\text{(i) p.d.f. of }Y\text{ will be f}(y)=k\left(\frac{1}{2}-\frac{y}{8}\right)\text{ for }0\leqy\leq1 \text{ or }2\leqy\leq3 \text{ and 0 elsewhere}

Unparseable latex formula:

\text{so }k\int_0^1\left(\frac{1}{2}-\frac{y}{8}\right\text{d}y +k\int_2^3\left(\frac{1}{2}-\frac{y}{8}\right)\text{d}y=1 \Rightarrow k\left[\frac{y}{2}-\frac{y^2}{16}\right]_0^1+ k\left[\frac{y}{2}-\frac{y^2}{16}\right]_2^3=1

Unparseable latex formula:

\Rightarrow\left(\frac{1}{2}-\fracx{1}{16}+\frac{3}{2}-\frac{9}{16}-1+\frac{4}{16}\right)k=1 \Rightarrow \frac{5}{8} k=1 \Rightarrow k=\frac{8}{5}

\text{p.d.f. Mof }Y \text{ is thus f}(y)=\frac{8}{5}\left(\frac{1}{2}-\frac{y}{8}\right)\text { for }0\leq y\leq1 \text{ and }2\leq y\leq3, 0\text{ elsewhere}

\text{ See attached diagram for sketch}

\text{E}[Y]=\int_0^1 \frac{8}{5}\left(\frac{1}{2}-\frac{y}{8}\right)\text{d}y+ \int_2^3\frac{8}{5}\left(\frac{1}{2}-\frac{y}{8}\right)\text{d}y= \frac{8}{5}\left[\frac{y^2}{4}-\frac{y^3}{24}\right]_0^1+\frac{8}{5}\left[\frac{y^2}{4}-\frac{y^3}{24}\right]_2^3

=\frac{8}{5}\left(\frac{1}{4}-\frac{1}{24}+\frac{9}{4}-\frac{27}{24}-1+\frac{8}{24}\right)=\frac{8}{5}\times\frac{16}{24}=\frac{16}{15}

\text{(ii) E[Number of seeds falling on farmland] }=96\times\frac{3}{16}=18

\text{so number of new fever trees on farmland is expected to be B}\left(96,\frac{3}{16}\right)

\text{which we approximate by N}(18,14.625)

\text{Pr(23 or more new trees)}=\text{P}\left(z\geq\frac{22.5-18}{\sqrt{14.625}}\right)=\text{P}(z\geq1.177)=0.12

\text {So there is reason to doubt that the ancient traditions have been followed properly}

STEP 1989 .14.jpg9.8KB

Reply 122

brianeverit

1989 STEP I Question 15

\text{Basic time on each route is }

A;25\text{ minutes }B:20\text{ minutes and }D:10\text{ minutes}

\text{Route }A: Pr(X\geq27)=1-\text{P(held up for less than }2\text{ minutes) }=0.8

\text{Route }B: Pr(XZ\geq27)=1-\text{P(I talk for less than }7\text{ minutes )}

=1-\text{P(I talk to fewer than 3 friends )}=1-e^{-4}-4e^{-4}-\frac{4^2e^{-4}}{2!}=1-3e^{-4}=1-0.238=0.762

\text{Route }D: Pr(X\geq27)=\text{P(fewer than 4 detours)}

=1-\left(5\left(\frac{4}{5} \right)^4\frac{1}{5}+\left(\frac{4}{5}\right)^5\right)=1-\frac{1280+1024}{5^5}=0.263

\text{P(Route B }\mid \text{more than 27 minutes)}=\frac{\text{P(Route B and more than 27 minutes)}}{\text{P(more than 27 minutes)}}

=\frac{\frac{1}{3}\times0.762}{ \frac{1}{3}(0.8+0.762+0.263)}=0.418

(i) \text{mean time by route A}=25+5=30\text{minutes}

Unparseable latex formula:

\text{for route B}=20+3\times4=32\text{minutes }}

\text{for route D, mean distance}=2+4=6\text{km so mean time}=30\text{ minutes }

\text{so he should use route A or D}

(ii) \text{Probabilities for less than 32 minutes are }

A:0.7, B:Pr\text{(I talk to fewer than 4 friends)}=0.4335\text{ and}

D:Pr\text{(fewer than 5 diversions)}=1-\left(\frac{4}{5}\right)^5=0.672

\text{so I should choose route A}

Reply 123

brianeverit

1989 STEP I Question 16

(i) \text{If A guesses 1 then B wins £2 if he guesses 3 and loses £1 if he guesses 2 so he should guess 3 }

\text {If A guesses 2 then B wins £1 if he guesses 1 but loses £1 if he guesses 3 so he should guess 1 }

\text{If A guesses 3 then B wins £1 if he guesses 2 but loses £2 if he guesses 1 so he should guess 2}

(ii) \text{B's return if A guesses 1 is £}(2b_3-b_2)\text{ If A guesses 2, it is £}(b_1-b_3)

\text{ and ifA guesses 3 it is £}(b_2-2b_1)

Unparseable latex formula:

\text{B's expected winnings are £}\left(\frac{1}{3}(2b\-3-b_2+b_1-b_3+b_2-2b_1)\right)=\frac{1}{3}(b_3-b_1)

\text{Clearly his expected wnnings aremaximsed by taking }b_1b_2=0 _3=1

(iii)\text{A wins £2 with probability }a_3b_1,\text{£1 with probability }a_2b_1+a_1b_2

\text{he loses £1 with probability }a_3b_1+a_1b_3 \text{ and loses £2 with probability} a_1b_3

\text{hence, A's expected winnings are }2a_3b_1+a_1b_2+a_2b_3-a_3b_2a_2b_1-2a_1b_3

=(b_2-2b_3)a_1+(b_3-b_1)a_2+(2b_1-b_2)a_3

\text{ so }X=b_2-2b_3,Y=b_3-b_1\text{ and }Z=2b_1-b_2

\text{For }X,Y,Z \text{ to be non-negative we require }b_2 \geq 2b_3,b_3 \geq b_1 \text{ and }2b_1 \geq b_2

2b_1 \geq b_2 \text{ and }b_2 \geq 2b_3 \Rightarrow b_1 \geq b_3 \text{ but } b_3 \geq b_1 \text{ so we must have }b_1=b_3

\text{so we may take } b_1=b_3 = \frac{1}{6} \text{ and } b_2= \frac{1}{3}

\text{ when }b_1+b_2+b_3=1 \text{ and }X,Y,Z \text{ are non-negative as required}

\text{Whatever values of }a_1,a_2,a_3 \text{ are chosen, expected winnings are zero so he loses nothing in the long run}

Reply 124

ArchimedesSpiral

The answer worked out nicely so I think this is correct, but I'd appreciate checking.

STEP II (A) -- Question 10
Rearranging:

\mathbf{a} - \mathbf{b} = \mu v - \lambda u

Three cases:

•

no solutions

•

one solution

•

line of solutions (infinite)

There is a line of solution or no solution if v and u are parallel (no solution if

\mathbf{a} - \mathbf{b}

isn't parallel to v or u; infinite solutions otherwise.)

There is exactly one solution, then, if

\mathbf{v} \neq t\mathbf{u}

for any scalar t and

\mathbf{a} - \mathbf{b}

lies in the plane described by

\mu v - \lambda u

for

\mu, \lambda \in \mathbb{R}

.

For next part, label the equations (1) and (2). Multiply (1) by b:

bx + by + b^2z = b^2 + 2b

Subtracting:

z(b^2 - 1) = b^2 - 1

For

b \neq \pm 1

z = 1

. Substituting into (1):

x + y + b = b + 2 \iff x + y = 2

Into (2):

b(x + y) + 1 = 2b + 1 \iff b(x + y) = 2b

So if

b \neq \pm 1

there is a line of solutions

z = 1, x + y = 2

.

If

b = \pm 1

, then for

b = 1

:
(1):

x + y + z = 3

(2):

x + y + z = 3

which intersects in a plane, not a line, as the equations are identical.

If

b = -1

x + y - z = 2 - 1

z - x - y = 1 - 2

which are again linearly dependent, so intersect in a plane.

Hence

b = 1, b = -1

are the two values for which the planes do not intersect in a line.

Otherwise:

r = \begin{pmatrix}{1 \\ 1 \\ 1}\end{pmatrix} + \lambda\begin{pmatrix}{1 \\ - 1 \\ 0}\end{pmatrix}

For the last part, note that the above line contains all possible points of intersection provided

b \neq \pm 1

. Considering this case first, then want one point of intersection between the two lines. Letting:

\mathbf{a} = \begin{pmatrix}{1 \\ 1 \\ 1}\end{pmatrix}, \mathbf{b} = \begin{pmatrix}{0 \\ 0 \\ c}\end{pmatrix}, \mathbf{u} = \begin{pmatrix}{1 \\ -1 \\ 0}\end{pmatrix}, \mathbf{v} = \begin{pmatrix}{1 \\ d \\ 0}\end{pmatrix}

Condition 1:

\mathbf{u} \neq t\mathbf{v}

. This is true iff.

d \neq -1

.

Condition 2: Assuming condition 1 holds,

\mathbf{a} - \mathbf{b}

must lie in the plane described by the linear combination of

\mathbf{u}

and

\mathbf{v}

. Considering:

\lambda\mathbf{u} - \mu\mathbf{v} = \begin{pmatrix}{\lambda - \mu \\ -\lambda - \mu d \\ 0}\end{pmatrix}

. As condition 1 holds (by assumption), the linear combination is the plane [latex]\begin{pmatrix}{x \\ y \\ 0}\end{pmatrix}, so condition 2 requires

c = 1

.

So

b \neq \pm 1, d \neq -1, c = 1

is one condition.

If

b = \pm 1

, then

(*)

describes the plane:
b = 1:

x + y + z = 3

or:
b = -1:

x + y - z = 1

Condition for line intersecting a plane exactly once in three dimensions: direction vector does not lie in the plane.

Vector equation of plane for b = 1:

r = \begin{pmatrix}{1 + \lambda \\ 1 - \lambda + \mu \\ 1 + \mu}\end{pmatrix}

. Equate:

\begin{pmatrix}{1 \\ d \\ 0}\end{pmatrix} = \begin{pmatrix}{\lambda \\ -\lambda + \mu \\ \mu}\end{pmatrix}

.

Clearly

\mu = 0

; then solution only possible with

\lambda = 1, d = -1

.

Hence for

b = 1

, intersects at one point iff.

d \neq -1

.

Working for

b = -1

proceeds similarly, and gives same result.

All cases now considered.

(edited 11 years ago)

Reply 125

ArchimedesSpiral

Original post by squeezebox

STEP II Question 3

Regarding the sketch, I think the elipse becomes a vertical straight line from v = -1 to v = 1, and the hyperbola becomes the line v = 1.

I get the same result for

x = 0

but a different result for

y = \frac{\pi}{2}, 0 \leq x \leq a

. My working:

Let

y = \frac{\pi}{2}

Unparseable latex formula:

u + iv = \frac{1}{2}\left(e^(i\frac{\pi}{2}e^x - e^{-x}e^{-i\frac{\pi}{2}}\right) = \frac{1}{2}\left(ie^x - e^{-x}(-i) = i\frac{e^x + e^{-x}}{2} = i\cosh x

Equating real and imaginary parts:

u = 0, v = \cosh x

for

0 \leq x \leq a

, so

1 \leq v \leq \cosh a

.

I'd appreciate if anyone can spot an error in my working, or say if they think it's okay?

Reply 126

bogstandardname

Just collating a solution for STEP I Q1 I think Fahran was correct for one bit but not the second, so I'll post only the second.

STEP I 1989 Q1 Part 2.docx11.9KB

Reply 127

Nick_

Original post by *bobo*

Step III (2)

I'm confused by the last part of the question. You let the z axis be the vertical height relative to an origin on the ground. Then you said that when the seam is on the surface z=0, but since the surface is inclined I don't think that holds.

Reply 128

Nick_

Since there is no solution for STEP 3 question 13 I will briefly outline mine:
Consider the 5 tensions acting on the particle as complex numbers and the displacement x relative to the origin as a complex number also. The Tension from each string is (a*e^i theta - x)*(Y/a) (theta being the angle of the hole through which the string comes relative to a fixed initial line), summing the tensions we find that, since the sum of e^i*theta (when theta= 0, 2pi/5, 4pi/5, 6pi/5,8pi/5 ) is 0 using geometric series, we find that total tension is just -(5Y/a) *x , so acceleration =-(5y/ma) *x, T=2pi root(ma/5Y).
For the second part the sum Tensions is the same as before but subtracting the old tension from A and adding the new one. So T=-(5Y/a)*x - Y/a(1-x)+kY/a(1-x), giving Total tension =-(Y/A)(4+k)x + C for some constant C.
so T'=2pi*root(ma/Y(4+k))
For T'=T/2 we have 4+k= 5*4, so k=16

Reply 129

davros

Original post by Nick_

Is this the question about the coal seam? I don't think bobo's posted on here for a good few years now!

You might want to compare notes with

Original post by DJMayes

...

I think he was asking about this question recently, but no-one was brave enough to take it on :biggrin:

Reply 130

Nick_

Original post by davros

Is this the question about the coal seam? I don't think bobo's posted on here for a good few years now!

You might want to compare notes with

I think he was asking about this question recently, but no-one was brave enough to take it on :biggrin:

It would be nice to see what someone else got because i'm not sure my method was sound. I defined the height of of plane as being relative to the other using the three points and found the relationship between x and y when that height=0. Gave an answer very quickly but I'm not entirely convinced that its a valid method.

Reply 131

mikelbird

Original post by Nick_

I hope we have the same answer.....

Step1989Paper3Question2.pdf83.4KB

Reply 132

mikelbird

While I am looking at 1989 Step 3 here is what i hope to be a solution to q3

Step1989Paper3Question3.pdf90.5KB

Reply 133

Nick_

Original post by mikelbird

I hope we have the same answer.....

I don't keep my solutions so I can't remember if I got if mine was positive or negative but I definatelty remember getting (+ or -) arctan(30/31) as my final answer.

Reply 134

mikelbird

Just to add to our solutions....

Step1989Paper3Question5.pdf83.9KB

Reply 135

mikelbird

and another one....

(edited 10 years ago)

Step1989Paper3Question6.pdf74.0KB

Reply 136

mikelbird

Just for completeness.....

Step1989Paper3Question4.pdf94.6KB

Step1989Paper3Question7.pdf70.6KB

Reply 137

mikelbird

I know there have been several comments about III Q8 but I still don't think the given answers are right. I have checked these against the initial conditions and whether they actually satisfy both equation, without finding an error. As my answer disagrees with all submitted so far I will give the working....