Tag Archives: gcse maths

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


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.