# How do I solve the Hannahs sweets question from the 2015 GCSE paper?

Here is a question from the 2015 edexcel GCSE paper that was considered particulary challenging.

**There are n sweets in a bag. 6 of the sweets are orange. The rest of the sweets are yellow.**

**Hannah takes a random 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 1/3.**

**Show that n² – n – 90 = 0.**

I will go through how to solve this question step by step.

Firstly we note that we have 6 orange sweets, and as we have n sweets in total therefore at the start we have a 6/n chance of choosing an orange sweet, because 6/n of the total sweets are orange.

After Hannah has eaten an orange sweet the probability changes. We now have n-1 sweets left in the bag, 5 of which are orange, since one orange sweet is missing from the bag. therefore the probablity the second sweet is orange, after one orange sweet has already been eaten, is 5/(n-1).

Therefore by the laws of probability the total probability that Hannah has chosen 2 orange sweets is 6/n x 5/(n-1). This is because when we want the probability that two events happen (in this case event one is that the first sweet is orange, event two is that the second sweet is orange) we multiply together the probability that each event occurs.

Now we know that the probability is equal to 1/3 from the question so now we can set up an equation

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

times together the denominators and numerators

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

expand through the brackets

30/(n^{2}-n) = 1/3

times both sides by 3

90/(n^{2}-n) = 1

multiply each side by n^{2}-n

90 = n^{2}-n

Now simply rearrange to find the solution we are seeking

n^{2 }-n - 90 = 0

We have now solved part a of the question.

As a bonus, if we wanted to solve this quadratic equation this is how we would do it:

A quadratic fuction is of the form an^{2}+bn+c

In our example a=1 b=-1 and c=-90

We now want to find two numbers y and z such that x+y=b=-1 and xy=c=-90. We can start by listing factor pairs of 90 until we find a pair with a difference of -1

-90 + 1 = -89

-45 +2 = -43

-30 + 3 = -27

-18 +5 = -13

-15 + 6 = -9

-10 + 9 = -1 We have found our pair

Therefore (n-10)(n+9) = n^{2 }-n - 90 = 0

So if (n-10)(n+9)=0 either (n-10) =0 ie n=10 or (n+9) = 0 ie n=-9.

So 10 and -9 are our solutions. However in this case we know that we cannot have a negative number of sweets in a bag therefore we deduce that there are 10 sweets in the bag.