# STEP I, II, III 2010 Solutions

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Index of STEP solutions; 2010 STEP papers.

Thanks to

1 – Solution by jasbirsingh

2 – Solution by gcseeeman

3 – Solution by matt2k8

4 – Solution by miml

5 – Solution by miml

6 – Solution by jj193

7 – Solution by jj193

8 – Solution by miml

9 – Solution by matt2k8

10 – Solution by matt2k8

11 – Solution by brianeverit

12 – Solution by curious_lemma

13 – Solution by matt2k8, Get me off the £\?%!^@ computer

1 – Solution by Unbounded, Salavant

2 – Solution by Unbounded, Farhan.Hanif93

3 – Solution by Unbounded, matt2k8

4 – Solution by Unbounded, matt2k8

5 – Solution by matt2k8

6 – Solution by jj193

7 – Solution by matt2k8

8 – Solution by ziedj

9 –

10 –

11 – Solution by gcseeeman

12 – Solution by curious_lemma

13 – Solution by metaltron

1 – Solution by Unbounded

2 – Solution by matt2k8

3 – Solution by Unbounded

4 – Solution by Unbounded; correction by Dalek1099

5 – Solution by Get me off the £\?%!^@ computer

6 – Solution by Get me off the £\?%!^@ computer

7 – Solution by matt2k8

8 – Solution by matt2k8

9 – Solution by AnonyMatt

10 – Solution by Pyoro

11 –

12 – Solution by Pyoro

13 –

Thanks to

*miml*for originally running this thread.*4 solutions left to fill.***STEP I**1 – Solution by jasbirsingh

2 – Solution by gcseeeman

3 – Solution by matt2k8

4 – Solution by miml

5 – Solution by miml

6 – Solution by jj193

7 – Solution by jj193

8 – Solution by miml

9 – Solution by matt2k8

10 – Solution by matt2k8

11 – Solution by brianeverit

12 – Solution by curious_lemma

13 – Solution by matt2k8, Get me off the £\?%!^@ computer

**STEP II**1 – Solution by Unbounded, Salavant

2 – Solution by Unbounded, Farhan.Hanif93

3 – Solution by Unbounded, matt2k8

4 – Solution by Unbounded, matt2k8

5 – Solution by matt2k8

6 – Solution by jj193

7 – Solution by matt2k8

8 – Solution by ziedj

9 –

10 –

11 – Solution by gcseeeman

12 – Solution by curious_lemma

13 – Solution by metaltron

**STEP III**1 – Solution by Unbounded

2 – Solution by matt2k8

3 – Solution by Unbounded

4 – Solution by Unbounded; correction by Dalek1099

5 – Solution by Get me off the £\?%!^@ computer

6 – Solution by Get me off the £\?%!^@ computer

7 – Solution by matt2k8

8 – Solution by matt2k8

9 – Solution by AnonyMatt

10 – Solution by Pyoro

11 –

12 – Solution by Pyoro

13 –

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#5

(Original post by

The second and third "here" don't open fully?

**boromir9111**)The second and third "here" don't open fully?

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#6

(Original post by

They work for me, but I'll upload it onto mediafire.

**miml**)They work for me, but I'll upload it onto mediafire.

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#7

Can I shotgun II 1 please? Second-easiest question on paper (4 being the easiest <3) so shouldn't involve much pain to type up.

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#9

I/12

Hopefully I've made it clear enough, couldn't get latex working.

Hopefully I've made it clear enough, couldn't get latex working.

Spoiler:

Show

E(X) = 1*P(X=1) + 2*P(X=2) + 3*P(X=3) + ... = sum of r*P(X=r), where r ranges over the positive integers.

The given sum is equal to (P(X=1) + P(X=2) + P(X=3) + ...) + (P(X=2) + P(X=3) + ...) + (P(X=3) + ...) + ...

------------------------

Notice that P(X=>4) = 1 - P(X<=3), and that P(X=1) equals zero.

For X to equal 2, we can either have MD, or DM, which has a total probability of 2pq.

For X to equal 3, we can either have MMD, or DDM, which has a total probability of p^2q + pq^2 = pq(p+q) = pq(1) = pq.

Therefore P(X=>4) = 1 - (pq + 2pq) = 1 - 3pq = 1 - 3p(1-p)

= 1 - 3p + 3p^2 = (1 - 3p +3p^2 -p^3) + p^3 = (1-p)^3 + p^3 =

------------------------

Now notice that for n > 1, for X to take some value n, we must have DDD....M (n-1 Ds and 1 M) or MMM...D (n-1 Ms and 1 D). Therefore, P(X=n) = qp^(n-1) + pq^(n-1). If we sum this, using the formula for a GP, we find that P(X=>n) = p^(n-1) + q^(n-1).

Using the formula for E(X) which we derived earlier, we see that

E(X) = SUM[p^(r-1) + q^(r-1)] - 1, where r ranges over the positive integers.

(the -1 is to take account for the fact that the formula for P(X=>n) gives P(X=>1) as 2, when it should in fact be 1).

So E(X) = 1/(1-p) + 1/(1-q) -1 = 1/p + 1/q -1 = (p+q)/pq -1 =

Now, since pq = p(1-p) = p - p^2 = 0.25 - (p - 0.5)^2, we can see that pq has a maximum value of 0.25, and so 1/pq has a minimum value of 4. From this it follows that

The given sum is equal to (P(X=1) + P(X=2) + P(X=3) + ...) + (P(X=2) + P(X=3) + ...) + (P(X=3) + ...) + ...

**which simplifies to E(X).**------------------------

Notice that P(X=>4) = 1 - P(X<=3), and that P(X=1) equals zero.

For X to equal 2, we can either have MD, or DM, which has a total probability of 2pq.

For X to equal 3, we can either have MMD, or DDM, which has a total probability of p^2q + pq^2 = pq(p+q) = pq(1) = pq.

Therefore P(X=>4) = 1 - (pq + 2pq) = 1 - 3pq = 1 - 3p(1-p)

= 1 - 3p + 3p^2 = (1 - 3p +3p^2 -p^3) + p^3 = (1-p)^3 + p^3 =

**p^3 + q^3**------------------------

Now notice that for n > 1, for X to take some value n, we must have DDD....M (n-1 Ds and 1 M) or MMM...D (n-1 Ms and 1 D). Therefore, P(X=n) = qp^(n-1) + pq^(n-1). If we sum this, using the formula for a GP, we find that P(X=>n) = p^(n-1) + q^(n-1).

Using the formula for E(X) which we derived earlier, we see that

E(X) = SUM[p^(r-1) + q^(r-1)] - 1, where r ranges over the positive integers.

(the -1 is to take account for the fact that the formula for P(X=>n) gives P(X=>1) as 2, when it should in fact be 1).

So E(X) = 1/(1-p) + 1/(1-q) -1 = 1/p + 1/q -1 = (p+q)/pq -1 =

**1/pq -1**Now, since pq = p(1-p) = p - p^2 = 0.25 - (p - 0.5)^2, we can see that pq has a maximum value of 0.25, and so 1/pq has a minimum value of 4. From this it follows that

**E(X) => 3.**
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#10

OK, here we go, II 1, LaTeX isn't fantastic but that's because I'm still learning how to use LyX. (I really hope this is right, but I used a grapher, looks right, circle osculates nicely). Question 11 on II I also remember being easy but only if you can draw a diagram... hard to do on computer. I'd tackle a III, but would have to start from scratch on questions and it's a little late.

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#17

(Original post by

I'm about to do II- q4

**matt2k8**)I'm about to do II- q4

(Original post by

Won't be happy...

**ziedj**)Won't be happy...

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#19

(Original post by

Woops didn't see that post haha

**matt2k8**)Woops didn't see that post haha

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#20

II/12

I did this in a hurry so there might be a few mistakes!

I did this in a hurry so there might be a few mistakes!

Spoiler:

Since f is a probability density function, the integral of f(x) between 0 and 1 will be equal to 1. This means that [INT(a) between 0 and k] + [INT(b) between k and 1] will equal one. Carrying out the integration and rearranging, we find that

Now k > 0 => (1-b)/(a-b) > 0 => 1-b > 0 =>

and k < 1 = (1-b)/(a-b) < 1 => 1-b < a-b =>

------------------------------

E(X) = [INT(ax) between 0 and k] + [INT(bx) between k and 1]

= (ak^2)/2 + b(1 - k^2)/2

and if we substitute in our expression for k, factorise/expand/cancel factors, we arrive at

------------------------------

We have have two possible scenarios: either m < k, or m > k. Consider the first: we must have [INT(a) between 0 and k] > 0.5, and so ak < 0.5. If we substitute in for k and rearrange we see that this implies that a+b > 2ab. To find m in this case we must solve

[INT(a) between 0 and m] = 0.5

i.e. am = 0.5, and so

The second possible situation, where m > k, occurs when [INT(a) between 0 and k] < 0.5. Carrying out the integration gives the condition that a+b < 2ab. We can now solve for m as before:

[INT(a) between 0 and k] + [INT(b) between k and m] = 0.5

=> ak + bm - bk = 0.5

=> k(a-b) + bm = 0.5

=> 1 - b + bm = 0.5

=> b(m-1) = -0.5

=>

-------------------------------

Suppose first that m = 1/(2a). Then, after some rearranging and factorising, E(X) - m = (1 - 2b + ab)/2(a-b) - 1/(2a) = (b(a-1)^2)/2(a-b) > 0. Therefore E(X) > m.

Now suppose that m = 1 - 1/(2b). Then E(X) - m = (1 - 2b + ab)/2(a-b) - 1 + 1/(2b) = (a(b-1)^2)/2(a-b) > 0, and so E(X) > m.

Thus, whatever the value of m,

Show

Since f is a probability density function, the integral of f(x) between 0 and 1 will be equal to 1. This means that [INT(a) between 0 and k] + [INT(b) between k and 1] will equal one. Carrying out the integration and rearranging, we find that

**k = (1-b)/(a-b).**

Now k > 0 => (1-b)/(a-b) > 0 => 1-b > 0 =>

**b < 1**

and k < 1 = (1-b)/(a-b) < 1 => 1-b < a-b =>

**a > 1**

------------------------------

E(X) = [INT(ax) between 0 and k] + [INT(bx) between k and 1]

= (ak^2)/2 + b(1 - k^2)/2

and if we substitute in our expression for k, factorise/expand/cancel factors, we arrive at

**E(X) = (1 - 2b + ab)/2(a-b)**which is what we wanted.

------------------------------

We have have two possible scenarios: either m < k, or m > k. Consider the first: we must have [INT(a) between 0 and k] > 0.5, and so ak < 0.5. If we substitute in for k and rearrange we see that this implies that a+b > 2ab. To find m in this case we must solve

[INT(a) between 0 and m] = 0.5

i.e. am = 0.5, and so

**m = 1/(2a)**The second possible situation, where m > k, occurs when [INT(a) between 0 and k] < 0.5. Carrying out the integration gives the condition that a+b < 2ab. We can now solve for m as before:

[INT(a) between 0 and k] + [INT(b) between k and m] = 0.5

=> ak + bm - bk = 0.5

=> k(a-b) + bm = 0.5

=> 1 - b + bm = 0.5

=> b(m-1) = -0.5

=>

**m = 1 - 1/(2b)**

-------------------------------

Suppose first that m = 1/(2a). Then, after some rearranging and factorising, E(X) - m = (1 - 2b + ab)/2(a-b) - 1/(2a) = (b(a-1)^2)/2(a-b) > 0. Therefore E(X) > m.

Now suppose that m = 1 - 1/(2b). Then E(X) - m = (1 - 2b + ab)/2(a-b) - 1 + 1/(2b) = (a(b-1)^2)/2(a-b) > 0, and so E(X) > m.

Thus, whatever the value of m,

**E(X) > m**.

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