VCE Methods Question Thread!

[S_R_K]

Hi, I'm having some issues with this problem from my textbook (Methods 1 and 2, chapter 7H).
I would really appreciate some help.

A transformation is defined by the matrix
[0 2]
[-3 0]
Find the equation of the image of
the straight line with equation y = 2x + 3 under this transformation.

Same method as usual for linear transformations, just multiply a generic point (x, y) by the 2x2 matrix, and that gives the coordinates of the image point (x', y') in terms of the pre-image (x, y).

\(\begin{bmatrix}
x' \\
y'
\end{bmatrix} = \begin{bmatrix}
0 & 2 \\
-3 & 0
\end{bmatrix} \begin{bmatrix}
x \\
y
\end{bmatrix} \)

[amyzzwq]

Hi, can someone help me identify why the way I did this question is wrong? I used binomial on my CAS and since they said scored exactly 15 goals given that he scored at least 12 shouldn't I use binomialcdf with lower bound 12 and upper bound 15 to find Pr(X=15 intersect X<=12)?

The answer didnt find the intersection and used binomialpdf to find Pr(X=15)

[wingdings2791]

Hi, can someone help me identify why the way I did this question is wrong? I used binomial on my CAS and since they said scored exactly 15 goals given that he scored at least 12 shouldn't I use binomialcdf with lower bound 12 and upper bound 15 to find Pr(X=15 intersect X<=12)?

The answer didnt find the intersection and used binomialpdf to find Pr(X=15)

Hi, I think I may know what's wrong.
Your working out, using \(Pr(X=15)\cap Pr(X\ge12)\), is correct. Where you've gone wrong is in trying to find the probability of between 12 and 15 goals being successful \(Pr(12\le X\le15)\), which is a subset of \(Pr(X=15)\cup Pr(X\ge12)\). Using \(binomialPDf\) to find \(Pr(X=15)\) is the correct approach because exactly 15 goals being scored is one event (which doesn't include the chances of scoring 12/20, 13/20, 14/20 goals) and it's the event that's being investigated. It's easiest to think of it this way: scoring 15 goals definitely means scoring more than 12 goals. However, scoring more than 12 goals does not necessarily mean scoring precisely 15 goals.

Hopefully that makes sense!

[fun_jirachi]

Hi, can someone help me identify why the way I did this question is wrong? I used binomial on my CAS and since they said scored exactly 15 goals given that he scored at least 12 shouldn't I use binomialcdf with lower bound 12 and upper bound 15 to find Pr(X=15 intersect X<=12)?

The answer didnt find the intersection and used binomialpdf to find Pr(X=15)

Just adding on to wingdings' excellent response: To better identify incorrect answers, you need to recognise when the numbers seem a bit off. Making this 100% clear (in case you couldn't infer from wingdings' reply) \(P(X = 15 | X \geq 12) =  \frac{P(X=15 \ \cap \ X \geq 12)}{P(X\geq 12)} = \frac{P(X=15)}{P(X \geq 12)}\). From here, you need to see if your values make enough sense for a correct answer (which relies on some intuition). Given that the distribution is \(X \sim B(20, 0.7)\), the fact that you essentially have \(P(X=15) \approx 0.649\) should be ringing some alarm bells (as should a final result of roughly 0.73! The chance shouldn't be this high, even with the restricted range of results you're analysing). This should tell you that you've either punched in the numbers wrong or your working is wrong. In this case, it's just as a result of not being the most careful with your sets, but other times it might be something different; that's why it's important to look for telltale signs like this.

Hope this helps

[amyzzwq]

Hi, I think I may know what's wrong.
Your working out, using \(Pr(X=15)\cap Pr(X\ge12)\), is correct. Where you've gone wrong is in trying to find the probability of between 12 and 15 goals being successful \(Pr(12\le X\le15)\), which is a subset of \(Pr(X=15)\cup Pr(X\ge12)\). Using \(binomialPDf\) to find \(Pr(X=15)\) is the correct approach because exactly 15 goals being scored is one event (which doesn't include the chances of scoring 12/20, 13/20, 14/20 goals) and it's the event that's being investigated. It's easiest to think of it this way: scoring 15 goals definitely means scoring more than 12 goals. However, scoring more than 12 goals does not necessarily mean scoring precisely 15 goals.

Hopefully that makes sense!

Oh, so I always have to decide base on what the question is asking, I always thought it was part of the formula to find Pr(12<=X<=15).

[fun_jirachi]

Oh, so I always have to decide base on what the question is asking, I always thought it was part of the formula to find Pr(12<=X<=15).

Generalising, \(P(A|B) = \frac{P(A \cap B)}{P(B)}\), given two events A, B. It is always part of the formula to find \(P(A \cap B)\), but in this case, the expression you've deduced for \(A \cap B\) is incorrect (you deduced that \(A \cap B = 12 \leq X \leq 15\), when it is in fact only \(X=15\)). You should always decide based on what the question is asking regardless of the topic, not doing so will lead to making mistakes here and there.

[amyzzwq]

Alright, thank you so much for the explanation!

Generalising, \(P(A|B) = \frac{P(A \cap B)}{P(B)}\), given two events A, B. It is always part of the formula to find \(P(A \cap B\), but in this case, the expression you've deduced for \(A \cap B\) is incorrect (you deduced that \(A \cap B = 12 \leq X \leq 15\), when it is in fact only \(X=15\)). You should always decide based on what the question is asking regardless of the topic, not doing so will lead to making mistakes here and there.

[thenuttyprofessor]

Hey everyone,

I've started doing methods exam prep now, and Exam 1's are going pretty smoothly so far dropping no more than 2-3 marks per exam and steadily been improving, but with regards to Exam 2 I feel like a lot of the time I am just not getting the pathway of working out the difficult p.s questions despite spending time on them. As a result my results so far for those have been fluctuating a lot and I haven't been seeing any real consistent improvement. I review my mistakes and figure out how to do those questions, but each time there is a completely new p.s question I get stuck and my past experience doesn't seem to help me. How can I actually see more improvement in Exam 2 and problem solving? I don;t want to spend the next 2 months mindlessly doing the same thing with no results.
Also, what exam scores should I be looking at for a 45 in Methods?

Thanks

[(Noname):):):)]

Hi AN!!,


With VCE exams looming any closer, I have started to go thru some past papers (after all the sooner you start preparation, the easier to see where your weaknesses are!)

This was from a 2016 Exam 1 paper, but I was wondering, how to do part c)? I know it is Binomial and I even tried using the formula, but I just don't understand how the solutions say it is (16/81)^6. From my interpretation, using the formula, I got 6C0 multiplied by (16/81)^0 multiplied by (1-16/81)^6 which = (65/81)^6...

It would be great if someone can tell me what I am doing wrong here 

[fun_jirachi]

The probability that you choose a non-tagged sheep is 2/3 (note there is replacement). You do this four times a day, for six days for a total of 24 selections. Hence, the solution is just \(\frac{2}{3}^{24} = \left(\frac{16}{81}\right)^6\).

I just don't understand how the solutions say it is (16/81)^6. From my interpretation, using the formula, I got 6C0 multiplied by (16/81)^0 multiplied by (1-16/81)^6 which = (65/81)^6...

It would be great if someone can tell me what I am doing wrong here 

You've just mixed up the exponents. It should be \(\binom{6}{0} \times \left(\frac{16}{81}\right)^6 \times \left(\frac{65}{81}\right)^0\) not \(\binom{6}{0} \times \left(\frac{16}{81}\right)^0 \times \left(\frac{65}{81}\right)^6\). Remember, the 'success' in this case is when on a particular day you pick only untagged sheep. Your working out implies you're looking for the probability that on six consecutive days you pick at least one tagged sheep on each day.

[(Noname):):):)]

Well it's the same thing, you've just mixed up the exponents. It should be \(\binom{6}{0} \times \left(\frac{16}{81}\right)^6 \times \left(\frac{65}{81}\right)^0\) not \(\binom{6}{0} \times \left(\frac{16}{81}\right)^0 \times \left(\frac{65}{81}\right)^6\). Remember, the 'success' in this case is when on a particular day you pick only untagged sheep. Your working out implies you're looking for the probability that on six consecutive days you pick at least one tagged sheep on each day.

Makes sense now, thanks!

[redset8]

Hey AN,

Just wondering if someone could step me through this question. I'm a bit lost.

Cheers!

[fun_jirachi]

You're given that \(P(\text{hybrid apple}) = 0.4\) and \(P(\text{hybrid apple} \ | \ \text{altitude} > 1000\text{m}) = 0.15\).

Hints:
- From the information given, can you find the probability that an apple is grown at an altitude of greater than 1000m? You can then deduce the probability that an apple is grown at an altitude of less than 1000m.
- Can you use this probability somehow to find what the question asks? You need to find \(P(\text{hybrid apple} \ | \ \text{altitude} < 1000\text{m})\).

[redset8]

You're given that \(P(\text{hybrid apple}) = 0.4\) and \(P(\text{hybrid apple} \ | \ \text{altitude} > 1000\text{m}) = 0.15\).

Hints:
- From the information given, can you find the probability that an apple is grown at an altitude of greater than 1000m? You can then deduce the probability that an apple is grown at an altitude of less than 1000m.
- Can you use this probability somehow to find what the question asks? You need to find \(P(\text{hybrid apple} \ | \ \text{altitude} < 1000\text{m})\).

Well, P(H | A>1000) = Pr (H and A>1000) / Pr (A>1000). And likewise for Pr( H | A<1000).
I'm stuck from here...

[redset8]

Actually it's all good I realised that Pr(A>1000)=1/3 is given from a previous part.
Thanks anyway jirachi!!