The Maths Thread

[Dan]

Did my HSC half-yearlies last week.

Advanced: 79%, making me around 3rd overall. One girl killed it with 98%.
Extension 1: Hopefully around the same mark. Were given a near-impossible circle proof (as usual), but the calculus and differentiation sections weren't too bad.
Extension 2: If I get every mark for every question I put down an answer for (which won't happen), I'd end up with 42.5%. Had to leave out over half of the test. Polynomials section was easy, I couldn't answer anything in the hyperbola section and the ellipse section was difficult too. No 'easy' marks in the test, compared to the sample tests I'd seen from other schools.

[Spark]

No easy marks in MX2 full stop. Get past papers from the 90s and just hammer away.

[Turbinator]

Originally Posted by Spark

How'd you work it out, ss?

Expected distance the ant travels is x

Expected distance from three corners next to the ant is y

Expected distance from three corners next to the spot diagonal of the ant is z

If we assume that the ant is placed at z, there is a 1/3 probability that it will travel 1 meter and reach the spot diagonal from it. With the remaining 2/3 probability that it will travel the distance y and then 1 meter.

Therefore, z = (1/3)(1) + (2/3)(1+y) = [1 + (2/3)(y)]

y = (1/3)(1+x) + (2/3)(1+z) = [1 + (1/3)(x) + (2/3)(z)]

And if we assume that the ant is placed at x, it will always be 1 meter + y. Therefore, x = (1+y).

So now we solve for x through simple substitution.

x = 1 + [1 + (1/3)(x) + (2/3)(z)]
x = 3 + z = 4 + 2/3(y)
x = 4 + 2/3(x-1)
x = 10 meters

Don't know how well I explained that, but I think you guys get the idea.

[Dan]

Originally Posted by Spark

No easy marks in MX2 full stop. Get past papers from the 90s and just hammer away.

Yeah, very true. Reportedly other local schools were getting 'find the foci, vertices, directrices and sketch' questions for 4 marks total in a 40-50 mark test, whereas we got 1 mark for questions requiring a page of working and adapting what we knew, rather than just sketching.

Intending to download a stack of past papers at some stage. Hard finding the time around EX2 and whatnot though.

[silentstriker]

Originally Posted by Spark

How'd you work it out, ss?

I'm a visual person, so I just drew a cube and worked backwards. And found probability of getting to the spot from each of the corners. So, assuming the attached image...

Ant is at 1, you need to get to 8.

From 1, you can go to 2, 3, 5. From either 2, 3, 5, the ant can either crawl back to 1 (1/3 chance), or it can go to a spot that is one closer to '8'. Meaning, from 2, it can go back to either 1 (1/3 chance), 4 (1/3 chance), or 6 (1/3 chance). For our purposes, 4 and 6 are the same since they are the same distance from 8. So 1/3 chance of going 'back', and 2/3 chance of going 'forward'. From that spot, the same formula applies again - if you get to 6, you have a 1/3 chance of going 'forward' (and thus arriving at 8), or 2/3 chance of going 'back' to either 5 or 2. Now you can generalize those distances.



So:

Expected number of steps from 4, 6, 7 (lets call this x) = 1/3 (1) + 2/3 (y+1)

So, if you are at corners 6, 7, or 4, you have a 1/3 chance of making it in one step, or 2/3 chance of making it in 'y+1' steps. y is going to corners 2,3,5, which are two steps away (minimum) from your destination. Why 'y+1'? Because you go backwards to another corner, thereby going '1 away' from your destination, so you've inevitably added one step.

Simplify:
x = 1/3 (1) + 2/3(y+1)
x = 1/3 + 2/3y + 2/3
x = 1 + (2/3)y


Now we have to figure out 'y'. From y (corners 2, 3, 5) you have a 2/3 chance of getting closer as we said, and 1/3 chance of being yet farther.

y = 2/3 (1+x) + 1/3 (1 + z). z being the steps from the starting corner.

Now z is simply whatever steps y is plus 1 since you HAVE to get to y with 100% chance from z (z is starting point).

z = y + 1

So your equations are:

x = 1 + (2/3)y
y = 2/3 (1+x) + 1/3 (1 + z)
z=y+1

At this point, it's plug and chug.

x turns out to be 7, y turns out to be 9, and z turns out to be 10.

Now if you do problems like this, you can sometimes just 'see' that y has to be 9, which makes it a lot faster, but that's not a good enough answer if you need to show work .

[8ankitj]

Nice approaches those. I will post my approach (based on state transition diagrams ) later tonight.

[silentstriker]

Well, if you take the cube, you could also do:



Another way of looking at what I did (sorry for using a different cube):

1) A is starting, G is ending
2) H = F = C since they are all equidistant from G = call this State X
3) D = B = E (for the same reason) = Call this State Y

So what are the chances of going from B, D or E to G (eg from State Y to G)? Well, you have a 2/3 chances of going from Y --> X. And once at X, you have 1/3 chance of going from X--> G.

So you have 2/3 * 1/3 = 2/9 IF YOU TOOK THE SHORTEST ROUTE. The probability of NOT doing that (e.g NOT TAKING the shortest route) is 1 - (2/9) = 7/9 (and you have to use 2+x steps again)

So x = 2(2/9) + (2+x)(7/9)

x= 9

Since it ALWAYS takes 1 step (eg 100% probability) to go from A -->Y, you add 1: 9+1 = 10.

[Spark]

So it was a lot easier than I thought. **** I hate probability

[silentstriker]

Originally Posted by Spark

So it was a lot easier than I thought. **** I hate probability

Haha yea, sometimes the simplest problems are annoying. Like how I didn't think that .9r = 1 for like 20 minutes and I was trying to show it before having a 'doh!' moment.

[smalishah84]

Originally Posted by Spark

So it was a lot easier than I thought. **** I hate probability

This

[smalishah84]

Originally Posted by Turbinator

Expected distance the ant travels is x

Expected distance from three corners next to the ant is y

Expected distance from three corners next to the spot diagonal of the ant is z

If we assume that the ant is placed at z, there is a 1/3 probability that it will travel 1 meter and reach the spot diagonal from it. With the remaining 2/3 probability that it will travel the distance y and then 1 meter.

Therefore, z = (1/3)(1) + (2/3)(1+y) = [1 + (2/3)(y)]

y = (1/3)(1+x) + (2/3)(1+z) = [1 + (1/3)(x) + (2/3)(z)]

And if we assume that the ant is placed at x, it will always be 1 meter + y. Therefore, x = (1+y).

So now we solve for x through simple substitution.

x = 1 + [1 + (1/3)(x) + (2/3)(z)]
x = 3 + z = 4 + 2/3(y)
x = 4 + 2/3(x-1)
x = 10 meters

Don't know how well I explained that, but I think you guys get the idea.

nice........what major are you? math?

and I am guessing that SS is an engineer?

[silentstriker]

Engineer? WTF? Dire.

[smalishah84]

lol....then?

[Spark]

Originally Posted by silentstriker

Haha yea, sometimes the simplest problems are annoying. Like how I didn't think that .9r = 1 for like 20 minutes and I was trying to show it before having a 'doh!' moment.

I just don't like probability. Something I've never got the "knack" of in the way that makes me happy and something where I'm never entirely comfortable with whether I got the question right or not which irks me to no end.

Give me an analysis proof to do any day.

[smalishah84]

Originally Posted by Spark

I just don't like probability. Something I've never got the "knack" of in the way that makes me happy and something where I'm never entirely comfortable with whether I got the question right or not which irks me to no end.

Give me an analysis proof to do any day.

awta.

I think probability is one of those things that you get it or you don't.

I think I am in the latter group.