MYTUTOR SUBJECT ANSWERS

1046 views

Show that n^2 -n - 90 = 0

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 = n- n 

Subtract 90 on both sides

n2 - n - 90 = 0

You have solved the questions and shown full working.

Dylan N. GCSE Maths tutor, GCSE Law tutor, Mentoring -Personal Statem...

2 years ago

Answered by Dylan, a GCSE Maths tutor with MyTutor


Still stuck? Get one-to-one help from a personally interviewed subject specialist

536 SUBJECT SPECIALISTS

£22 /hr

Lucy B.

Degree: Anthropology (Doctorate) - Cambridge University

Subjects offered:Maths, Science+ 5 more

Maths
Science
Drama
Chemistry
Biology
-Personal Statements-
-Oxbridge Preparation-

“Enthusiastic Cambridge grad, with experience in Biology and Maths, passionate about helping students reach their potential and achieve the grades they deserve. ”

MyTutor guarantee

£18 /hr

Liam M.

Degree: Mathematics (Bachelors) - St. Andrews University

Subjects offered:Maths, Science+ 2 more

Maths
Science
Physics
Chemistry

“I have a real passion for Maths and Physics and would relish the opportunity to pass this passion on to others! ”

PremiumWilliam A. A Level Law tutor, GCSE Law tutor, A Level Economics tutor...
£36 /hr

William A.

Degree: Law (Bachelors) - Cambridge alumni University

Subjects offered:Maths, Law+ 3 more

Maths
Law
Economics
Chemistry
.LNAT.

“I recently graduated from the University of Cambridge in Law and completed my A Levels in 2012 with 4 A*s in Maths, Economics, Physics and Chemistry.I am a friendly and approachable person and have previously tutored A Level Econom...”

About the author

Dylan N.

Currently unavailable: for regular students

Degree: Law (Bachelors) - Oxford, St Catherine's College University

Subjects offered:Maths, Law+ 2 more

Maths
Law
Economics
-Personal Statements-

“I am a law student at Oxford University and looking forward to working with you. I have done a vast amount of mentoring whilst at school so I am extremely friendly and approachable. Questions are the best way to learn so never be afra...”

MyTutor guarantee

You may also like...

Other GCSE Maths questions

The Curve C has the equation 2x^2-11+13. The point Q lies on C such that the gradient of the normal to C at Q is -1/9. Find the x-co-ordinate of Q

Show that (4+√12)(5-√3)= 14+6√3

Solve the simultaneous equations: 3x + 2y = 9 and x + 7y = 22.

How do I solve equations with unknowns in the denominators?

View GCSE Maths tutors

We use cookies to improve your site experience. By continuing to use this website, we'll assume that you're OK with this. Dismiss

mtw:mercury1:status:ok