Bazza's 3/4 Question Thread

[b^3]

nah i know that

but my answer (and cas when simplified) had (1+x) ^ 3/2
instead of just (1+x)

down the bottom

Ok I'm confused about what the problem is here now?
Are you trying to say that CAS gives a different answer to Soyme's?
Soyme's answer is a different form from what the CAS gives, but its the same thing


Does that clear it up?

EDIT: beaten, nvm

[Deleted User]

OH
Wow
jeez i'm dumb
fml
so i would get full marks for that question
but in the next one i sort of skirted over the fact the derivative i got wasn't what i expected (worked backwards essentially), so may lose 1 there

Is that Christian Eriksen?

[WhoTookMyUsername]

Viktor Fischer lol

[WhoTookMyUsername]

For anyone who was wondering, my answer was the same as somyes, i slightly misremembered ( sorry! ) so i had the right answer, but i thought it was wrong, so i didn't do the next question correctly ( worked backwards ) and ( rightfully ) lost a mark lol


Also
Struggling with this probabiliity question, can someone please explain their reasoning?

A basketball team has a 30% chance of winning a match, a 50% chance of drawing a match and a 20% chance of losing a match. The team plays a 4 - match tournament

Determine the probability that the team won one, drew one and lost two

[nerdgasm]

My approach to this question was to work out the number of different ways that this could occur, and multiply by the probability that each way could occur (so I guess it's more of a 'competition maths' based approach, though I think you learn to use the binomial theorem to calculate various probabilities, which is based on the same approach).

So, how many different ways could you win once, draw once, and lose twice in four games? The key here is to realise the losses are indistinguishable, so L(1)L(2)WD = L(2)L(1)WD. It can be seen that there are 12 different ways this can occur.

(E.g. there are 4 places to place the W, then 3 places to place the D, and then you're forced into using the two L's, so there are 4*3 = 12 ways to do it. Alternatively, try writing down all the combinations by hand, maybe starting by placing the two L's, since that's the most likely spot where people go wrong).

Either way, there are 12 equivalent ways to obtain one win, one draw and two losses, and each has a probability of (0.5)(0.3)(0.2)^2, therefore your final probability is 12(0.5)(0.3)(0.2)^2.

[TrueTears]

Essentially the question breaks into: how many ways can you permute WDLL. This is a direct application of the Mississippi "formula", which generalising it has many uses:

[WhoTookMyUsername]

thanks for the help
the solutions went
4C2 x 2C1 x 1C1 x (3/10)(1/2)(1/5)^2

OR
4C1 x 3C1 x 2C2 x (3/10)(1/2)(1/5)^2

which i think is essentially what amalgam did,
but is there a "shorter" common sense approach?
I can't see what the answers were doing exactly, can anyone please explain?

[Somye]

4C2 is the number of ways you will lose 2 out of 4
2C1 is the number of ways you will draw 1 out of the remaining 2
1C1 is the number of ways you can win 1 game
Probabilities are self explanatory, all they've done is multiplied 3 separate equations together as they must all occur for the desired outcome to occur 

[WhoTookMyUsername]

thanks

for http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2006mmcas2-w.pdf

I'm struggling with ER 1c) and 2b)

They are not hard questions (although somehow i managed to completely screw up this exam, mistakes everywhere in ER) but in the corresponding report http://www.vcaa.vic.edu.au/Documents/exams/mathematics/mathsmethodscas2assessrptnov06.pdf

i don't understand the method i didn't use

1c) (m = 2pi + 2(sqrt3 - pi/3))
I did the second method (looking at intercepts then looking at difference between them) but i'm not sure why the above method works, i sort of understand their logic, but the points on the graph of the two tangents aren't 2pi apart (?) and i'm not sure where the 2 in 2(sqrt3 - pi/3) has come from.



2b) writing down long term probability as 0.4 / (0.4+0.7)

I did achieved the (exact) answer in about 10 lines of working, but i have no idea where the logic behind "directly" (VCAA) plugging these numbers comes from



Can someone please explain the logical reasoning and thought process behind these questions? THanks

[Jenny_2108]

thanks

for http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2006mmcas2-w.pdf

I'm struggling with ER 1c) and 2b)

They are not hard questions (although somehow i managed to completely screw up this exam, mistakes everywhere in ER) but in the corresponding report http://www.vcaa.vic.edu.au/Documents/exams/mathematics/mathsmethodscas2assessrptnov06.pdf

i don't understand the method i didn't use

1c) (m = 2pi + 2(sqrt3 - pi/3))
I did the second method (looking at intercepts then looking at difference between them) but i'm not sure why the above method works, i sort of understand their logic, but the points on the graph of the two tangents aren't 2pi apart (?) and i'm not sure where the 2 in 2(sqrt3 - pi/3) has come from.



2b) writing down long term probability as 0.4 / (0.4+0.7)

I did achieved the (exact) answer in about 10 lines of working, but i have no idea where the logic behind "directly" (VCAA) plugging these numbers comes from



Can someone please explain the logical reasoning and thought process behind these questions? THanks

1c) you see from the graph, the distance of x-intercepts between 2 gradients are: 2pi+ sqr3-pi/3 + sq3-pi/3

2b) this is formula of long-term Markov chain. If you use essential textbook, see 16C

[pi]

Worth committing the long term formulas for Markov chains to memory (and/or write on bound ref). Handy stuff.

[Jenny_2108]

Worth committing the long term formulas for Markov chains to memory (and/or write on bound ref). Handy stuff.

I just feel how strangely MHS doesnt teach this formula :O Its quite important imo

[pi]

Worth committing the long term formulas for Markov chains to memory (and/or write on bound ref). Handy stuff.

I just feel how strangely MHS doesnt teach this formula :O Its quite important imo

We got taught that formula in class.

Some teachers may not know of it I suppose haha

[Jenny_2108]

We got taught that formula in class.

Some teachers may not know of it I suppose haha

Well, if the teachers dont teach, you can read the textbook
I assume MHS teachers are better than other schools 

[WhoTookMyUsername]

Thanks for the help, though for 1, that's the way i did it ( vcaa method 2) but i don't see the logic behind method 1 D:



Like where does that extra 2 come from? Adding 2pi takes to the next period, not the one indicated on the graph!?