[Doctor.]

It's just about a month till this exam so I decided to get this thread started nice and early

Post up your revision notes, questions and generally anything related to S1!!!!!!!

I am very active on the forum so hopefully can help too


Last of all, GET REVISING!!!!!!!!!!! (or still learning in my case....)

PS: I hate this paper its sooo just...annoying The permutations and combinations rubbish annoys me greatly!

[Millyshyn]

I'm struggling with it at the moment, my teacher isn't very motivating at all but I should really start doing something about it now. We finished the syllabus really early thankfully, otherwise it would've been a bit of a disaster. Hopefully this thread helps a bit I'll post my notes when I have a bit of spare time!

[kurdishboy94]

look at s1 in this perspective i have gotten A's in all my core exams all i have is D1 and S1 and C4.
S1 retaked twice and still got a B, its not even a high B i might aswell of gotten a C

[Doctor.]

(Original post by kurdishboy94)
look at s1 in this perspective i have gotten A's in all my core exams all i have is D1 and S1 and C4.
S1 retaked twice and still got a B, its not even a high B i might aswell of gotten a C

Ergh I know, this exam is just...annoying tbh.

Only bit I really struggle on is the 'permutations and combinations' section but they make it so bloody complex by adding in little bits. I honestly cant work my head around them!

The statistical bit Isnt too bad. I can do It, just make silly mistakes sometimes. Problem Is I don't really get the probability section which sucks because its over half of the exams!

Need some helpful revision hints on that haha

(Original post by Millyshyn)
I'm struggling with it at the moment, my teacher isn't very motivating at all but I should really start doing something about it now. We finished the syllabus really early thankfully, otherwise it would've been a bit of a disaster. hopefully this thread helps a bit I'll post my notes when I have a bit of spare time!

Neither was my teacher I don't think many teachers do teach It well to be honest

I would really apprecite the probability revision notes....I'm awful at it

[Doctor.]

So how is everyone finding S1? I am getting pretty bored of it all

Decided on revising other subjects for a little before I get back to S1

[Doctor.]

Need some help with the attached questions. I don't really understand how they get the answer

I have attached the questions and my answers, the ones I got wrong/don't understand are underlined in pencil (on the question paper). I would really appreciate it if someone explains to me how to get the correct answer.


Will give +rep and a big hug !!

[Quip]

(Original post by Doctor.)
Need some help with the attached questions. I don't really understand how they get the answer

I have attached the questions and my answers, the ones I got wrong/don't understand are underlined in pencil (on the question paper). I would really appreciate it if someone explains to me how to get the correct answer.


Will give +rep and a big hug !!

OK, starting with 2b.
The seating can only be WMWMWMWM or MWMWMWMW (M=man, W=woman).
So, taking each option in turn, four women can only seat in four specific places, so treat that like part a) but just four women can sit anywhere in four places. This can happen for each option of seating for the men, so square that to get the total variations for WMWMWMWM. Then multiply by 2 to get your final answer.

Does this make sense? (I hope it's right.)

[Doctor.]

(Original post by Quip)
OK, starting with 2b.
The seating can only be WMWMWMWM or MWMWMWMW (M=man, W=woman).
So, taking each option in turn, four women can only seat in four specific places, so treat that like part a) but just four women can sit anywhere in four places. This can happen for each option of seating for the men, so square that to get the total variations for WMWMWMWM. Then multiply by 2 to get your final answer.

Does this make sense? (I hope it's right.)

Sorry only just seen this was doing more questions

So that would be: 4! For each option.

Then we would have to x2, since there are two options. Although I don't understand why we are squaring? sorry

The answer was 1152 BTW


Thanks for helping btw, will +rep as soon as I get back to my Laptop!!

[QwertyG]

omg S1 is so boring I was doing a past paper earlier and gave up half way because I just mind blanked now im C2 and im firing at all the questions

[Doctor.]

(Original post by QwertyG)
omg S1 is so boring I was doing a past paper earlier and gave up half way because I just mind blanked now im C2 and im firing at all the questions

I don't understand how some people willingly do S2 lol. I mean S1 is boring enough, S2 must be like death

[QwertyG]

(Original post by Doctor.)
I don't understand how some people willingly do S2 lol. I mean S1 is boring enough, S2 must be like death

I got to take S2 in my A2 year, words cannot explain how much I hate the maths department for doing this

[Quip]

(Original post by Doctor.)
Sorry only just seen this was doing more questions

So that would be: 4! For each option.

Then we would have to x2, since there are two options. Although I don't understand why we are squaring? sorry

The answer was 1152 BTW


Thanks for helping btw, will +rep as soon as I get back to my Laptop!!

Yes it would be 4! and you square it because there are 4! possible ways for the women to sit and 4! ways for the men to sit.
For example, you could have
W1,M1,W2,M2,W3,M3,W4,M4 or W1,M1,W2,M2,W3,M4,W4,M3.
The women are in the same order, but the men have changed. There are 4! variations of the men's seating for each arrangement of the women's seating and as the women have 4! different ways (each way has 4! variations due to the men), in total you have to square it.

Then, as you have said multiply by two to account for both options.

[Doctor.]

(Original post by Quip)
Yes it would be 4! and you square it because there are 4! possible ways for the women to sit and 4! ways for the men to sit.
For example, you could have
W1,M1,W2,M2,W3,M3,W4,M4 or W1,M1,W2,M2,W3,M4,W4,M3.
The women are in the same order, but the men have changed. There are 4! variations of the men's seating for each arrangement of the women's seating and as the women have 4! different ways (each way has 4! variations due to the men), in total you have to square it.

Then, as you have said multiply by two to account for both options.

Ah I understand what you mean Thank you for clearing that up!!!

It always seems so simple once someone explains it to me lol

Wanna help with the others?

Been trying to work them out again, but don't seem to get the right answers

[Quip]

(Original post by Doctor.)
Ah I understand what you mean Thank you for clearing that up!!!

It always seems so simple once someone explains it to me lol

Wanna help with the others?

Been trying to work them out again, but don't seem to get the right answers

Not sure about question 4. I'll think about it.

For question 7, I'm probably missing something but, it doesn't say that anyone has to be in a particular car (does it?) so can't you just treat it as 10 places in any order? (Do you have the answers?)

For question 8,you need to multiply the number of arrangements when the days chosen are day 1 and 2 by 6 because the two flavours that need to be consecutive could be day 1 and 2, or day 2 and 3, or day 3 and 4 or day 4 and 5 or day 5 and 6 or day 6 and 7 (ie 6 options).

[Doctor.]

(Original post by Quip)
Not sure about question 4. I'll think about it.

For question 7, I'm probably missing something but, it doesn't say that anyone has to be in a particular car (does it?) so can't you just treat it as 10 places in any order? (Do you have the answers?)

For question 8, you've found the probability that they will be eaten on day 1 and 2 but it could be day 4 and 5. So probability for day 1 and 2 = 2/7 x 1/6, which I think you put. However, you also have to add the probability of it being day 2 and 3 = 2/6 x 1/5, etc until day 6 and 7. I'm not sure if there's a quicker way to do it, but I think this will work (hopefully).

Question 7: 210 (thats the only answer given)
Question 8: 210 and 2/7

[Quip]

Ok I've changed my mind about question 8.
You need to find how many ways are there of putting the strawberries together. As I pointed out in my amended post above, there are 6 different combination of days which will make them consecutive. On each day, for example if the days were day 1 and 2, there are 10 different ways of arranging the other five. To be honest I can't remember the mathematical way of working this out, I just counted them (I did S1 last year and can't remember how to use the C symbol).

So total ways of arranging the yoghurts is 6 x 10 = 60. Then divide this by the total ways of arranging the five which you've already calculated (210). This gives 2/7 - the correct answer (thankfully).

[Doctor.]

Ahh just worked out question 8!

The first part was simple once you realise the items are NOT INDEPENDENT:

7!/(2!x3!x2!) =210

Second part was an absolute pain. Too much working out but eventually got there, its very long winded.

[Chaarl]

S1 confuses me! why bring intergration and differentiation into stats? why cant they just stay in core?

[Doctor.]

(Original post by Chaarl)
S1 confuses me! why bring intergration and differentiation into stats? why cant they just stay in core?

What paper did they do that in?

Guess they got tired of S1 being the 'easy paper' I honestly dont get why people call it the easy paper

[Chaarl]

(Original post by Doctor.)
What paper did they do that in?

Guess they got tired of S1 being the 'easy paper' I honestly dont get why people call it the easy paper

its in the pdf and cdf
might be a different exam board to you though
me either, i can only do about 2 questions: choose and trees haha