This was actually a difficult question. My solution is very briefly written here. Parts 1 and 2 were easy, but part 3 was difficult to notice, because the 'trick' is quite clever.(Original post by 13 1 20 8 42)
Questions like this illustrate why people should check out that statistics section more often.
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STEP 2016 Solutions
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Why bother with a post grad course  waste of time?  17102016 

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 15062016 10:47

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 15062016 10:59
(Original post by student0042)

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 15062016 11:00
(Original post by Zacken)
I left my answer was ((7^7 + 1)^3  sum of 7's)((7^7 + 1)^3 + sum of 7's). 
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 15062016 11:01
(Original post by student0042)
That probably would have been better. Oops. 
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 15062016 11:03
(Original post by Zacken)
I don't think it would have... I'll probably get docked a mark or two. Ahhhh. 
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 15062016 11:04
(Original post by student0042)
Really? Did it want the answer like this? I don't remember what form it said, I thought it was just some form with [math] 7^7 [/math] in it. 
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 15062016 11:06
(Original post by Zacken)
It said "factors written as a sum of various powers of 7", I'm not sure if (7^7 + 1)^3 counts. 
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 15062016 11:08
(Original post by Zacken)
Just attaching it in big size for you:
(side note, this was a gift question!)
(other side note, I forgot sin 4 was positive...)Last edited by rohanpritchard; 15062016 at 11:10. 
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 15062016 11:09
(Original post by rohanpritchard)
How many marks do you recon lost for not including where it was discontinued? I put the y values but I did get what it meant.. Which was kind of stupidly obvious now... That's in all parts I didn't do it 
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 15062016 11:18
Hey Zacken, what do you think is roughly going to be a grade 2 based on this year's paper difficulty? Thanks

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 15062016 11:28
(Original post by kelvin1338)
Hey Zacken, what do you think is roughly going to be a grade 2 based on this year's paper difficulty? Thanks 
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 15062016 11:59
I did not do as well on this as I wanted aha. I need a 2 for my insurance which I think I should have managed, but I wanted a 1. Then again it's probably my fault for not doing enough practice for it...
On question 2 I goofed on part ii and couldn't work out what to substitute oops, and on question 3 I didn't include the dots for the discontinuity... 
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 15062016 12:17
STEP 1 Q9 solution. I couldn't remember for SURE which way round they put the coefficients of friction but I'm pretty sure it's this way, doesn't actually make any difference for the answer anyway. I realised I got the last bit wrong in the actual exam because I didn't account for the force due to friction at the rail also changing direction (I only changed it at the wall) _. This is correct here though.
Last edited by rohanpritchard; 15062016 at 16:05. 
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 15062016 12:35
Btw, e simplifies to sqrt 5  2. Its just a much more pleasing answer, with the symmetry with lambda = sqrt 5 + 2.Last edited by Mathemagicien; 15062016 at 12:36. 
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 15062016 12:36
Post rating:1 
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 15062016 12:54
(Original post by Zacken)
Yep, couldn't be bothered in the exam though. As it is, I spent 15 minutes checking everything because I was convinced my answer wasn't correct, with all the sqrt(5) shiz.
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 15062016 14:33
(Original post by Ecasx)
STEP I 2015 Q12
(i) Alice tosses a fair coin two times and Bob tosses a fair coin three times. Find the probability that Bob obtains more heads than Alice.
(ii) Alice tosses a fair coin three times and Bob tosses a fair coin four times. Find the probability that Bob obtains more heads than Alice.
(iii) Alice and Bob both throw a fair coin n times. The probability that they obtain an equal number of heads is p1, and the probability that Bob obtains more heads is p2. If Bob now throws the coin n+1 times, find the probability (in terms of p1 and p2) that Bob throws more heads than Alice. Hence generalise your results to parts (i) and (ii).
Spoiler:Show
Let A be the number of heads that Alice obtains, and let B the number that Bob obtains.
(i) There are many ways of doing this. For example, calculate P( B>A  B = r) in the four cases r = 0,1,2,3. Then P(B>A) is the sum of these probabilities. The answer is 1/2.
(ii) Same as above, with just one more case to consider. The answer is again 1/2. See a pattern here...?
(iii) In the n, n+1 case, let B' be the number of heads that Bob obtains in his first n throws, and Bi be the number he obtains in his last ( n+1 'th) throw. So B = B' + Bi. Now P(B>A) = P(B' > A) + P( B' = A and Bi = 1) = p2 + 1/2 * p1. In the n, n case, P(A>B' = P(A<B' = p2 by symmetry. So the total probability is 1 = P(A>B' + P(A<B' + P(A=B) = 2p2 + p1. Therefore p2 + 1/2 * p1 = 1/2.
Generalisation: if Alice throws a fair coin n times, for any positive integer n, and Bob throws it n+1 times, the probability that Bob throws more heads than Alice is 1/2.
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 15062016 14:56
Q7
Spoiler:ShowPretty straight forward;
SUT; all positive integers leaving remainder 1 or 3 = all positive odd integers
SnT; a number can't leave two remainders so this is an empty setSpoiler:ShowAny number in S is of form (4m + 1)
(4m+1)(4s+1) = 4(4ms + m + s) + 1
which is also an element of S
take an example 3+3 = 9 which is not an element of T, also can be done similarly using (4m+3) to show always an element of SSpoiler:ShowIntegers in T are odd and so all prime factors are elements of either S or T
Assume all prime factors of n are element of S, multiply out the factors 2 at a time each product being an element of S which would imply n is an element of S. therefore by contradiction at least one prime factor is an element of TSpoiler:Showa)
If we show none can be a product of an even number of Tprimes then the result follows
(4t+3)(4m+3) = 4(4tm + 3t + 3m + 2) + 1 which is an element of S
Therefore if n is a product of an even number multiply out pairwise giving you a product of elements of S implying n is an element of S > contradiction.
b)
just look for S primes that can be factorised as two numbers both element of T. simply swap the factors and done. take 3,3,7,11 gives (9,77)(21,33) both equal to 693Last edited by KingRS; 15062016 at 17:04. 
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 15062016 15:20
(Original post by Zacken)
I left my answer was ((7^7 + 1)^3  sum of 7's)((7^7 + 1)^3 + sum of 7's). 
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 15062016 15:28
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Updated: July 25, 2016
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