Expert: Michael Tadesse - 8/19/2010

Question
Hi Michael,

I live in the UK, and I'm trying to achieve an A* in  my igcse. I will list my
topics and what I find hard, however I still have more topics to learn. It
would be great if you tell me how to learn them effectively.
By the way, I don't have to learn the formulas in the exam as it's included
on the paper and my exam is a calculator test which I will sit in less than
a year away.

non-right angle triangles (sine and cosine rules)- sort of challenging
Indices- easy
Forming and solving equations-quite easy
Factors and Multiples- very easy
Averages (including cumulative frequency graphs/tables and other tables,
like the ones where you have to find the midpoints and times it by the
frequency, and then divide it by the total of the frequency)
Percentages-very easy
Area and volume- easy
Standard form- very easy
Straight line graphs-easy
Circle Theorems- quite hard as although I know all the rules, I can't
always see where a particular rule can be applied to solve a problem.
Constructions- easy
Transformations-easy
Probability- quite easy (see my question below)
Inequalities- easy
Trig function graphs- very hard
Quadratics- easy
Proportion- very hard
Vectors- quite challenging
Segments, cones and spheres- quite easy
Sets- easy
Quadratic function graphs- very hard
Trigonometry and Pythagoras- very easy.

How can I improve on this? Thanks so much! I hope I can count on you
Micheal.

Ryan

P.s. I'm stuck on this question. Could you help please?

Jack has a fair dice. 
One of the dice has 6 faces numbered from 1 to 6.
The other dice has 4 faces numbered from 1 to 4
Jack is about to throw the two dice
He will add the scores to find out the total.

Work out the probability that he will get:

A) a total of 7
B) a total of less than 5

Thank you so much!

Answer
Hello Ryan
I can see that you have a big exam ahead and I want to help you. There are some topics you should work on very hard like Vectors, quadratic functions Trigonometric functions and Circles. I can’t give you all the highlights today, time consuming and will be boring. So if think that I should give you some notes and exercises at a certain time interval like in two days, I can do it through email. But think about it and tell me next time you get the answer to your question.

Here is what you do to solve the probability question.
I reckon that a die has 6 faces.
The first one has all its faces numbered from 1-6
The second has its 4 faces numbered but the other 2 are not numbered representing 0.
So he loaded the dice and you have to determine the total outputs of the load
[1, 2, 3, 4, 5, 6]
[1, 2, 3, 4, 0, 0]
The above set explains the possible output of each die.
First how many combination are there?
Each number be related to the other 6 outputs of the other die so
6*6 = 36 outputs or combination.
But you’re required to find the probability that the sum of the numbers on the dice is exactly 7
[3, 4], [4, 3], [5, 2], [6, 1] are the only possible conditions when the sum is 7.
So the probability is 4/36 = 1/9
The second one asks the probability that the sum is less than 5
[1, 1] [1, 2] [1, 3] [2, 1] [2, 2] [1, 0] [2, 0] [3, 0] [4, 0] [0, 1] [0, 2] [0, 3] [0, 4] [1, 0] [2, 0] [3, 0] [4, 0] [0, 1] [0, 2] [0, 3] [0, 4] are the only possible conditions. It seems I repeated some terms but it’s because there are two 0s.
So the probability is 21/36 = 7/12
Thankz
Michael