2011 hsc mx1 marathon (1 Viewer)

[slyhunter]

Find a system that works for you, it's different for everyone.

 

[apollo1]

but i am confused about how to allocate the rest of my time, another hour for Q. 6-9 n then 30 minutes for 10??

yeh that sounds bout right. you want about 30 minutes at least for mistakes and stuff.

 

[COLDBOY]

Find the probability that of 5 cards chosen from a 52-card deck that 2 will be of the same suit.

Would the answer happen to be 0.27 (2 d.p)

 

[deswa1]

Wouldn't the answer be 1 as there are only four different suits so by the fifth card you have to double up on at least one suit

 

[apollo1]

Wouldn't the answer be 1 as there are only four different suits so by the fifth card you have to double up on at least one suit

this is correct.

 

[someth1ng]

Wouldn't the answer be 1 as there are only four different suits so by the fifth card you have to double up on at least one suit

Sorry man, I alredy stated it was 1...xD

 

[nightweaver066]

To keep the ball rolling, here's a new question.

A factory has 7 machines: 4 of model A which are in use, on average, 80% of the time and 3 of model B which are in use, on average, 60% of the time. If the foreman walks into the factory at a randomly selected time, what is the probability:
i) he will find 2 machines of model A and 1 of model B in use?
ii) he finds 2 machines in use?

 

[barbernator]

i) 0.0442 (4dp)
ii) 0.0179 (4dp)

conformation of correct answer please

OK new question....

 

[nightweaver066]

i) 0.0442 (4dp)
ii) 0.0179 (4dp)

conformation of correct answer please

OK new question....

Correct.






New question,
Prove that,

 

[barbernator]

are you sure you have denoted your C's right?

 

[nightweaver066]

are you sure you have denoted your C's right?

Hope that clears things up.

 

[barbernator]

you got me! what's the answer?

 

[apollo1]

you got me! what's the answer?

lol i couldnt get it either. but usually when they have squares of choose that means equating coefficients so i think nightweaver isnt giving us the expansion. the sly bastard.

 

[barbernator]

well if u consider (1+x)^n(1+1/x)^n, that gives you square coefficients for x^n, but i dont know where the (n+1) comes from, because that would hint that there is a differential, but then it should be n not (n+1) gahh

 

[apollo1]

well if u consider (1+x)^n(1+1/x)^n, that gives you square coefficients for x^n, but i dont know where the (n+1) comes from, because that would hint that there is a differential, but then it should be n not (n+1) gahh

ill post an easier binomial in a second

 

[apollo1]

prove the above using the expansion of (1+x)^n-1

 

[barbernator]

ah ok, might take me a little while, im new to this cool fx syntax

 

[apollo1]

ah ok, might take me a little while, im new to this cool fx syntax

lol you can just describe what you did and ill say yay or nay.

 

[muzeikchun852]

prove the above using the expansion of (1+x)^n-1

got everything except the minus 1.

first you expand the binomial. integrate. let x = 7

 

[barbernator]

expand then integrate
[/tex]

let x=0 to evaluate C, C=-1
let x=7 LHS= as u have shown
RHS = ((8)^n)/n -1 =((2^3)^n)/n -1 as required

yeh i gave up on the syntax