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What software was used to draw the diagram for question 14?

That's perfectly valid, just slightly longer.

For the probability question, since player A starts first, player B's chance of winning is always dependent on player A failing the turn before player B selects a marble. This immediately forces the condition: \frac{w}{w+y} < \Big (\frac{y}{w+y}\Big )^{2} \Longrightarrow \frac{y}{w} - 1 -...

Yes, they are equivalent.

After checking with Wolfram Alpha, I can confirm that my general formula is correct.

You're welcs. Does that now mean that everyone who sat the paper gets a free mark for it or nah?

For question 3, shouldn't n and 2m-1 be coprime?

My formula gives 42C9 exactly, but you originally told me that the answer was 42C7 :confused:

Is my solution not correct or something?: \sum\limits_{k=0}^{N-2n+1}(N-2n+2-k)\cdot\binom{n-2+k}{k}

Hm, I don't think that's quite right, but I could be wrong. Here's my formula for the general case (I worked it out as a probability): \textbf{Conjecture:} \text{ Suppose that } N \text{ cards are each labelled with a distinct number from } 1 \text{ to } N, \text{ inclusive. If } n \text{ cards...

Do you have the answer? I have a formula for the general case but I'm not fully sure that it is correct.

I think your mistake is that you are forcing the first person to only sit at table A, and discounting all of the arrangements where he/she sits at table B. To follow your approach in full completion, you can consider when the first person starts off at table B, and count the corresponding...

Yes, you are correct. I don't quite think it is in the syllabus, I just didn't see a way to answer the question without it :p

\text{Let } A \text{ be the event that the child has the disease, and } B \text{ be the event that the mother has the disease. Then, by the given information, we have} P(A|B) = 0.5, P(B) = 0.5, P(A|{B}^{c}) = 0. \text{ Now,} \begin{align*} P(A) &= P(A\cap{B}) + P(A\cap{B}^{c}) \\ &=...

My method is similar but slightly different. \begin{align*} I &= \int_{0}^{a}f(x)dx \\ &= af(a) - \int_{0}^{f(a)}f^{-1}(x)dx \\ &= 1\cdot\frac{\pi^{2}}{36} - \int_{0}^{\frac{\pi^{2}}{36}}\Big (\frac{1}{1-\sin\sqrt{x}}-1\Big )dx \\ &= \frac{\pi^{2}}{18} - \int_{0}^{\frac{\pi^{2}}{36}}\Big...

I did my HSC in 2012, and at that time there was like a footnote corresponding to my 4U mark that read, "Mathematics (Extension 2) students are not required to sit for the 2 Unit Mathematics examination. If this student had sat for the 2 Unit Mathematics examination it is likely that he/she...

Thanks for your advice

Well I just don't have my own interactive website in order to do that though :L Unless I can approach an existing site like that and they'll agree to include my book on there. Though I plan on self-publishing so I'm not sure how that'll work..

Hm, well is there a way I can provide a PDF copy which bars people from copying it to other devices or uploading it somewhere where people can download? Cause I have no idea how to do that.