**Probability Spaces**

When working out the probability of two things occurring, it can often be helpful to draw a probability space. This can help you visualise the problem you are given.

**Example 1**

You have two fair dice. What is the probability of throwing a total score greater than 8?

There are 6 possibilities for the first dice, and another 6 for the second, so we can draw ourself a 6×6 probabibity square.

We now need the total number that add up to greater than 8. So the lowest number for your first throw must be a 3, followed by a 6 on your second throw. Next is a 4 for your first throw, followed by a 5 or 6 on your second throw, etc.

Following this method, we fill in our probability space like such:

Now we can read off the total number of ways to get more than 8. There are 10 coloured squares, out of a possible 36.

Hence, the probability of throwing a score greater than 8 with two fair dice is to 2dp.

Drawing our probabilities like this can really help in visualising a problem. However probability are limited to two events, such as flipping a coin twice or rolling a dice twice. We’ll not take a look at a way of representing more than two events.

**Probability Trees**

Another way of representing multiple events is a probability tree.

A probability tree starts with a different branch (hence the name tree) representing each different option. Then for the second set of options, each branch has each option again branching off it.

For example, if it was the score of a fair dice we were interested we were interested in, then there would be initially 6 branches, each representing the numbers 1-6. Each of these 6 branches would then have another 6 new branches coming off them, each representing the number of the second roll. We’ll take a look at an example now to try and make things a bit more understandable.

**Example 2**

You flip a fair coin two times. What is the probability of getting a different results on each flip, (ie getting either HT or TH)?

Now this is a relatively easy question and I’m sure you could answer it without using a probability tree. But bare with me and we’ll see how useful a probability tree can be!

Firstly we draw our tree, and it should look something like this:

Notice that at each stage, the **total sum of probabilities is always 1**.

From the first set of branches, we can see that the .

It also follows that .

Also note that It also follows that .

So we can see that our answer of getting different results is .

Now this may seem trivial, using a probability tree when we could just count the results on our hands. But what if we flipped the coin 5 times instead of 2. Or what if it was rolling a dice with six options at each “branch”. The beauty of a probability tree is that it makes all these options easy to visualise, not matter how many times you repeat the event.

]]>A lot of the time you will be considering the probabilities of two or more events occurring. When dealing with these multiple probabilities you will need to consider whether these events are **independent** or **dependent**.

**Independent Events** – Multiple events are considered to be independent if the outcome of one event *does not effect* the outcomes of any others.

For example, if you roll three dice, the chances of getting a 6 on the third roll is not effected by what you have rolled on the first and second go.

**Dependent Events** – As the name suggests, multiple events are dependent if the outcome of 1 *effects the outcome* of the others.

For example, if you have a bag containing 5 black balls and 5 white balls. The probability of picking a white ball on the second pick depends on what you have picked first. If you picked a white ball first, then the chances of picking a white ball on the second go would be . However if you picked a black ball first, then the chances of getting a white ball on your second go would be .

**Example 1**

You roll a fair dice three times. What is the probability that on each roll you will roll a number greater than 4 on the first two rolls, and then a number less than 4 on the third roll?

The probability of rolling a a number greater than 4 is equal to the probability of rolling a 5 or a 6, which is . Likewise, the probability of rolling a number less than 4 is .

To find the total probability we multiply these individual probabilities together, resulting in:

.

We already know that if two events are independent, then the outcome of one does not effect the outcome of the other.

We now introduce a new term, **Mutually Exclusive**, (sometimes called disjoint). Two events are mutually exclusive when they **can’t occur at the same time**. For example, if you flip a coin once, the events ‘get a head’ and ‘get a tail’ are mutually exclusive, as you can only get one or the other.

** The AND and OR Rules**

When the events A and B are **independent**:

.

When two events, A and B, are **mutually exclusive**:

and

.

Remember these **only** work for the specific examples where the events are independent or mutually exclusive. If you are not told that the events are either independent or mutually exclusive, **don’t** use these formulas!

**Example 2**

A bag contains 20 balls, 12 are black, and 8 are white. A ball is taken from the bag, it’s colour written down and then replaced. A second ball is then taken from the bag, what is the probability that the two balls are different colours?

We try to break the question down and find out exactly what it’s asking us. For the two balls to be different colours, we need to find;

.

Because the first ball is put back into the bag before the second ball is chosen, then these two events are independent. Thus, from the formula given above, we can replace the ‘*and*‘ with a symbol.

Similarly, either the first ball is black and the second white, or the first ball is white and the second is black. These two events cannon occur at the same time; hence, from the formula above, we can replace the ‘*or*‘ with a symbol.

So:

becomes;

.

Which we can again simplify to:

.

Remembering our BODMAS rules, this gives:

.

You’v now seen how using the **AND OR Rules** correctly can simplify difficult sounding questions down to a few simple sums.

*Now take a look at our article on Probability Spaces and Probability Trees* for some further examples of multiple events.

Probability is widely used across many areas today such as finance, statistics, gambling and even physics!

For an specified event **A**, we often write “**The probability of A**” as .

For example, the probability of rolling a dice and getting an even number would be the number of possible successes, **2**, **4** and **6**, divided by the total number of possible outcomes, **1**, **2**, **3**, **4**, **5** and **6**. So .

If something is **certain to happen**, we say this has a probability of **1**.

If something is **impossible**, then we say this has a probability of **0**.

It follows then that the probability of something not happening is **1 minus the probability that it will happen**.

**Did you know?** – The exact history of probability is uncertain, but we know that the study of chance has been linked to gambling for thousands of years!

**Example 1**

There are 20 balls in a bag; 7 are red, 5 are yellow, 6 are green and 2 are blue. What is the probability of **not** picking a green ball?

Using the formula above, the number of green balls is 6, and total number of balls is 20, so .

**Example 2**

There’s another bag full of coloured balls, red, yellow, green and blue. It is not known how many of each colour are in the bag. The balls are picked out at random and replaced. You do this 1000 times and get the following results:

- 500 red balls picked
- 200 yellow balls picked
- 150 green balls picked
- 150 blue balls picked

i) What is the probability of picking a yellow ball?

ii) If there are 200 balls in the bag, how many are likely to be blue?

i) For the 1000 balls picked out, 200 were yellow. This gives us .

ii) Although we don’t know the exact number of balls in the bag, we can predict the numbers from the ones we have already picked out. We know that . So if there’s 200 balls in the bag, the expected number of blue balls is .

*Now take a look at our article on Probability – Multiple Events*.

**The Sine Graph**

We can tell a great deal from the graph of the Sine function. Clearly the function is a **bounded function** between **1 and -1**. This means that will be no more than 1, and no less than -1 for *any* value of x. Feel free to check this on your calculator!

We can also see that the Sine function is an **odd function**. This means that, for example, . This is clear for example, from the value of ; with and .

Furthermore, Sine is a **periodic function**, meaning that it repeats itself every set period. For the Sine function, this is every 360°. For any angle , , (honestly, try it if you don’t believe me).

**Did you know?** – Because of the periodicity of the Sine function, it is commonly used to model phenomena such as sound and light waves, sunlight intensity and day length, and average temperature variations throughout the year.

**The Cosine Graph**

At first sight, the graph of the Cosine function is very similar to that of the Sine function. In fact it’s actually the Sine graph, shifted to the right by 90°.

Similarly, the Cosine functions is a **bounded function** with limits **-1 and 1**, and also a **periodic function** with period of 360°. You can notice again that, for any angle ,

Unlike the Sine function however, the Cosine function is an **even function**. This means that for any value . Another way of visualising this is that an even function is symmetrical about the y-axis.

**The Tangent Fuction**

The graph of the Tangent is the blue function on the graph above. The red lines are not part of the graph, but they are lines called asymptotes. These are used to show where the function “shoots off” to infinity.

The reason for this is that . Because you can’t divide by zero, is undefined for values of such that , (90°, 270°,…). It follows that for the same values that .

The Tan function is also a **periodic function** with a period of 180°, however it is **not bounded**, unlike that of Sine and Cosine.

Another trigonometric formula for **any** type of triangle is the **Cosine Rule**:

or alternatively

or

Where the uppercase letters correspond to the angle, and the lower case letter is the side opposite it.

The different formulas are provided for use depending on what sides/angles you are given. I recommend choosing just one, learning that and changing the sides/angles accordingly.

The Cosine Rule is used when you have either **two sides and an angle** or **three sides**.

**Did you know?** – In France, the Cosine Rule is called Théorème d’Al-Kashi (Theorem of Al-Kashi) after the Persian mathematician Jamshīd al-Kāshī, who was one of the first to provide a formula of the rule.

**Example 1**

Find the angle **A** in the above diagram.

The angle we are dealing with is **A**, so using the formula;

,

we get:

.

Rearranging this results in:

, so .

So, using the Inverse Cosine Function, we get:

to 2dp.

**Example 2**

Again using ,

we get:

.

This gives .

Taking the square root gives us the length of **a** as .

It’s important to remember that the questions you may be asked in an exam might not be quite as obvious to answer as these examples. Often it is up to you to understand and interpret what the question in asking you. Remember, before trying to answer any question of this sort, **DRAW A DIAGRAM**!

See also the Sine Rule.

]]>As well as the usual trigonometric formulas for right angled triangles, there’s also a handy little formula that you can use for * any* triangle. This is known as the

or alternatively

Where the uppercase letters correspond to the angle, and the lower case letter is the side opposite it. It doesn’t matter which one you use, but I recommend the first one if you’re working out an angle, and the second one if you’re working out the length of a side. This is just to simplify your calculations.

The Sine Rule can be used when you have either **two sides with an angle opposite to one of the sides** or **one side and any two angles**.

**Example 1**

Find the angle **y** in the above diagram,

Using the first of the formulas above, we get .

Rearranging gives .

We then use the inverse sine to get to 2dp.

**Example 2**

Find the length **x** in the above diagram.

This time we are going to be using the second of the above formulas. Subbing in the above values we get .

Rearranging gives to 2dp.

You have now seen how to find both sides and angles using the sine rule.

See also the Cosine Rule

]]>Luckily trigonometry has the answer, with the formula Area = .

Where **C** is the angle between the sides **a** and **b**, (see diagram above).

**Example**

Find the area of the above triangle (not to scale) using the information given.

Using the above formula, we get the following equation:

So to two decimal places.

**Note** – It does not matter which sides or angles you choose in your formula, just as long as you choose the angle enclosed by the two chosen sides.

Where a, b and c are the sides of a **right angled triangle**, with c being the **hypotenuse**, (the longest side, opposite the right angle).

**Did you know?** – Pythagoras is said to have had a golden thigh, which – for some unknown reason – he exhibited in the Olympic games.

As the diagram above hopefully helps illustrate, you can see that if you add the square of the two shorter sides together, this is equal to the square of the hypotenuse.

**Example 1** – A ship sails **10km** east, and then **7km** North. Find the straight line distance from it’s starting point to 2dp.

First thing we need to do with these types of problems is to draw a diagram.

You can see from the diagram that the path traveled by the ship and the distance from it’s starting point form a right angled triangle.

Using Pythagoras’s Theorem, we can deduce that , a little rearranging gives , so . So to two decimal places, the distance traveled by the ship is **12.21km**.

Pythagoras’ Theorem can also be extended for use in 3 dimensional problems.

**Example** – Consider then cuboid with sides of length **12km**, **7km** and **5km**.

Let’s say we have a plane that travels from point A to point B. That is, it travels **12km** forwards, **7km** left and **5km** upwards. What is the total distance traveled?

The three dimensional version of Pythagoras’ Theorem works in just the same way. To find the distance traveled we take the square the 3 distances, and then take the square root of the sum. So in our example:

to 2 decimal places.

So the distance traveled by our plane is approximately 14.76km.

**Note** – This is an extra article for those who are interested in how we got to the quadratic formula, you do not need to know this for your exam!

We have our general quadratic in the form:

We are now going to **Complete the Square** to deduce the quadratic formula. You should remember from the Completing the Square section in the Solving Quadratics article, that we first need to get the on it’s own, so we divide by

The next thing we need to do is to move the number in it’s own to the left hand side of our equation, giving us

Now this is probably the trickiest bit of the whole operation. We take our plus half the coefficient of our original , and square it. Remembering to subtract half the coefficient squared.

*If you can’t see how I got to this point, I advise you to take a look at the article mention before, where I go through the steps in more detail.*

Now adding to both sides gives us

What we want now is to combine these two fractions on the left hand side. To do that we need to get them both over a common denominator, so multiplying the numerator and denominator of the last fraction by gives us

The left hand side can now be simplified.

Now taking the square root of both sides gives us

We need to remember here that taking the square root gives you both a positive and negative result. The left hand side can again be simplified to

We now subtract the from both sides to give a value for

As both the fractions on the left hand side have common denominator, they can again be simplfied.

Congratulations, you have just proved the Quadratic Formula!

I must add again that this is **not** needed for your exam, nor do you need to remember the formula as it is given you. This is just a little extra article for those with an interest in maths.

In the article on Factorising we touched briefly on how to factorise Quadratic Equations.

We’ll now go a bit more in-depth and talk a bit more about solving Quadratic Equations using various techniques.

**Quadratic Equations** are one of the most useful equations in Mathematics. Although they may seem confusing at the moment, quadratic equations are vital to pretty much every branch of Mathematics, from Fluid Mechanics to Population Studies.

Let’s start from the basics. Quadratic Equations are equations in which one of the terms involves an . They may be of the form:

**1)**

or

**2)**

or

**3)**

Here the letters **a**, **b** and **c** are just used to represent numbers.

Whatever form your quadratic equation is in, the first thing you need to do in solving it is to get a zero on one side of the brackets. It’s also a good idea, if you have a minus number in front of your term, to multiply everything by **-1**, as this will make your calculations easier. Here’s a couple of examples of type **2)** and **3)**.

**Example 1**

First we move everything over to the left hand side (LHS) of the equation:

Now notice that both terms have a common factor of , so we factorise like so:

You should know that if two things multiplied together are equal to zero, then one them is going to be zero.

Hence we have either , or , so:

, or

**Example 2**

Because this only contains one term and one number, this is pretty simple to solve.

First we multiply both sides by -1 to get , and then simple take the square root of both sides.

The simply means “plus or minus root 7”, this is because when you take the square root, you always get a positive and negative value.

**Warning** – It’s important to remember that you can’t take the square root of a negative number!

Now you’ve learn now to solve the two easier types of quadratic equation, now let’s move on to the hardest.

For quadratic equations of the form , there are generally 3 methods we can use to solve them:

I won’t go into detail here about how to factorise quadratic equations, as this is discussed in the article on factorising. Let’s do an example of factorising a quadratic equation to solve it.

**Example 3** – Solve the equation by factorising.

First we need to rearrange our equation to get one side equal to zero. It shouldn’t be hard to see that the above equation can be rearranged to . Now by using the methods discussed in the article on factorising, we can write our equation as .

If two things multiplied together are equal to zero, then one of our brackets must be equal to zero. Hence we have our solution, or

Completing the Square is a method used to solve quadratic equations that can’t easily be factorised or solved by other methods. At first it can seem a little complicated but after a bit of practice you’ll be an expert in to time.

**Example 4** – Solve the equation by completing the square.

Before we start anything else, when completing the square, we need to make sure that the number in front of the is one, so we divide our equation by 2 to get

The next thing we do is to take number not involving and ‘move it over’ to the other side.

Now the next step is a little confusing, but what we do is take plus half the coefficient of our original , and square them.

Now the more observant readers amongst you may have noticed that this is not quite right. If we expand the brackets, . To compensate for this extra , we place a in our equation, giving:

Still with us? Good. Now what we do is to move all of the numbers over to the left hand side, giving:

Now we take square roots of our equation

, (because , and remembering that taking the square root leaves you with a positive and minus value).

Now we simple add to both sides of our equation and we have our answer.

or

Now this method is a little tricky to get to grips with. If you’re having problems then the best thing will be to ask your teacher for some extra work sheets on completing the square and practice on those.

For a quadratic equation in the form , the following formula give you the solutions:

**Example 5** – Solve the equation using the quadratic formula.

The first thing we do is to write down the values of . We have , and .

Now we use our quadratic formula to find the values of

Which you will notice is the same result that we got by Completing the Square.

**Warning** – You will remember that you can only take the square root of a positive number. In some cases, quadratic equations have no solutions, this is when you end up with a square root of a minus number in your solution.

If you look back at the quadratic formula, this only occurs with b^2 – 4ac < 0[/latex]. [latex]b^2 - 4ac[/latex] is called the **discriminant** of the quadratic, and can be used to tell things about the roots (solutions) of the quadratic.

If the discriminant is **greater than zero**, then your equation will have **2 distinct (different) roots**.

If the discriminant is **equal to zero**, then your equation will have **1 repeated root**.

If the discriminant is **less than zero**, you end up with the square root of a minus number, which is impossible, so your equation has **no solution**.

You don’t have to remember the Quadratic Formula as it is given to you in the formula sheet in your exam. For those of you who are interested, there’s an article on how the quadratic formula is derived.

So there you have it. Which method to use depends on the question. Sometimes they will ask you to use a certain method, sometimes it comes down to practice which one to use.

Remember the only way to be good at anything in maths, especially quadratic equations, is lots of **practice**!