Recent content by aa180

  1. aa180

    BoS Maths Trials 2019

    What software was used to draw the diagram for question 14?
  2. aa180

    Maths Extension 2 predictions/thoughts?

    That's perfectly valid, just slightly longer.
  3. aa180

    Maths Extension 2 predictions/thoughts?

    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 -...
  4. aa180

    help on abstract perms and combs

    Yes, they are equivalent.
  5. aa180

    help on abstract perms and combs

    After checking with Wolfram Alpha, I can confirm that my general formula is correct.
  6. aa180

    BoS Maths Trials 2019

    You're welcs. Does that now mean that everyone who sat the paper gets a free mark for it or nah?
  7. aa180

    BoS Maths Trials 2019

    For question 3, shouldn't n and 2m-1 be coprime?
  8. aa180

    BoS Maths Trials 2019

    My formula gives 42C9 exactly, but you originally told me that the answer was 42C7 :confused:
  9. aa180

    BoS Maths Trials 2019

    Is my solution not correct or something?: \sum\limits_{k=0}^{N-2n+1}(N-2n+2-k)\cdot\binom{n-2+k}{k}
  10. aa180

    help on abstract perms and combs

    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...
  11. aa180

    help on abstract perms and combs

    Do you have the answer? I have a formula for the general case but I'm not fully sure that it is correct.
  12. aa180

    Perms and combs help

    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...
  13. aa180

    Conflicting Probability Question

    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
  14. aa180

    Conflicting Probability Question

    \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}) \\ &=...
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