The STEP Thread - Nos morituri te salutamus
Maths and statistics discussion, revision, exam and homework help.
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Ouch.
That was certainly harder than last year's. I managed to do Q5 (the final answer didn't quite come out) and (for some reason) Q13 pretty much completely. I also did bits of Q1,2,7, which I know won't get me many marks. In the last 5mins or so I realised that Q2 was probably a bit easier than I'd first thought and I tried to scrawl as much down as I could to get a few extra marks. Hopefully they'll push the grades down slightly from last year, in which case I'm, at best, borderline 2/3 (I need a 2). All in all... *******s. -
How'd did you do q. 13?(Original post by el GaZZa)
Ouch.
That was certainly harder than last year's. I managed to do Q5 (the final answer didn't quite come out) and (for some reason) Q13 pretty much completely. I also did bits of Q1,2,7, which I know won't get me many marks. In the last 5mins or so I realised that Q2 was probably a bit easier than I'd first thought and I tried to scrawl as much down as I could to get a few extra marks. Hopefully they'll push the grades down slightly from last year, in which case I'm, at best, borderline 2/3 (I need a 2). All in all... *******s.
Did you get (1-p)^3/1-p^3 for part (a) and something like 3 p^(2(r-1)) (1-p)^r for part (b)? -
Most of you are tlaking about step III i take it...
i sat I today, just "for fun" (stupid teachers) i was suprised how nice the paper was really, considering i'd not looked at any past papers. I thought it was much.. kinder than the P5 paper this year :/
anyway, if anyone sat it, i compelted the first two mechanics questions... one other (which i can't remember) and just kinda spashed around on another three... i figured that whoever saw the front of my paper where it had "questions attempted, 1, 3, 5, 9, 10, 6" would think "at least he tried" and maybe take pity on me
... or not, as the case may be. The question that started
A: 1, 6, 11, 16
B: 2, 7, 12, 17
C: 3, 8, 13, 18
D: 4, 9, 14, 19
E: 5, 10, 15, 20
"Find the general formula for each sequence..." easy enough.. "and thus prove that any term from B added to any term from C forms a term from E" looked really easy... but i bailed, just couldn't get it to come out lol, i must've been doing something drastic.
Only extension physics left to do now \o/
GL to anyone sitting STEP II on friday -
****e that looks well easy(Original post by Stubbsy)
Most of you are tlaking about step III i take it...
i sat I today, just "for fun" (stupid teachers) i was suprised how nice the paper was really, considering i'd not looked at any past papers. I thought it was much.. kinder than the P5 paper this year :/
anyway, if anyone sat it, i compelted the first two mechanics questions... one other (which i can't remember) and just kinda spashed around on another three... i figured that whoever saw the front of my paper where it had "questions attempted, 1, 3, 5, 9, 10, 6" would think "at least he tried" and maybe take pity on me
... or not, as the case may be. The question that started
A: 1, 6, 11, 16
B: 2, 7, 12, 17
C: 3, 8, 13, 18
D: 4, 9, 14, 19
E: 5, 10, 15, 20
"Find the general formula for each sequence..." easy enough.. "and thus prove that any term from B added to any term from C forms a term from E" looked really easy... but i bailed, just couldn't get it to come out lol, i must've been doing something drastic.
Only extension physics left to do now \o/
GL to anyone sitting STEP II on friday -
Yes, I got something very similar to that. All I remember is that all of them could be expressed as geometric series (or one part of a product of the answer could), parts b and c involved the P(AnB)=P(A)P(B|A) rule - where A is one of them finishiing first (btw you got the 3C1=3, nice one) and B|A is both finsihing at the same time given the other finishes first (easy to work out). Part c: the grand prize can be awarded in 2 ways- two falling out first, and one drop, then second drop (first part of this has been worked out in b), then add the totals to get a nice-ish exprssion. Actually, I broke one of my golden rules- to ignore the stats section, lucky I didn't otherwise it would've been a guaranteed U. Anyways, pretty drunk so I be back on Saturday.(Original post by beauford)
How'd did you do q. 13?
Did you get (1-p)^3/1-p^3 for part (a) and something like 3 p^(2(r-1)) (1-p)^r for part (b)? -
Surely part (c) was just 1 - part(a) - part(b) summed over n >= 2 and then the case n=1 taken into account?(Original post by el GaZZa)
Yes, I got something very similar to that. All I remember is that all of them could be expressed as geometric series (or one part of a product of the answer could), parts b and c involved the P(AnB)=P(A)P(B|A) rule - where A is one of them finishiing first (btw you got the 3C1=3, nice one) and B|A is both finsihing at the same time given the other finishes first (easy to work out). Part c: the grand prize can be awarded in 2 ways- two falling out first, and one drop, then second drop (first part of this has been worked out in b), then add the totals to get a nice-ish exprssion. Actually, I broke one of my golden rules- to ignore the stats section, lucky I didn't otherwise it would've been a guaranteed U. Anyways, pretty drunk so I be back on Saturday.