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Re: 2015 permutation X2 marathon Ahhhhh - no. For starters, no restrictions gives you 8! / (2!)^4 All pairs together is 4!. And ... you do know that "all pairs together" and "no pairs together" are not opposites?

And ... if you'll allow me to be even more picky (for the sake of OP, not you): In the 2nd we need k not equal to 1 (if we really want a circle). And for the last one, you probably want an absolute value around the Im(z), unless you specify Im(z_1) > 0 for an ellipse, or unless you want...

Re: 2015 permutation X2 marathon I think I am going to have to give up on this one. I now understand what you just explained, but it is not helping me to see the resulting calculation.

Re: MX2 2015 Integration Marathon She doesn't need to look it up - I guarantee she knows a lot more maths than you (or me). This is the integration marathon - she is setting a question for people to attempt, not asking how to do it. And your answer was the exact type of response she asked...

For your 3rd example, I don't think k>0 is sufficiently restrictive in order to get an ellipse.

Yes, any condition that reduces the sample space is conditional probability.

Yes, it's a possibility, but it hasn't been seen yet.

It is not specifically in the syllabus. But neither the HSC question nor the one here needs Bayes theorem to work them out. This one involves just removing one black (the one in his hand), so 4 of the remaining 9 jellybeans are black. All conditional probability involves a reduction in the size...

Re: HSC 2015 3U Marathon The answer in the first calculation is being used in the second. 3/8 is the probability of a specific family having 2 boys and 2 girls, so 5/8 is the probability that they don't.

Nothing can happen if the problem is not fixed for 10 years. The marks stand. Even at the time, we didn't know there was a problem until well after the results came out, because we were unaware of what their answer was until solutions came out. I always wonder if there were markers that...

It is the same issue as in the 1997 2 unit HSC Q9a(iv). The correct answer is 1/3. At the time the answer that was marked correct was 1/5, and this incorrect answer still appears in many published solutions. It took MANSW 10 years to finally issue the correct solution to this question.

The answer is NOT 2/7, it is 4/9. The answer of 2/7 is the probability that both are black given that one is black. This question is asking for the probability that both are black given that a SPECIFIC one is black. Bayes Theorem would confirm this answer.

Given that we were discussing differentiating, it is extremely unlikely we would want to differentiate an expression involving variable eccentricity with respect to the eccentricity. We would have to be asked to examine the rate at which some property of the ellipse is changing as e changes.

Example?

I realise you would never think that. My comment was directed at HSC students with less understanding than you. It is amazing how many people believe d/dπ (π) = 1 is a valid statement.

Just making sure that everyone realises that d/dπ is meaningless. You can never say that d/dπ (π) is 1.

Re: HSC 2015 3U Marathon I know you understand what you have said, but just to clarify the wording for anyone reading it: The first calculation is the probability of one particular family having two boys, not the probability of exactly one of the four families having two boys.

You are going to have to write this more clearly. Is there one modulus or two? And if it is a locus, it must be an equation. Where is the equal sign?

Re: 2015 permutation X2 marathon I'm trying to digest this, but without much success. What I DO understand is: * the colouring of the beads * the alternating colours * "adjacency <=> committee membership" What I don't understand: * why are you considering bracelets and not just a ROW of...

So you were originally aiming for 200 - that means you've got 6 and a half weeks to fit in the extra 150.