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It depends on how you work the question. The repeated arrangments of double/triple letters need to be factored in for probability because the repetition of an identical arrangment increases the chance that it will occur. But, as I said, it depends on how you go about calculating it.

That's the way I did it. I tried a few similar methods also which gave 4/7. So what is the correct answer then?

i) The probability that no accidents occur in a year is equal to P₀ which is equal to e^-2.6 ii) Think of this as a binomial probability question [P₀ + (1 - P₀)]¹⁰. P(zero crashes for at least two years) is equal to 1 - (P(zero crashes one...

If things had been right I should have been. They didn't really put me in a class. I had a couple periods a week with Mr Niven ontop of my normal 3U classes instead.

Using [cosy-cos(y+2q)] / 2sinq= sin(y+q) you get the identity: [cosy-cos(y+2q)] = 2sinqsin(y+q) You assume true for k so: sinq + sin3q + sin5q+...+sin(2k-1)q= [(1-cos2kq) / 2sinq] then you add sin(2k+1)q to both sides... LHS = sinq + sin3q + sin5q+...+sin(2k-1)q + sin(2k+1)q RHS =...

It's clicking into place, thanks for bearing with me :p.

This is my main hang up ^2ⁿC₂ = 2^n-1(2ⁿ - 1) In the first round you have 2^n-1 pairs, then 2^n-2 in the next and so on until you have 1 pair in the last round. If you sum these you get (2ⁿ - 1), i.e. the...

I'm afraid not. I don't think my brain is made to understand your logic. Part of it is that I'm not sure what you're refering to with all thes things. The number of people paired against each other is 2^n, but becomes 2^(n-1) in round two and so on... Saying that the number of matches (2^n - 1)...

i) P(A wins first draw) = 2/5 (2 green marbles in a bag of 5) ii) For B to win A must first draw a red (i.e. not win) then B must draw a red: P(B wins first draw) = (3/5)(3/5) = 9/25 iii) P(A wins in less than four turns) = P(A wins 1st turn) + P(A wins 2nd turn) + P(A wins 3rd turn)...

Just one :p. ^XC_Y is the number of size-y subsets of a size-x set (more or less) so then ^2ⁿC₂ represents the number of unique pairings that can be made from 2ⁿ people. What I still don't understand is why dividing the number of...

Interesting, I'd like to know your reasoning as well. For example, how does ^(2ⁿ)C₂ represent the 'number of pairs'?

Someone's been exposed to too much postmodernism :rolleyes:.

Hey guys, this is a question I just did from a past HSC paper. I'm looking at the last part of the question ( the 2^1-n bit) in particular but I've written up the rest of it as a lead in. I got the answer by extrapolating the pattern, factorising and summing but the way the answer...

Yeah, emphasis on "did'. I had a concept I liked for a major work (I was wanting to write a stage play) but the problem with english ext. 2 is that you don't need immediate results so as soon as you have an assessment task for another subject it gets pushed to the side. Before I knew it I was at...

It's not that you have to do 4 "2 unit" subjects, it's just that you have to do 4 different subjects. English is a subject and mathematics is a subject.

I got caught into that with 4U english and 4U mathematics. It's annoying because you can only drop from the top. 4U maths wasn't going to go so in the end my 4th unit of english got cut :( (i.e. I'm currently doing 11 units).

When you combine all that it seems to make a pretty good case for not going.

Students unite to protect our fair nation!

I'd be intereted in seeing how you would write the anwer without scientific notation :p.

Haha, I hope that stays there for a while :p.