VCE Methods Question Thread!

[ringring]

hey everyone!

I'd really appreciate some help with the attached question.

[VanillaRice]

hey everyone!

I'd really appreciate some help with the attached question.

Sketch the graph, and then draw in the rectangles.
What does the integral represent?

Spoiler

The area under the graph from x=0 to x=2

How does this relate to the rectangles?

Hope this helps

[TheCommando]

http://imgur.com/a/jbS6h
Hey how do i do q7

Also since there are 2 e(x) formulas, whe would i use e(x)=np over the sum of x multiply pr(x) and vice versa

[VanillaRice]

http://imgur.com/a/jbS6h
Hey how do i do q7

Also since there are 2 e(x) formulas, whe would i use e(x)=np over the sum of x multiply pr(x) and vice versa

I'm not quite sure what you mean by there being '2 e(x) formulas' - there is only one mean value in the question.

Since this is a binomial random variable, recall the definitions of mean and variance
E(X) = np and Var(X) = np(1-p)
We also know E(X)=12 and Var(X)=9. From here, have a go at finding n and p.

Once you've done that, you can then find Pr(X=7) using the binomial distribution.

Hope this helps

[Shadowxo]

http://imgur.com/a/jbS6h
Hey how do i do q7

Also since there are 2 e(x) formulas, whe would i use e(x)=np over the sum of x multiply pr(x) and vice versa

Great response by VanillaRice, I'll just answer the second part of your question:
Since it's binomial the probability p is always the same, so you'd use E(x)=np
You'd use E(x)=x1*p1 +x2*p2 etc when the probability is changing, like if you selected 3 coloured balls. You'd have a different probability for 0, 1, 2 etc
E(x)=np is the same as the above formula, only with a constant p.

[TheCommando]

Oh ok
I was reffering in general terms but i see that ex =np is for binomial only

And we just solve simultaneously to find n and p right

[gnaf]

Do we need to know transition matrix? Esp in solving 2009 Exam 2 Q3g (VCAA)

[Rieko Ioane]

Hi everybody,I'm having trouble with this question http://imgur.com/a/UXtBk
b) why is Janet not considered a friend? Why do we count for her probability by multiplying by 9/20?

Many thanks 

[VanillaRice]

Oh ok
I was reffering in general terms but i see that ex =np is for binomial only

And we just solve simultaneously to find n and p right

Correct!

Do we need to know transition matrix? Esp in solving 2009 Exam 2 Q3g (VCAA)

Nope! Out of the study design as of 2016.

Hi everybody,I'm having trouble with this question http://imgur.com/a/UXtBk
b) why is Janet not considered a friend? Why do we count for her probability by multiplying by 9/20?

Many thanks 

The probability is want is Janet and (exactly) one other friend. Technically speaking, she is a 'friend', and the question is indeed asking for two friends, but with one exception - one 'friend' must be Janet. So, we consider Janet on her own, plus any one out of the other four friends (doesn't matter which).

Hope this helps

[Rieko Ioane]

Correct!
Nope! Out of the study design as of 2016.
The probability is want is Janet and (exactly) one other friend. Technically speaking, she is a 'friend', and the question is indeed asking for two friends, but with one exception - one 'friend' must be Janet. So, we consider Janet on her own, plus any one out of the other four friends (doesn't matter which).

Hope this helps

I see...thanks. However, I'm having trouble seeing how this is any different to part a? For part b then, couldn't we have nCr(5,2) as the first part of the binomial formula?

[VanillaRice]

I see...thanks. However, I'm having trouble seeing how this is any different to part a? For part b then, couldn't we have nCr(5,2) as the first part of the binomial formula?

We are being specific with one of our selections (i.e. Janet). The first person we want winning is Janet. What's the probability of Janet winning? 0.45 (19/20). Also, we want one more friend to win (but it doesn't matter who). But, now that we have already selected Janet, she is no longer a part of the group of 5 'friends' i.e. there are only 4 left to pick from! This is why we use nCr(4,1), since we want one more friend from the remaining four (doesn't matter which one though). The reason we cannot use nCr(5,2)... instead is because this will give us the probability of picking any two people. However, the question tells us that Janet must be one of the two.

Hope this clarifies things

[Opengangs]

I know it's already answered, but here is how I see part (a) and (b).

Part (a) asks for any two friends to win the prize; we're not necessarily saying who exactly will win the prize.
Consider these five people:
A, B, C, D, and E.
A is just as likely to win as B; B is just as likely to win as A, etc.

So, we know this is a binomial probability, selecting two of the five available people to be a winner.

Now, part (b) can be considered a conditional probability, in that we know one winner already. This means that Janette has a 100% chance at being chosen, yet she still has that 0.45 chance at winning. It might seem confusing at first, but when we specify that Janette won, we're only interested in choosing Janette and A, or Janette and B, or Janette and C, or Janette and D.

This means that there are four ways in selecting the second person to win (ie. this becomes a binomial probability in itself), which is what the solution says.

[ringring]

Thanks!
I'm also stuck on q3 

[MightyBeh]

Thanks!
I'm also stuck on q3 

What about the question is giving you trouble? Solutions are in the spoiler but I'd encourage you to have another try before checking. If you don't know where to start, you need to find the x-intercept to find the area between the curve and the x-axis.

Spoiler

[kiki.]

When finding the sample space for binomial distributions, how do I write it in CAS? For example...

Pr(x=1)+Pr(X=0) < 0.05,  where p=0.2

I use solve but it never gives me a n=... answer D: