(1)∙(2x)^3∙y^0+(3)∙(2x)^2∙y+(3)∙(2x)∙y^2+1∙(2x)^0∙y^3\\ &=9x^3+12x^2 y+6xy^2+y^3\\ \end{align*} 
As we can see here, using the Pascal’s Triangle has dramatically reduced the time it takes to undertake binomial expansion. But is there a quicker way?
We use the binomial coefficient since Pascal’s Triangle becomes tedious to derive for large values of \(n\). Remarkably, it turns out that these binomial coefficients exactly correspond to the numbers inside the Pascal’s Triangle. So, this is a much faster way of deriving our coefficients!
This gives us the generalised Binomial Expansion formula:
\begin{align*} (a+b)^n &={n \choose 0} a^n+{n \choose 1} a^{n1} b+{n \choose 2} a^{n2} b^2+⋯+{n \choose n1}ab^{n1}+{n \choose n} b^n\\ &= \sum_{k=0}^n {n \choose k}a^{(nk)} b^k\\ \end{align*} 
Using our generalised Binomial Expansion formula above, it is apparent that we can determine the \((k+1)\)th term:
\(T_{k+1}={n \choose k} a^{nk} b^k\),for \(0≤k≤n\)
Example: Write down the 5th term of the expansion \((3+2x)^14\).
We note here that we are trying to find \(T_5\). As such, \(k+1=5,k=4\).
So, the 5th term of the expansion is:
\({14 \choose 4} (3)^{144} (2x)^4={14 \choose 4}∙(3)^10∙(2)^4∙x^4\) 
Using the binomial theorem to find coefficients of specific terms:
Suppose we wanted to find the coefficient of a particular term in our expansion. The following 3step procedure can be utilised:
Example: Find the coefficient of \(x^2\) in the expansion of \((2x5)^7\)
Step 1: Write down the binomial expansion using sigma notation (binomial theorem) and simplify to isolate the \(x\):
\((2x5)^7=\sum_{k=0}^7{7 \choose k} (2x)^{7k} (5)^k =\sum_{k=0}^7 {7 \choose k}(2)^{7k}∙(5)^k ∙(x)^{7k} \) 
Step 2: Find the required value of \(k\)
As we are trying to obtain the coefficient of \(x^2\), we can find the value of \(k\) by equating the power of \(x\) inside the sigma with \(2\).
\begin{align*}
&7k=2\\
&k=5\\
\end{align*}
Step 3: Write down the coefficient of \(x\) by substituting the value of \(k\)
\({7 \choose 5}∙(2)^2∙(5)^5=262500\)Example: Find the coefficient of \(\frac{1}{x^2}\) in the expansion of \((2x+\frac{1}{x})^{10}\)
Step 1: Write down the binomial expansion using sigma notation (binomial theorem) and simplify to isolate the \(x\):
\begin{align*} (2x+\frac{1}{x})^10 &= \sum_{k=0}^{10}{10 \choose k} (2x)^{10k} (\frac{1}{x})^k\\ &=\sum_{k=0}^{10}{10 \choose k} (2)^{10k} (x)^{10k} (\frac{1}{x})^k\\ &=\sum_{k=0}^{10}{10 \choose k} (2)^{10k} (x)^{10k} (x^{1} )^k\\ &=\sum_{k=0}^{10}{10 \choose k} (2)^{10k} (x)^{102k}\\ \end{align*} 
Step 2: Find the required value of \(k\)
As we are trying to obtain the coefficient of \(\frac{1}{x^2}\) , we can find the value of \(k\) by equating the power of \(x\) inside the sigma with \(2\).
\begin{align*}
&2=102k\\
&2k=12\\
&k=6\\
\end{align*}
Step 3: Write down the coefficient of \(x\) by substituting the value of \(k\)
\({10 \choose 6} (2)^{106}=3360\)\(243x^5810x^4 y+1080x^3 y^2720x^2 y^3+240xy^432y^5\) 
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