This is the full question:

There are n sweets in a bag. 6 sweets are orange. The rest of the sweets are yellow. Hannah takes a sweet out of the bag at random. She eats the sweet and then takes another at random. She eats the second sweet. The probability that Hannah eats two orange sweets is 1/3. Show that n² - n - 90 = 0

This is the question that caused much upset and pain during this years GCSE exam. But hopefully with my explanation we can work out what is going on.

Always in wordy question extract your information:

- n sweets (n could be any number)

- 6 orange

- picks 2 sweet one at a time and eats them

The fact that she eats them is important because in probability we call this non-replacement (she is not putting the sweets back in so the probability of picking an orange/yellow sweet is always changing).

- probability of orange and orange again is 1/3

So because we are not replacing the sweets we know that to get 2 orange, she picks an orange first and then another.

If there are 6 to start of with, the chance of me picking one is 6/n. (n is the total number of sweets)

I eat this sweet so there is one less sweet in the bag: n-1 (n was the original total, now there is one less)

Therefore the probability of me taking another orange providing I have already picked one is 5/n-1

We were told in the question the probability of picking 2 orange is 1/3.

So:

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

We know that when we multiply factions - you can do so instantly - no need to change anything so you get:

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

Lets multiply both sides by 3

90/n(n-1) = 1

Now lets multiply both sides by n(n-1)

90 = n(n-1)

Expand the brackets

90 = n²- n

Subtract 90 on both sides

n² - n - 90 = 0

You have solved the questions and shown full working.

Simplify the following: 5(x+2) - 3(6-x)

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