Year 12 Maths Help Thread

I do, but it's almost always wrong, especially when it's like 'explain why binomial/normal/etc is/isn't appropriate here'
The textbook doesn't explain it as such either so I kinda just learnt what to do but not why if that makes sense.
I need to ace stars because I failed c1 :/

to get an A for a level maths, will it matter if I exceed the 240/300 ums for AS?
as in will it allow me to lose more marks in c34

There is no requirement to get 80%+ in AS.

But yes, the more you get in AS, the less you need to get in A2 to get 480+.

Ah okay, for an A* is it the same case?

Can somebody explain to me how to come up with the answers to perm and comp questions such as Q6: http://www.ocr.org.uk/Images/65851-question-paper-unit-4732-probability-statistics-1.pdf.
I can get the answers to some of them but most of the mark schemes just include a bunch of factorial stuff- which I know is the basis behind perms and comps but I would rather have a method where I can just put the number of possibilities on the bottom and a few perms/comps on the top.

Can somebody explain to me how to come up with the answers to perm and comp questions such as Q6: http://www.ocr.org.uk/Images/65851-question-paper-unit-4732-probability-statistics-1.pdf.
I can get the answers to some of them but most of the mark schemes just include a bunch of factorial stuff- which I know is the basis behind perms and comps but I would rather have a method where I can just put the number of possibilities on the bottom and a few perms/comps on the top.

The first one is just n factorial = 7 factorial= 5040 arrangements
This is because for the first place there are 7 ways to arrange the people. for the second chair there are 6 possible ways and so on.. so 7x6x5x4x3x2x1= 7!

the second one:
You have tom and jerry already seated so now you need to chose for the other 5 places. this would be 5! ways. But there are 6 positions assuming Tom or jerry have already been allocated a seat. Hence the number of ways this is possible is 6 multiplied by 5!. now to get probability just divide this value by the answer in (a)

Try the rest on your own.

Well I've already done this paper but I lost all the marks on part ii. Can't quite remember the answer though so for no 2 boys to sit together the order must be BGBGBGB. Would that then be 4!3!/5040 then?

Probability help please
I've done all parts of this question but I don't understand 5iii

3rd ticket on 9th attempt - see link between ii and that question for that one.

Then, there is only one way from there (after winning 2 in 8, then 3rd on 9th, and 4th one on 12) and that probability is found by.. (look what they've multiplied by)

Thank you, but where do the 2 failures come from (1-0.27)^2 for the 4th ticket on the 12th attempt?

3rd one is won on the 9th. Then if the 4th one is won on the 12th, then you can't win any on the 10th and 11th, and there is only one way of arranging this (think of it being the first win on the third go).

For stats 1, how do you get those lengthy description questions right? These usually add up and cost me 10-15 marks 😒 I can't afford to get anything below 70 in stats because of my core 1. My initial plan was to do well in c1&2 but after c1 flopped I need to make up for it with stats/: I can do stats, it's just all them bloody 'explain why this happens' or 'explain why this girl won't buy ice cream' etc and I'm just guessing and usually get it wrong. The maths is generally ok for me (still need to work on scaling and probability but it's slowly getting there)