## Monday, June 8, 2015

### 4th June 2015's Edexcel exam paper: The Probability of Hannah eating two orange sweets

The probability problem in 4th June 2015's Edexcel exam paper by exam board Edexcel has left students feeling glum and helpless.

This is the problem in full:

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 then takes at random another sweet from the bag. She eats the sweet. The probability that Hannah eats two orange sweets is $\dfrac{1}{3}$.

a. Show that $n^2-n-90=0$.

b. Solve $n^2-n-90=0$ to find the value of $n$.

Here is how to tackle the problem using the tree diagram.

A tree diagram is simply a way of representing a sequence of events. Tree diagrams are particularly useful in probability since they record all possible outcomes in a clear and uncomplicated manner.

Let us consider a simple example for illustration purpose.

Says we want to draw $2$ balls from a bag containing $8$ green balls and $6$ purple balls and we're asked to find the probability of drawing $2$ purple balls. Here, there are two sequential events we need to consider:

1. The draw of first ball,

2. The draw of the second ball.

In this example, the first branch represent the first draw. For the first ball, it's either green or purple. At the beginning of the tree diagram, there is the initial dot. Trace a branch up for green ball, label the probability of the first ball being green,

\begin{align*}\text{P(drawing first green ball)}&=\dfrac{\text{the number of green balls we have in the bag}}{\text{the total number of balls we have in the bag}}\\&=\dfrac{8}{8+6}\end{align*},

and label "Green" at the end of the branch, as shown in the first diagram below.

Likewise, trace a branch down for purple ball in the first draw with probability of drawing a purple ball, is, \begin{align*}\text{P(drawing first purple ball)}&=\dfrac{\text{the number of purple balls we have in the bag}}{\text{the total number of balls we have in the bag}}\\&=\dfrac{6}{8+6}\end{align*}

on it and label "Purple" at the end of the branch, like in the second diagram below:

Next, perform the next step in this two sequential events of drawing the second ball from the bag that now has $8+6-1$ balls left.

The second ball is either green or purple.

Therefore, at the "Green" end from first branch, draw one branch up for drawing the second green ball with the probability,

\begin{align*}\text{P(drawing second green ball)}&=\dfrac{\text{the number of green balls left in the bag}}{\text{the total number of balls that are left in the bag}}\\&=\dfrac{8-1}{8+6-1}\end{align*}

on it, and label it as "Green", and it should look like this:

Draw one branch down for drawing the second purple ball with probability, \begin{align*}\text{P(drawing second purple ball)}&=\dfrac{\text{the number of purple balls left in the bag}}{\text{the total number of balls that are left in the bag}}\\&=\dfrac{6}{8+6-1}\end{align*}

on it, label it "Purple". We have:

We're not done yet. We want to repeat the process that for the purple branch from the first step.

Upon finishing, our tree diagram should look like the following:

This is easy stuff to draw a completed tree diagram, isn't it?

We're asked to find the probability of drawing 2 purple balls. This means we must look for the nodes of the branches with "Purple-Purple" label, i.e. the purple ball on the first drawing and also another purple ball on the second drawing.

What's the probability of that? You wonder...

In a tree diagram to find a probability of an outcome we:

• multiply along the branches to find the product of all probabilities along a branch on a tree,
• add vertically to find the probability of more than one outcome (one branch).
In our case, we just multiply the possibilities of those two up:

\begin{align*}\text{P(drawing2purple balls)}&=\dfrac{6}{8+6}\times \dfrac{6-1}{8+6-1}\\&=\dfrac{6}{14}\times \dfrac{5}{13}\\&=\dfrac{15}{91}\end{align*}

Now, back to the problem of probability about Hannah eats two orange sweets:

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 then takes at random another sweet from the bag. She eats the sweet. The probability that Hannah eats two orange sweets is $\dfrac{1}{3}$.

If you try to draw the respective tree diagram based on the illustrated example above, you will get:

Question (a): Show that $n^2-n-90=0$.

From the tree diagram, we can conclude that:

\begin{align*}\text{P(Hannah eats two orange sweets )}&=\dfrac{6}{n}\times \dfrac{6-1}{n-1}\\&=\dfrac{6(5)}{n(n-1)}\end{align*}

Since we're told that the probability Hannah eats two orange sweets is $\dfrac{1}{3}$, we set the above probability that is represented by the variable $n$ as $\dfrac{1}{3}$, we get:

$\dfrac{6(5)}{n(n-1)}=\dfrac{1}{3}$

Simplifying we have:

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

$n^2-n-90=0$

Question (b): Solve $n^2-n-90=0$ to find the value of $n$.

The quadratic equation above can easily be factored into the form:

$n^2-n-90=0$

$(n+9)(n-10)=0$

Therefore, $n=-9$ or $n=10$.

Bear in mind that $n$ represents the total quantity of sweets, which means it must be a positive integer.

We can strike out the first answer, and therefore $n=10$ is what we're after.