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# STEP 2016 Solutions watch

1. (Original post by 13 1 20 8 42)
Questions like this illustrate why people should check out that statistics section more often.
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.

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2. (Original post by student0042)
I left my answer was ((7^7 + 1)^3 - sum of 7's)((7^7 + 1)^3 + sum of 7's).
3. (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).
That probably would have been better. Oops.
4. (Original post by student0042)
That probably would have been better. Oops.
I don't think it would have... I'll probably get docked a mark or two. Ahhhh.
5. (Original post by Zacken)
I don't think it would have... I'll probably get docked a mark or two. Ahhhh.
Really? Did it want the answer like this? I don't remember what form it said, I thought it was just some form with 7^7 in it.
6. (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.
It said "factors written as a sum of various powers of 7", I'm not sure if (7^7 + 1)^3 counts.
7. (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.
I'm not sure either, but (7^7 + 1)^3 looks a lot nicer.
8. (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...)
How many marks do you recon lost for not including where it was discontinued? I put the y values but I didn't get what it meant.. Which was kind of stupidly obvious now... That's in all parts I didn't do it
9. (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
Probably 4/5 marks?
10. Hey Zacken, what do you think is roughly going to be a grade 2 based on this year's paper difficulty? Thanks
11. (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
My honest opinion is that it'll be about 62 for a 2.
12. 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...
13. 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.
Attached Images
14. STEP1 2016 Q9.pdf (700.1 KB, 212 views)
15. (Original post by Mathemagicien)
Yay! That's the answer I got.

Btw, e simplifies to sqrt 5 - 2
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.
16. (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.
I did exactly the same, thought it would be a nice fraction so spent ages checking.

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17. (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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Oh damn that's quite an easy question.
18. Q7
Spoiler:
Show
Pretty 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 set
Spoiler:
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Any 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 S
Spoiler:
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Integers 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 T
Spoiler:
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a)
If we show none can be a product of an even number of T-primes 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 693
19. (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).
what do you think the mark weighting will be for parts 1 and 2
20. (Original post by smartalan73)
STEP I, Question 3

Attachment 550379
I stupidly marked the coordinates of the points at the beginning and ends of the horizontal lines. Given that I still got all of the shapes right, how many marks would I have lost do you think?
21. (Original post by jahovasfitness)
I stupidly marked the coordinates of the points at the beginning and ends of the horizontal lines. Given that I still got all of the shapes right, how many marks would I have lost do you think?
If you still got the coordinates right, I don't see why you should lose any marks?

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