OCR (Not MEI) S1 Wednesday 8th June 2016

madberry1

How's everyone feeling for this?

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Reply 1

CBMFM99

Not too good. Haven't revised it in a while now and I've just attempted a past paper 😁

Reply 2

duncant

I bet coding using summarised data will come up. Its usually a 6 mark question and last came up in June 2012.

Reply 3

gmanijn592

Does anyone know why the xtreme past papers website is down? Desperately trying to find ocr Solomon press s1 past papers...I know they exist! PLEASE

Reply 4

duncant

Original post by gmanijn592

Does anyone know why the xtreme past papers website is down? Desperately trying to find ocr Solomon press s1 past papers...I know they exist! PLEASE

They only exist for Edexcel, not for OCR

Reply 5

madberry1

Same here😂 Not looking so good

Reply 6

Inker

terrified. these past papers are hard AND boring.

Reply 7

S.G.

I'm only getting around 85 UMS. Hate perms and combs. Always lose those marks

Reply 8

scrlk

Anyone feeling ready?

Reply 9

LordHeskey

Original post by scrlk

Anyone feeling ready?

Sort of, but after the bloodbath of C1 who knows what we'll get :cry2:

Reply 10

B.reynolds1999

Original post by LordHeskey

Sort of, but after the bloodbath of C1 who knows what we'll get :cry2:

I'm thinking it will be pretty hard because core 2 wasn't that bad. Hoping there are no arrangement questions in it though :biggrin:

(edited 7 years ago)

Reply 11

LordHeskey

Original post by B.reynolds1999

I'm thinking it will be pretty hard because core 2 wasn't that hard. Hoping there are no arrangement questions in it though :biggrin:

Yeah same, really not sure what to expect, definitely the Maths paper I'm least ready for...

Reply 12

B.reynolds1999

Original post by LordHeskey

Yeah same, really not sure what to expect, definitely the Maths paper I'm least ready for...

All i know is that stats need to boost my grade up from the tragic core 1. I'm hoping the probability questions aren't wordy and i have a feeling they'll put some dodgy expectation and variance questions in. And like 2010 and 2013 probably a sum to infinity question with geometric distributions.

(edited 7 years ago)

Reply 13

scrlk

Original post by LordHeskey

Yeah same, really not sure what to expect, definitely the Maths paper I'm least ready for...

IIRC Jan 2013 was by far the hardest paper they've done (49 for an A) so if you can do well on that I think you'll be prepared.

Reply 14

physicskid123

A washing-up bowl contains 6 spoons, 5 forks and 3 knives. Three of these 14 items are removed atrandom, without replacement. Find the probability that(i) all three items are of different kinds, [3](ii) all three items are of the same kind.

I don't get why for the last one we don't multiply by 3?

Reply 15

scrlk

Original post by physicskid123

3! as you can get different arrangements of forks, spoons and knives

Reply 16

physicskid123

Original post by scrlk

3! as you can get different arrangements of forks, spoons and knives

I get that, I'm talking about question ii.

Reply 17

scrlk

Original post by physicskid123

I get that, I'm talking about question ii.

Sorry, my mistake. :redface:

Don't know if it helps at all but I visualised it as a tree diagram - there's no need to multiply it by 3.

[P(X] + [P(X] + [P(X]

Reply 18

physicskid123

Original post by scrlk

Sorry, my mistake. :redface:

Don't know if it helps at all but I visualised it as a tree diagram - there's no need to multiply it by 3.

[P(X] + [P(X] + [P(X]

my knight in shining armour, thank you!!

Reply 19

duncant

Original post by physicskid123

The probability of P(S,S,S) is going to be the constant, despite the fact that the order in which the spoons were taken from is different. You 3! (like you should have done in i)) something when the order in which you pick something will change. For example, part i) had 3! arrangements (S,F,K F,S,K K,S,F ...) however part ii) has 3!/3! arrangments (which =1). You can only arrange S S S one way.

Heres a good example, say you want to find how many ways you can rearrange MISSISSIPPI. You would do 11!/(4! 4! 2!), with repeats being factorialised(?) on the bottom. Its the same as spoons, 3! arrangements with 3 repeats, therefore 3!/3!=1