# Hannah Sweets Maths Problem – Edexcel (June 2015)

In this post I’ll be demonstrating how to solve the “Hannah Sweets” maths problem which was posed in an Edexcel exam paper around June 2015. This problem is in fact one which is related to probabilities and goes like this:

There are (n) sweets in a bag;

6 of the sweets are orange;

The rest of the sweets are yellow;

Hannah takes at random a sweet from the bag… She eats the sweet;

Hannah takes at random another sweet from the bag… She eats the sweet;

The probability that Hannah eats two orange sweets is 1/3;

Show that: n^2 – n – 90 = 0.

[Source: Edexcel]

At first sight, this question could look a bit tedious, but upon further inspection – we notice that it can be broken into parts and solved fairly easily. Let me explain how…

Firstly, because of the information provided above, we can say that the probability of finding an orange sweet in the bag (under initial conditions) is 6/n. Note that we were told that there are 6 orange sweets in the bag (to begin with) and (n) sweets in the bag in total.

Now, as the rest of the sweets in the bag are yellow, what we have under initial conditions is an (n-6)/n chance of finding a yellow sweet in the bag. This expression happens to be irrelevant though, because we are told: 1. The probability that Hannah eats two orange sweets is 1/3. 2. Show that n^2 – n – 90 = 0.

What we really need is to find the chances of finding a second orange sweet after taking one orange sweet out of the bag… The chances of this happening would be 5/(n-1) because if you take one orange sweet out of the bag, 5 will be left and (n-1) sweets will be left.

Using our most basic probabilistic theorems we can deduce that the probability of selecting two orange sweets out of the bag in a row would be:

(6/n) * (5/(n-1))

Since we’ve been told that the probability that Hannah eats two orange sweets in a row is 1/3, we can say that:

(6/n) * (5/(n-1)) = 1/3

This would give us the equation:

30/(n(n-1)) = 1/3

Which would in turn give us:

90/(n(n-1)) = 1

Then:

n(n-1) = 90

Therefore we’d get:

n^2 – n = 90

And finally:

n^2 – n – 90 = 0.

From here on we’d only have to solve this quadratic equation to figure out how many sweets were in the bag to begin with. It turns out that n=10.

# Multiplying Even and Odd Numbers

Today I’m going to be showing you what would happen if you were to multiply:

a) An even number by an even number;

b) An odd number by an odd number;

c) An even number by an odd number.

Firstly, let us define what an even number is:

An even number can be described using the expression $2n$, whereby (n) would be a whole number ranging from 0 upwards.

Next, let us define what an odd number is:

An odd number can be described using the expression $2n+1$, and similarly (as is the case with even numbers), (n) would be a whole number ranging from 0 upwards.

Now, since we’ve defined how both even numbers and odd numbers can be described in terms of mathematical expressions, let’s focus our attention on multiplying even numbers by even numbers, odd numbers by odd numbers and even numbers by odd numbers…

Multiplying even numbers by even numbers:

Let’s produce two whole numbers which could be equal to one another or not equal to one another… Let’s call these numbers ${ n }_{ 1 }$ and ${ n }_{ 2 }$.

Using these two whole numbers we can multiply two unknown even numbers by each other in such a manner:

$2{ n }_{ 1 }\cdot 2{ n }_{ 2 }$

This would invariably give us the result $4{ n }_{ 1 }{ n }_{ 2 }=2\cdot 2{ n }_{ 1 }{ n }_{ 2 }$. Now, the product of ${ n }_{ 1 }{ \cdot n }_{ 2 }$ would be a whole number and since this is the case, you would have to say that an even number multiplied by an even number would produce an even number. Let’s not forget that even numbers are multiples of 2.

Multiplying odd numbers by odd numbers:

Once again, let’s come up with two whole numbers which could be equal to one another or not equal to one another… These whole numbers will be ${ n }_{ 3 }$ and ${ n }_{ 4 }$.

This would mean that two odd numbers being multiplied by one another would produce an expression as such:

$\left( 2{ n }_{ 3 }+1 \right) \left( 2{ n }_{ 4 }+1 \right)$

And if we expand the expression above, we’ll get:

$4{ n }_{ 3 }{ n }_{ 4 }+2{ n }_{ 3 }+2{ n }_{ 4 }+1$

Now if we re-arrange the expression above, we can get:

$2\left( 2{ n }_{ 3 }{ n }_{ 4 }+{ n }_{ 3 }+{ n }_{ 4 } \right) +1$

Since the expression $2{ n }_{ 3 }{ n }_{ 4 }+{ n }_{ 3 }+{ n }_{ 4 }$ must be a whole number, you would be forced to conclude that an odd number multiplied by an odd number would produce an odd number.

Multiplying even numbers by odd numbers:

We will for the last time come up with two whole numbers ${ n }_{ 5 }$ and ${ n }_{ 6 }$.

An even number and an odd number being multiplied by one another could be shown using the mathematical expression below:

$2{ n }_{ 5 }\cdot \left( 2{ n }_{ 6 }+1 \right)$

Lazily, we could conclude that an even number multiplied by an odd number would produce an even number. This is because even numbers are all multiples of 2.

Ok… So, let’s summarise what we’ve discovered:

i) An even number multiplied by an even number would produce an even number;

ii) An odd number multiplied by an odd number would produce an odd number;

iii) An even number multiplied by an odd number would produce an even number.

Knowing this we can further strengthen our mathematical reasoning. 🙂

# Binomial Expansions on Python

The other day I was asked to solve a complex probability problem. This probability problem was related to a vocabulary quiz which was being circulated around the web. It is a vocabulary quiz which contains 15 questions with 2 two possible answers to each question. My goal was to simply state the chances of someone (with a terrible vocabulary) answering at least 7/15 questions in this quiz correctly without referring to Google or any other search engine… Their answers to these questions were going to be completely random.

Now, in order solve this probability problem I first had to acknowledge seven essential facts:

1) The person taking this quiz has a terrible vocabulary and just wouldn’t be able to answer any of the questions in this quiz with confidence;

2) The person answering these questions would be providing random answers to them;

3) The person taking this quiz isn’t allowed to use any tools/resources which would help him/her answer these questions correctly;

4) There were only two possible answers to each question (right (R) or wrong (W) answers);

5) Each answer to these questions were mutually exclusive, in other words, selecting the right answer to one question wouldn’t increase your chances of answering any other question correctly ;

6) There were 15 questions in total;

7) A binomial distribution  would be required to solve the problem as there were two mutually exclusive outcomes to each question being answered.

With these facts in mind I knew exactly what I had to do to solve this problem, however, expanding the expression (R+W)^15 was going to be a very tedious task indeed and would require plenty of manual labour.

Was there another way to solve this problem?? Could a computer program help me solve the problem more easily??

Well it turned out that I would be able to solve the problem more easily, but that I’d have to use Python (a high level general purpose programming language) including an “add on” called Sympy to achieve such a feat. So what I essentially did was install both Python(x, y) and Sympy on my Windows laptop.

*You can find out how to download both Python and Sympy by watching the video below:

I then searched “Python IDLE” on my laptop and opened up Python’s shell application. This is what I typed into the application:

If you look at the image above carefully you will see that this Python shell application (thanks to the help of the “add on” Sympy, including some handy code) was able to expand (R+W)^15 for me… In fact, it did it in a split second.

From the moment (R+W)^15 had been expanded, to solve the problem I had been asked to solve, all I had to do was:

i) Sum up the coefficients of the variables (R, W) included in the expansion;

ii) Place this sum (a) under the value of the sum of coefficients sitting beside R variables with exponentials greater or equal to 7 (b) – in a fraction b/a.

This fraction (b/a) would give me the probability / chances of someone randomly being able to answer at least 7/15 questions correctly in the vocabulary test under the conditions which had been set.

a = 2(1 + 15 + 105 + 455 + 1365 + 3003 + 5005 + 6435) = 32,768

b = 1 + 15 + 105 + 455 + 1365 + 3003 + 5005 + 2(6435) = 22,819

b/a = 22,819/32,768 = 0.70 (to 2 decimal places)

Solution to problem: 70% chance

Now the irony of the story is this… I gave myself this problem to solve because my “posh” vocabulary is horrendous. It turned out that I managed to get 9/15 questions in the vocabulary test correct and all my answers were guesses. 🙂

# Mathematical Art Work, including Isometric Drawings (Visualising Maths)

Over the past couple of months I’ve been drawing mathematical figures and logos. Quite recently I discovered that art can help me understand complex mathematics a little better, especially three dimensional problems. It turns out that infinitely many complex sketches can emerge out of isometric fields which can be created using pencils, rulers and compasses. Isometric fields are used by engineers and architects who model three dimensional structures. Isometric fields allow you to represent cubes and various different 3 dimensional shapes on 2 dimensional surfaces in quite spectacular fashion. Here are some pieces of work I produced using such fields…

Follow Mathematics Videos & Proofs’s board Mathematical Art & Beauty on Pinterest.

If you’d like to see more work that I’ve been able to produce, please visit my mathematics Pinterest page. I’ve thoroughly enjoyed drawing optical illusions and logos using isometric paper, and I could indeed start designing isometric logos and pieces of text to make some extra money whilst studying mathematics. It’s not that often you bump into sound business ideas, especially business ideas related to both mathematics and art.

If you are a mathematics student but have never used isometric paper before, I’d recommend downloading isometric paper via the links below and producing mathematical sketches. Isometric paper could potentially be very useful to those interested in learning more about vector geometry and extra dimensions.

And by the way, thanks for stopping by. Enjoy your new year celebrations! 🙂

# THE IRRATIONAL NUMBER VALUE INDEX

Want to derive the value of specific irrational numbers from scratch?

Just click on the surd you’d like to discover more about, and you will be directed to a page which will give you instructions on how to derive its value.

What we haven’t included on this list are the square roots of numbers which can easily be found. We’ve only found the value of surds up to the square root of 11.

 √2 √3 √5 √7 √11 (1+√5)/2

# 1.618 as a continued fraction

The value 1.618 is an approximation to the golden ratio, a number which is found extensively in nature. As it’s a very interesting number, let’s find out what it would look like as a continued fraction…

$1.618\\ \\ =1+\frac { 618 }{ 1000 } \\ \\ =1+\frac { 1 }{ \frac { 1000 }{ 618 } } \\ \\ =1+\frac { 1 }{ \frac { 618 }{ 618 } +\frac { 382 }{ 618 } } \\ \\ =1+\frac { 1 }{ 1+\frac { 382 }{ 618 } } \\ \\ =1+\frac { 1 }{ 1+\frac { 1 }{ \frac { 618 }{ 382 } } } \\ \\ =1+\frac { 1 }{ 1+\frac { 1 }{ \frac { 382 }{ 382 } +\frac { 236 }{ 382 } } }$

$\\ \\ =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 236 }{ 382 } } } \\ \\ =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ \frac { 382 }{ 236 } } } } \\ \\ =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ \frac { 236 }{ 236 } +\frac { 146 }{ 236 } } } }$

$\\ \\ =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 146 }{ 236 } } } } \\ \\ =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ \frac { 236 }{ 146 } } } } } \\ \\ =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ \frac { 146 }{ 146 } +\frac { 90 }{ 146 } } } } } \\ \\ =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 90 }{ 146 } } } } } \\ \\ =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ \frac { 146 }{ 90 } } } } } }$

$\\ \\ =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ \frac { 90 }{ 90 } +\frac { 56 }{ 90 } } } } } } \\ \\ =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+... } } } } }$

# How to find the square root of 2 using simple algebraic rules

Ever wondered how to discover the value of the square root of 2 using simple algebraic rules?

I’ve been through plenty of maths books and 99.9% of them haven’t demonstrated how this feat can be accomplished.

Today I’m going to share with you the methods which you can use to find the value of irrational numbers as continued fractions, and in the process you will learn how to write the value of the square root of 2 as a continued fraction.

See the mathematics below…

${ x }^{ 2 }=2\\ \\ { x }^{ 2 }+x=2+x\\ \\ x\left( x+1 \right) =\left( x+1 \right) +1\\ \\ \frac { x\left( x+1 \right) }{ \left( x+1 \right) } =\frac { \left( x+1 \right) }{ \left( x+1 \right) } +\frac { 1 }{ \left( x+1 \right) } \\ \\ x=1+\frac { 1 }{ x+1 } \\ \\ x=1+\frac { 1 }{ \left( 1+\frac { 1 }{ x+1 } \right) +1 }$

$\\ \\ x=1+\frac { 1 }{ 2+\frac { 1 }{ x+1 } } \\ \\ x=1+\frac { 1 }{ 2+\frac { 1 }{ \left( 1+\frac { 1 }{ x+1 } \right) +1 } } \\ \\ x=1+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ x+1 } } } \\ \\ x=1+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ \left( 1+\frac { 1 }{ x+1 } \right) +1 } } } \\ \\ x=1+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ x+1 } } } } \\ \\ x=1+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ \left( 1+\frac { 1 }{ x+1 } \right) +1 } } } }$

$\\ \\ x=1+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ 2+... } } } } \\ \\ x=\sqrt { 2 } \\ \\ \therefore \quad \sqrt { 2 } =1+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ 2+... } } } }$

# How To Add Latex Formulas, Equations and Expressions To Your Maths Answers On Brainly.com (For Google Chrome Users Only)

Ok, so now that I’ve offended you – you probably want to leave this page, however, please remember why you got here in the first place. You simply want to add Latex formulas, equations and expressions to your answers so that you can pick up extra “Brainliest” answers and have Brainly.com’s users massaging your ego. Since it most likely is the case, bear with me for one moment and let me explain how you can perform such a feat with a minimum amount of fuss…

Now, I’m assuming you are a Google Chrome user because this post has been written for “Google Chrome Users Only”. If you aren’t a Google Chrome user – then surely you like the sound of my voice, or more specifically my average writing skills – and for that I’ll praise you. It’s not often people spend seconds reading my posts let alone minutes or hours.

Oh, you’re still here? I guess you are a Google Chrome user after all, therefore I’ll hurry up and demonstrate how to add damn Latex code to answers.

You’re probably familiar with the “Chrome Web Store”. If you aren’t, here’s its link: https://chrome.google.com/webstore/category/apps . Now, what you must do when in this store is search for the app “Daum Equation Editor” because this app is spectacular and in a league of its own (the guys who developed this app haven’t paid me to say this). This app will let you produce mathematical formulas, equations and expressions in Latex and copy and paste them into your answer boxes on Brainly.com. Furthermore, you won’t have to produce any Latex code whatsoever to achieve your lifelong ambitions. Life can’t get any better than that can it? …Copying then placing mathematical codes in to answer boxes on Brainly.com.

Assuming you have downloaded this app thanks to this post,  play around with it and explore its degrees of freedom. If the expressions you have produced are satisfying, then copy and paste the relevant Latex code it provides you with (underneath your expressions) in to your beloved answer(s) on Brainly.com. Make sure it sits within the $$…$$ tags which can be produced by clicking on the “paste/edit equation” button that can be found on answer boxes on Brainly’s site.

If you have any questions concerning copying and pasting Latex codes in to your answer boxes, just message me on my Brainly.com profile page or comment on this post.

Bye, bye for now. Have fun answering maths questions. Hehe. 🙂

# How to differentiate y=arccosx

Below I’ll be demonstrating how to differentiate y=arccosx using implicit differentiation…

$y=\arccos { x } \\ \\ \cos { y=x } \\ \\ -\sin { y\cdot \frac { dy }{ dx } } =1\\ \\ \frac { dy }{ dx } =-\frac { 1 }{ \sin { y } }$

But…

$\sin ^{ 2 }{ y+\cos ^{ 2 }{ y=1 } } \\ \\ \sin ^{ 2 }{ y } =1-\cos ^{ 2 }{ y } \\ \\ \sin { y=\sqrt { 1-\cos ^{ 2 }{ y } } }$

Therefore…

$\frac { dy }{ dx } =-\frac { 1 }{ \sqrt { 1-\cos ^{ 2 }{ y } } } \\ \\ \therefore \quad \frac { dy }{ dx } =-\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } }$

# How to differentiate y=arcsinx

Below I’ll be demonstrating how to differentiate y=arcsinx using implicit differentiation…

$y=\arcsin { x } \\ \\ \sin { y } =x\\ \\ \cos { y } \cdot \frac { dy }{ dx } =1\\ \\ \frac { dy }{ dx } =\frac { 1 }{ \cos { y } }$

But…

$\sin ^{ 2 }{ y+\cos ^{ 2 }{ y } } =1\\ \\ \cos ^{ 2 }{ y=1-\sin ^{ 2 }{ y } } \\ \\ \cos { y=\sqrt { 1-\sin ^{ 2 }{ y } } }$

Therefore:

$\frac { dy }{ dx } =\frac { 1 }{ \sqrt { 1-\sin ^{ 2 }{ y } } } \\ \\ \therefore \quad \frac { dy }{ dx } =\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } }$