VCE Methods Question Thread!

[Zealous]

Now we can bring this into a binomial distribution.



We have 500 trials, and our probability of success is 0.16308 (taking a bag rejected to be "success").



So I got 0.0236 as well, might be a book error or we may have both made a seriously big mistake.

[Jason12]

how to anti differentiate 2x/x^2+1?

[Zealous]

how to anti differentiate 2x/x^2+1?

Please use brackets or LaTeX when you type up fractions next time, it can become very confusing to figure out what the actual expression is. There are two ways of interpreting it:

1:


2:
   A specialist student will easily do a 'u' substitution to integrate this, but for methods you can do this by recognition.



This means that:

So if we let     We can substitute that into the above.

Hence:

[skeletalclown]

Now we can bring this into a binomial distribution.



We have 500 trials, and our probability of success is 0.16308 (taking a bag rejected to be "success").



So I got 0.0236 as well, might be a book error or we may have both made a seriously big mistake.

Hmmm, this question is in the Continuous Random Variables chapter, rather than the Discrete Random Variables chapter, so maybe there's another method to do it without resorting to Discrete Random Variable techniques? To be honest that's all I can think of? I'll ask my teacher after school goes back I guess. Thanks so much for your help!!

[keltingmeith]

The two topics aren't mutually exclusive - using that logic, you'll never see conditional probability outside of discrete. Yet, VCAA (and textbooks) put conditional probability in EVERYTHING.

[skeletalclown]

The two topics aren't mutually exclusive - using that logic, you'll never see conditional probability outside of discrete. Yet, VCAA (and textbooks) put conditional probability in EVERYTHING.

Ahh, fair enough then! Im jsut confused, because I can't think of a continuous probability technique to solve this, and the binomial idea, although coming close to the actual answer is still wrong. But yeah, I'll definitely ask my teacher next term!! And thanks again for the help!

[soNasty]

'You have a jar which contains five white marbles, three red marbles and two blue marbles. You draw two marbles, with replacement, from the jar. Let X be the number of red marbles drawn."

a) What values can X take?
b) Draw up a table to show the probability distribution of X.

help please :\ (even tho this is probs a bloody easy question, but i cant see it)

[keltingmeith]

For a, X can be either white, red or blue, since those are the marbles on offer.

For b, your table should look like this:

x                         White                     Red                Blue

Pr(X=x)                1/2                        3/10               1/5

[soNasty]

For a, X can be either white, red or blue, since those are the marbles on offer.

For b, your table should look like this:

x                         White                     Red                Blue

Pr(X=x)                1/2                        3/10               1/5

Sorry but thats not the answer at the back of the book,
it says X can only take values of 0,1,2
and the corresponding Pr(X=x) values are 0.49, 0.42, 0.09 respectively

[keltingmeith]

Sorry but thats not the answer at the back of the book,
it says X can only take values of 0,1,2
and the corresponding Pr(X=x) values are 0.49, 0.42, 0.09 respectively

Oh shoot, I read the question wrong - very sorry about that! Let's try this again:

X represents the amount of red balls drawn, and you're going to draw two random balls with replacement. So, you could either draw two balls that aren't red, one ball that is red, or two that are red. So, X can be 0, 1 or 2 - that's part a done.

Now, you know what the table should look like, so instead I'll focus on the probabilities. If you've drawn no reds, you've either drawn two whites, two blues, or one white and one blue, so you get:



For one red, you've either drawn one red and one blue, or one red and one white, so you get:



Finally, for two reds, it's really easy - you must have drawn two reds, so you get:



Once again, sorry for the mix-up. n.n;

[soNasty]

Thanks man!

[alchemy]

Quick question guys: a^a = sqrt(2)/2. How do I solve for a without using CAS. There's a method to do this but I forgot

[Zealous]

Quick question guys: a^a = sqrt(2)/2. How do I solve for a without using CAS. There's a method to do this but I forgot

Such an odd question... you can find the solutions using algebraic manipulation, but there may be better methods which I'm unaware of.

[keltingmeith]

Such an odd question... you can find the solutions using algebraic manipulation, but there may be better methods which I'm unaware of.

Now that is cool - and possibly the only way you'll answer it at year 12 level. The only other method I can think of is through the Lambert W Relation - and if an examiner asked a year 12 student to even understand something like that, I'd personally bop him over the head. That or Newton's method, but unfortunately the only numerical method covered by year 12 maths is Euler's method in specialist...

(now that I've said that, |zxn| will probably come in and correct me, he's quite good at doing that )

[soNasty]

X~Bi(6,0.2)

How would I find the mode and median using cas?