VCE Methods Question Thread!

[@#035;3]

Quick questions, thanks in advance.

1) For events C and D, Pr (C')= 0.36, Pr(D)= 0.50 and Pr(C U D)= 0.8. Pr(C intersection D) is?

2) A and B are mutually exclusive events with Pr(A)= 0.7 Pr(B)= 0.2. Pr(A'U B) is?

3) On a week night, the probability that Sonja watches TV is 0.4. If she watches TV, the probability that she does her homework is 0.3. The probability that she neither watches TV nor does her homework is 0.2. The probability that on a week night Sonja does her homework is:

[bills]

1) When you say "there will be one tail", I'm assuming that you mean exactly one tail. We can work it out like this (T = tail, H = heads):
THH: 0.5 * 0.5 * 0.5 = 1/8
HTH: 0.5 * 0.5 * 0.5 = 1/8
HHT: 0.5 * 0.5 * 0.5 = 1/8
TTH = 1/8
THT = 1/8
HTT = 1/8
TTT = 1/8

As you can see, there are seven possibilities where at least one toss resulted in a tail. Of those seven, only three has exactly one tail (THH, HTH, HHT). Therefore, our final answer is:

(probability of exactly one tail) / (probability of getting at least one tail)

= (1/8 + 1/8 + 1/8) / (7/8)

= 3/7

[nerdgasm]

Hi for this question,
Could someone please confirm that I am right? (no answers)

Farmer Taylor of Burra has two tanks.  Water from the roof of his farmhouse is collected in a 100kL tank and water from the roof of his barn is collected in a 25kL tank.  The collecting area of his farmhouse roof is 200 square metres, while that of his barn is 80 square metres.  Currently, there are 35kL in the farmhouse tank and 13kL in the barn tank.
Rain is forecast and he wants to collect as much water as possible.  He should:
(A) empty the barn tank into the farmhouse tank
(B) fill the barn tank from the farmhouse tank
(C) pump 10kL from the farmhouse tank into the barn tank
(D) pump 10kL from the barn tank into the farmhouse tank
(E) do nothing.


I got A. Anyone else get it as well?

This is extraordinarily sad of me, but I remember when this question was on the Westpac Maths Competition at least 6 years ago - and I got it wrong then, because I had absolutely no idea what to do (so I said he should do nothing). Now, six years on, I look at this question and it is still puzzling - but here's my attempt at it: (tl;dr, I think the answer is D)

So people don't have to read a wall of text

Firstly, what on earth is the darned question about? Why on earth should it matter what Farmer Taylor does at all? What is the link between the volume and collecting area of his farmhouse roof and barn roof?. Answering these questions will let us go to the heart of the problem.

I think the key idea is to work out why the five options we have been given for Farmer Taylor make any difference at all. When I first saw this question, I thought, "No matter which water redistribution scheme he chooses, we still have 48kL of water combined over the two tanks. Why does it matter?". After all, if a torrential storm were to come by and dump the next Great Flood on the farm, both tanks would be filled and Farmer Taylor would have 125kL of water in total. And then, it hit me.

Suppose we could look at a certain moment when it is raining. If (at this moment) neither tank is full, then both tanks will collect water. But if one tank is full and one is not, then only the non-full tank collects water. Thus, if it were to rain a certain amount, more water will be collected when both tanks can collect water, compared to if only one tank can. As a result, the former case will lead to more water being collected than the latter case (because no water is being 'wasted' on a full collecting tank). Thus, we now understand why what Farmer Brown does can lead to a difference in the amount of water he collects.

Notice that I have not actually used a single number from the problem in the previous paragraph. Therefore, it is now time to apply the information from the problem to the thought process that was outlined. We now make the assumption that because the collecting area of the farmhouse is 2.5 times that of the barn, this means that the increase in the farmhouse tank water volume will be 2.5 times the increase in the barn tank water volume. Or if you prefer, for every 7 kL of water that rains, 5kL will go to the farmhouse tank, and 2kL will go to the barn tank. This calculation assumes, of course, that both tanks can be filled.

Let us define any time when both tanks can still be filled as 'optimal efficiency'. Also, let us define any time when exactly one tank can be filled as 'suboptimal efficiency'. Thus, our goal is to maximise the number of kL of water used in the period of optimal efficiency, while minimising the number of kL of water used in the period of suboptimal efficiency.

We now turn to examining the 5 options.

Option A:After the initial redistribution, we have 48kL in the farmhouse and 0kL in the barn. Now, it can be shown that the first 72.8 kL that rains will be at optimal efficiency. 52 of these kL go into the farmhouse, and 20.8 go into the barn. Thus, at the end of the period of optimal efficiency, there are 100kL in the farmhouse, and 20.8kL in the barn. That's pretty good - but note that in order to fill the remaining 4.2kL in the barn, you would need a total of 14.7 litres of water at suboptimal efficiency. You would require 72.8+14.7 = 87.5 litres of water to completely fill the two tanks.

It is now easy to see that Options B  and C are pretty poor choices. Option B instantly places us into the suboptimal efficiency period, meaning that we can only fill one tank - so we can never be as efficient as Option A. Option C only gives us 7 kL of optimal efficiency, at which point we have 30kL in the farmhouse and a full 25kL barn. Clearly, this is also vastly inferior to option A.

However, let's have a look at option D. After the initial redistribution, we have 45kL in the farmhouse and 3kL in the barn. Incredibly, this actually let us fill up both tanks completely at optimal efficiency! Note that the farmhouse has 55kL left, and the barn has 22kL left. Thus, after 77kL of rain, option D will let us have both tanks completely full. This is the most efficient way possible to collect rain, and thus option D is the answer. I'll leave the calculation of option E as an exercise to the reader  .

To give a 'frame by frame' account of options A and D (since these were the two most efficient options out of the five), here's a summary:
Let be the amount of rain that falls, in kL.
If , then in both options A and D, we will have  a combined total of kL across the two tanks. No difference between the two options here.
Also, if , then in both options A and D, the two tanks will end up completely full.
It is only when that Option D collects water that option A does not. But that's reason enough to choose option D (provided that it did not start raining while Farmer Taylor was trying to perform this calculation, of course ).

[bills]

3) On a week night, the probability that Sonja watches TV is 0.4. If she watches TV, the probability that she does her homework is 0.3. The probability that she neither watches TV nor does her homework is 0.2. The probability that on a week night Sonja does her homework is:

There are two possibilities for Sonja to do her homework, after watching TV, or after not watching TV. Since watching TV and not watching TV are mutally exclusive events, we can calculate that the probability for Sonja to not watch TV is 0.6. We can work out the probability that Sonja watches TV like this (T = Sonja watches TV, T' = Sonja doesn't watch TV, H = Sonja does her homework, H' = Sonja doesn't do her homework):

TH = 0.4 * 0.3
T'H = 0.6 * 0.7
TH' = 0.4 * 0.7
T'H' = 0.2

Pr(Sonya does her homework) = (probability of doing homework) / (probability of doing homework + probability of not doing homework)
= (TH + T'H) / 1
= ((0.4 * 0.3) + (0.6 * 0.7)) / 1
= 0.54

[Peanut Butter]

2) A target consists of four coloured concentric circles with radii r cm, 2r cm, 3r cm, 4r cm respectively. Assuming that anarcher hits the target every time and that they are equally likely to hit any point on the target, what is the probability that the arrow hits: 
a) the bullseye
b) The region between the circles with radii 3r cm and 4r cm?
c) The bullseye and the region between the cirlces with radii 3r cm and 4r cm

Not sure if this is right... but:

a) Pr(bullseye) = (area of bullseye) divided by (area of entire target)
                         = (pi x r^2) divided by (pi x (4r)^2)
                         = 1/16

b) The region between the circles with radii 3r cm and 4r cm:
    = (area of entire target) - (area of target with 3r cm)
    = (pi x (4r)^2) - (pi x (3r)^2)
    = 7 x pi x r^2
    Therefore, probability (region) = (7 x pi x r^2) divided by (pi x (4r)^2)
                                                     = 7/16

c) 7/16    x     1/16
    = 7/256


Again, I'm not sure if this is correct (I'm only doing methods 1&2 as well). Let me know if the answers are correct/incorrect.

If it doesn't make sense, let me know

[bills]

Not sure if this is right... but:

a) Pr(bullseye) = (area of bullseye) divided by (area of entire target)
                         = (pi x r^2) divided by (pi x (4r)^2)
                         = 1/16

b) The region between the circles with radii 3r cm and 4r cm:
    = (area of entire target) - (area of target with 3r cm)
    = (pi x (4r)^2) - (pi x (3r)^2)
    = 7 x pi x r^2
    Therefore, probability (region) = (7 x pi x r^2) divided by (pi x (4r)^2)
                                                     = 7/16

c) 7/16    x     1/16
    = 7/256


Again, I'm not sure if this is correct (I'm only doing methods 1&2 as well). Let me know if the answers are correct/incorrect.

If it doesn't make sense, let me know

a) and b) are correct, however, I'm pretty sure the actual question c) is worded incorrectly. Since the region between the 3r circle and 4r circle is mutually exclusive with the bullseye, it is impossible to hit both at the same time. Are you sure the question doesn't say "The bullseye or the region..."?

[Peanut Butter]

a) and b) are correct, however, I'm pretty sure the actual question c) is worded incorrectly. Since the region between the 3r circle and 4r circle is mutually exclusive with the bullseye, it is impossible to hit both at the same time. Are you sure the question doesn't say "The bullseye or the region..."?

That's exactly what I was thinking when I was solving it.

I just assumed that the question meant using two separate darts (sequentially).

However it makes complete sense for their to be an error in the question

[keltingmeith]

That's exactly what I was thinking when I was solving it.

I just assumed that the question meant using two separate darts (sequentially).

However it makes complete sense for their to be an error in the question

OR it's a trick question and the answer is 0.

[nerdgasm]

How would you do this question attached ?

where

A condensed summary: (normally I'd write more, but the MasterChef grand final is on, and I'm not missing that ).

It seems that we're being asked a question about the tangent to g(x) when x = 1, and where it intersects the graph of g(x) again. Also, when it intersects the graph of g(x) again, it's also the normal to g(x). This is somehow meant to let us deduce the value of a.

I came up with the following rudimentary plan:
Step 1. Work out what the equation of the tangent is when x = 1 (this is probably dependent on the value of a).
Step 2. Work out what the x-coordinate of the second intersection point with g(x) is.
Step 3. Hope it works out

So for Step 1, I first calculated g'(x). I get g'(x) = 3ax^2 - 16ax. Hence, g'(1) = -13a. Therefore, the gradient of the tangent at x = 1 is -13a.

 Now for the y-intercept. Let's try the formula y - y1 = m(x-x1). Here, we let (x1, y1) = (1, g(1)). Therefore, we need g(1), which is easily evaluated as -7a. Hence, we get y - (-7a) = -13a(x - 1). This simplifies to y = -13ax + 6a. That's the equation of the tangent when x = 1.

For Step 2, in order to work out when this tangent hits g(x) again, we let the equation of the tangent equal g(x):
-13ax + 6a = ax^2(x-8). Move this all to one side to get ax^2(x-8) + 13ax - 6a = 0.

Factorise the 'a' term, to get, a(x^2(x-8) + 13 x - 6) = 0. As a > 0, a can't be 0. Hence, we can divide by it and obtain
x^2(x-8) + 13x - 6 = 0. Expanding, we get x^3 - 8x^2 + 13x - 6 = 0. But how to solve this? Well, remember what the solutions of this equation represent. The solutions of this equation represent when the tangent at x = 1 intersects g(x).. But by definition, the tangent at x = 1 MUST intersect g(x) at x = 1! Therefore, we can say x = 1 is a solution to the cubic equation. Performing polynomial long division (or trial factorisation), we get (x-1)(x^2-7x+6) = 0.

Factorise by inspection to get (x-1)(x-1)(x-6) = 0. Thus, when x = 1 or x = 6, the tangent at x = 1 intersects g(x). But we want the value that isn't x = 1, therefore, x = 6.

Hence, we have concluded that the tangent of g(x) at x = 1 also intersects g(x) at x = 6.

Now, our question states that at this second intersection point (x = 6), the tangent line is actually the normal to the curve. Tangents and normals are related to the gradient of the function at that point, so let's work out what g'(6) is. We get g'(6) = 12a.

We're now up to Step 3. Do you remember what that was? .

Well, we know that the tangent at x = 1 is the normal at x = 6. So, we can say that the tangent at x = 1 is perpendicular to the tangent at x = 6. In other words, a straight line with a gradient of -13a is perpendicular to a straight line of gradient 12a.

If two straight lines are perpendicular, then their gradients must multiply to -1 (unless they are horizontal/vertical lines, which in this case is not possible as a is non-zero). Therefore, we can say -13a(12a) = -1. From there, algebra establishes that a^2 = 1/156. Hence, as we are told a > 0, we have a = 1/sqrt(156) = 1/2sqrt(39), or sqrt(13)/26sqrt(3), if you prefer.

[Escobar]

endpoints don't have gradients, do they?

[plato]

endpoints don't have gradients, do they?

Nope, they don't. The graph is not continuous at an end point.

[knightrider]

A condensed summary: (normally I'd write more, but the MasterChef grand final is on, and I'm not missing that ).

It seems that we're being asked a question about the tangent to g(x) when x = 1, and where it intersects the graph of g(x) again. Also, when it intersects the graph of g(x) again, it's also the normal to g(x). This is somehow meant to let us deduce the value of a.

I came up with the following rudimentary plan:
Step 1. Work out what the equation of the tangent is when x = 1 (this is probably dependent on the value of a).
Step 2. Work out what the x-coordinate of the second intersection point with g(x) is.
Step 3. Hope it works out

So for Step 1, I first calculated g'(x). I get g'(x) = 3ax^2 - 16ax. Hence, g'(1) = -13a. Therefore, the gradient of the tangent at x = 1 is -13a.

 Now for the y-intercept. Let's try the formula y - y1 = m(x-x1). Here, we let (x1, y1) = (1, g(1)). Therefore, we need g(1), which is easily evaluated as -7a. Hence, we get y - (-7a) = -13a(x - 1). This simplifies to y = -13ax + 6a. That's the equation of the tangent when x = 1.

For Step 2, in order to work out when this tangent hits g(x) again, we let the equation of the tangent equal g(x):
-13ax + 6a = ax^2(x-8). Move this all to one side to get ax^2(x-8) + 13ax - 6a = 0.

Factorise the 'a' term, to get, a(x^2(x-8) + 13 x - 6) = 0. As a > 0, a can't be 0. Hence, we can divide by it and obtain
x^2(x-8) + 13x - 6 = 0. Expanding, we get x^3 - 8x^2 + 13x - 6 = 0. But how to solve this? Well, remember what the solutions of this equation represent. The solutions of this equation represent when the tangent at x = 1 intersects g(x).. But by definition, the tangent at x = 1 MUST intersect g(x) at x = 1! Therefore, we can say x = 1 is a solution to the cubic equation. Performing polynomial long division (or trial factorisation), we get (x-1)(x^2-7x+6) = 0.

Factorise by inspection to get (x-1)(x-1)(x-6) = 0. Thus, when x = 1 or x = 6, the tangent at x = 1 intersects g(x). But we want the value that isn't x = 1, therefore, x = 6.

Hence, we have concluded that the tangent of g(x) at x = 1 also intersects g(x) at x = 6.

Now, our question states that at this second intersection point (x = 6), the tangent line is actually the normal to the curve. Tangents and normals are related to the gradient of the function at that point, so let's work out what g'(6) is. We get g'(6) = 12a.

We're now up to Step 3. Do you remember what that was? .

Well, we know that the tangent at x = 1 is the normal at x = 6. So, we can say that the tangent at x = 1 is perpendicular to the tangent at x = 6. In other words, a straight line with a gradient of -13a is perpendicular to a straight line of gradient 12a.

If two straight lines are perpendicular, then their gradients must multiply to -1 (unless they are horizontal/vertical lines, which in this case is not possible as a is non-zero). Therefore, we can say -13a(12a) = -1. From there, algebra establishes that a^2 = 1/156. Hence, as we are told a > 0, we have a = 1/sqrt(156) = 1/2sqrt(39), or sqrt(13)/26sqrt(3), if you prefer.

Wow awesome explanation nerdgasm 

Thanks so much 

[knightrider]

For this question.

If X is normally distributed with μ = 40 and σ = 7, find, correct to 4 decimal places:

 Pr(X ≥ 30) and Pr(25 ≤ X ≤ 45).

i got Pr(X ≥ 30)=0.9234 and Pr(25 ≤ X ≤ 45)=0.7464

Are my answers right?

Can anyone help with this ?
 

[Hutchoo]

^ seems legit

[plato]

Both probabilities look OK according to my Classpad.