VCE Methods Question Thread!

[garytheasian]

Drawing a tree might be the best way to approach this question

[garytheasian]

Thank you!

Does anyone know how I can graph binomial distribution graph on the CAS?

] Add Lists and Spreadsheet, name cell A "xvalues" and cell B "probx", move under cell A and type " =seq(x,x,0,number of trials)", now move under the B cell and type in "=binompdf(n,p,xvalues)"

Now we need to go back home and create a data and statistics page. Move your cursor to the x axis and add the variable "x values" and for the y-axis add the variable "probx" and then done .

[bills]

Can anyone help w 3a, please?

Q. 3a:
Pr(Different colours) = BW + WB
= (5/9 * 4/9) + (4/9 * 5/9)
= 40/81
= 49.4% (1 d.p)

[keltingmeith]

Is this what you were thinking of?
https://purplemath.com/modules/perimetr6.htm

Eh, not really a proof. See below for it, if you feel like spoiling the surprise:

Proof that the largest area of a quadrilateral with fixed perimeter is when it is a square

Okay, so, we know two things:

1. We have a quadrilateral.
2. This quadrilateral has some fixed perimeter, let's call it p.

So, we want to find out what values the length (l) and width (w) will give us the largest possible area. So, we know that:



Welp, let's try (as before) to sub l into the area expression:



Well, this is just a negative quadratic! So, the largest area is simply going to be where the turning point is - halfway between the two intercepts. Since our two intercepts are 0 and (1/2)p, halfway between them is p/4.

So, this means that the width of our quadrilateral is p/4 - to find the length, sub that into , and you'll get that the length is also p/4. Since the width and the length are equal, we have a square.

Remembering that we found this value as the width when the area is a maximum, this means that the largest area of any quadrilateral with fixed perimeter is when the quadrilateral is a square. QED.

EDIT: I should clarify, they certainly proved it, but the case was very specific, even if it doesn't look like it was.

[Floatzel98]

Hey guys, I'm finally doing probability and I'm finding it insanely difficult. I can't even seem to do most of the 'review' chapter and now i'm up to Discrete variables  and don't really understand them either. Can someone maybe walk me through this question with some explanation of what to do (mainly with part C)

1 A fair coin is tossed three times and the number of heads is noted.
   a List the sample space
   b List the possible values of the random variable X, the number of heads, together with the corresponding sample outcomes.
   c Find

[Cosec]

Hey guys, I'm finally doing probability and I'm finding it insanely difficult. I can't even seem to do most of the 'review' chapter and now i'm up to Discrete variables  and don't really understand them either. Can someone maybe walk me through this question with some explanation of what to do (mainly with part C)

1 A fair coin is tossed three times and the number of heads is noted.
   a List the sample space
   b List the possible values of the random variable X, the number of heads, together with the corresponding sample outcomes.
   c Find

Sample space is easy, just list all the possible outcomes in its variations.
So it could be (HHH) or (TTT) and you just have to go through and change them all until you cover each outcome.

Well, lets say X is the number of times a Head is thrown.
If we toss the coin 3 times we can have 0 heads, 1 head, 2 heads, or 3 heads.
So you just link them up.
If pr(x=0) that is no heads at all the sample space will be (TTT).
If pr(x=1) it is 1 head of the 3 tosses so the sample space would be (HTT), (THT), (TTH), etc.
And do the same for the rest of them.

Pr(X>=2) will be the probabilty of two heads occuring plus the probabilty of 3 heads occuring since it is greater then 2.
So lets assume its a fair coin. 50/50 chance. So each outcome is going to have (1/2*1/2*1/2) proabilty of occuring.
So Pr(X>=2) will be the sum of the probabilties for each outcome that statisfies it. Which would be the number of outcomes that have two or more heads and x it by the probabilty of the outcome occuring (which will be the same for each).

Hopefully it makes some sense.

[cosine]

Hey guys, I'm finally doing probability and I'm finding it insanely difficult. I can't even seem to do most of the 'review' chapter and now i'm up to Discrete variables  and don't really understand them either. Can someone maybe walk me through this question with some explanation of what to do (mainly with part C)

1 A fair coin is tossed three times and the number of heads is noted.
   a List the sample space
   b List the possible values of the random variable X, the number of heads, together with the corresponding sample outcomes.
   c Find

The number of heads you can get is: 0 (TTT), 1 (HTT) (TTH) (THT), 2 (THH) (HTH) (HHT) and 3 (HHH)

Find the probability that we get at least 2 heads, or more:

Pr(X=2) = 0.125 + 0.125 + 0.125 = 0.375
Pr(X=3) = 0.125



That basically means if you throw a coin three times, the probability of obtaining AT LEAST 2 heads in total of the three tosses will be a 50% chance.

[qwerty101]

can someone help me out on this question? - not methods but 'maths' i guess much appreciated!

[StupidProdigy]

The answer to this question is E. I didn't get that though..if it was E, shouldn't it be the graph y=e^-x rather than -e^-x? Thankya

[cosine]

The answer to this question is E. I didn't get that though..if it was E, shouldn't it be the graph y=e^-x rather than -e^-x? Thankya

Left end point of rectangles means that the rectanges intersect the graph as their left edge, as shown below. Now, there is a portion of the area of the rectangle that is not actually under the curve, and so if this area is included in the total area of under the graph, then this is an over-estimation.



I hope this helps you

[e^1]

can someone help me out on this question? - not methods but 'maths' i guess much appreciated!

The goal is to find a 3-digit number. A 3-digit number is a solution if it has no cows and 3 bulls, as suggested by the question.

The approach is kind-of wishy-washy, but from the table we see only two numbers only appear once: 4 and 3. Assume that these numbers are part of the answer.

In this approach, I make the chosen number to be bold. Imagine that I remove this number from each of the three guesses. That means 1 bull would be subtracted from the guess that has the chosen number in the desired digit placement, and 1 cow would be subtracted from the other rows/guesses that have this chosen number. If we arrive at a solution, then we should have 0 cows and 0 bulls in each row.

So, we have:

Numbers:	Cows:	Bulls:
7    5    1	1	1
5    7    4	1	0
3    1    7	1	0

So the second digit could be 5, 7 or 1. We also have 3 cows and 1 bull left in total. However,  if  we take 5, 7  or 1 out we get 2 cows and 0 bulls  left. But this would mean there are numbers which are in the wrong places, but we have no bulls. So something is wrong, and so we can't have 4 and 3 together as part of the answer. Thus we can have possibly 4 or 3, or neither.

I have left the rest of the working out below (in spoilers), if you want to do the rest by yourself.

Spoiler

With some (a lot?) luck and a bit of guesswork, I guess 4 to be the third digit of the answer. Then we get:

Numbers:	Cows:	Bulls:
7    5    1	1	1
5    7    4	1	0
1    7	1	1

(note that I made I removed the number 3 from this table, since I am assuming that 4 is part of  the solution, and we  can't  have 4 and 3 at the same time)

Since we have one cow in  the second row/guess of the table, then either 5 or 7 is part of the answer. We also note that in the third guess, we have 1 cow and 1 bull. Since 7 cannot be the third digit, we must have 1 as the second digit. Finally we have 7 as the first digit, since it removes both one cow from  third and second guess. Let's  see how this goes:

Numbers:	Cows:	Bulls:
7    5    1	0	1
5    7    4	1	0
1    7	1	0

Finally...

Numbers:	Cows:	Bulls:
7    5    1	0	0
5    7    4	0	0
1    7	0	0

It seems that we have 714 as an answer. Although, you can notice how un-methodical I have done it... so if anyone could find a better approach, that would be nice.

[kinslayer]

can someone help me out on this question? - not methods but 'maths' i guess much appreciated!

Here's my attempt

First note that a number that's a bull can't be a cow, so each set of three numbers contains two numbers which appear in the final set.

Starting from the first set, we can pick (a) 7,5; (b) 7,1; or (c) 1, 5. In scenario (a), the second set forces us to take 7,5 again, which leads us to a contradiction, because the third set requires us to pick 7,1 or 7,3; neither of which will work. In scenario (c), consider the third set. We must have either 1, 3 or 1, 7. 1, 7 is out, because we discarded it in the first set. 1,3 is out because neither 1 nor 3 appear in the second set -- another contradiction.

So two of our numbers are 7 & 1. In the second set, we're forced to pick 7 & 4, since we already know that 7,5 won't work. So our three numbers are 7, 4, and 1.

To get the order, either 7 or 1 is in the correct position in the first set. If 1 is in the correct position in the first set then it is in the wrong position in the third set, which means that 7 is in the correct position.

So in this scenario we have 1, 4, and 7 as our set of three numbers. But this means that 4 is a bull in the second set, which means that 7 is a cow, which is a contradiction.

So 7 is in the correct position in the first set, which means 1 is in the correct position in the third set, which means that the full set is

7 1 4.

[Apink!]

Why can't this be a probability distribution?

Apparently, it's because "it's not a function" --Itute

Please help me

[kinslayer]

Why can't this be a probability distribution?

Apparently, it's because "it's not a function" --Itute

Please help me

That table can be thought of as defining a probability mass function which assigns to each value in the support of the distribution a value x. If the probability mass function p(x) = Pr(X=x) then p(-2) = 0.2 but also p(-2) = 0.1. Functions can only assign one value per input, so it is not a function, so it is not a probability mass function, so it does not define a probability distribution.

In words, that table says that the probability that X = -2 is 0.1 but it also says that it is 0.2. It can't be two things at the same time so it cannot be a probability distribution.

[keltingmeith]

That table can be thought of as defining a probability mass function which assigns to each value in the support of the distribution a value x. If the probability mass function p(x) = Pr(X=x) then p(-2) = 0.2 but also p(-2) = 0.1. Functions can only assign one value per input, so it is not a function, so it is not a probability mass function, so it does not define a probability distribution.

In words, that table says that the probability that X = -2 is 0.1 but it also says that it is 0.2. It can't be two things at the same time so it cannot be a probability distribution.

Just to elaborate, try graphing it - you will very quickly see it fails the vertical line test.