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Re: 2015 permutation X2 marathon AND ..... you KNOW the host's strategy of only ever revealing a goat.

Re: 2015 permutation X2 marathon If the distribution wasn't even in one variable, you would have to integrate the distribution itself. If it was uneven in both variables, you would need a double integral. That is my understanding anyway - correct me if I'm wrong.

Re: 2015 permutation X2 marathon Still haven't had time to think about it properly. But you talk about THE result. There are two different results depending on the relative values of a and b.

Re: 2015 permutation X2 marathon Possibly. I'll think about it when I have more time.

Re: 2015 permutation X2 marathon I haven't thought about this much yet, but my suspicion is that the moment you take logs you are no longer picking numbers with equal probability over the intervals specified. We are not simply solving p^2 >4q - we are finding points in a particular region...

This can be answered with more conventional Ext 1 logic: 10 + 10P2 + 10C2 + 2(10C3) + 10C4 See if you can work out the logic behind that calculation.

You seem to have an issue with rich people. Why is that?

On the other hand, it could be a locus question. z.z* = |z| is the unit circle centred at the origin.

Firstly, please point out where in my comment I advised someone not to go there. Now - if you were talking face to face with a Turkish civilian, and you had been discussing the actions of the Turkish government/military, even if he was agreeing with you on that matter, you would have no...

Re: 2015 permutation X2 marathon Sorry - I thought you were the one who also posted the previous question. Anyway - in your question is there an assumption that men and women have equal ability?

Re: 2015 permutation X2 marathon p² - 4q must be positive. That is a region on the p-q Cartesian plane. The restrictions on p and q define a rectangular box on the Cartesian plane. So you basically just need to find the fraction of the box than contains the above region. The hitch is...

Re: 2015 permutation X2 marathon Is the aim here to block out my questions?

Re: HSC 2015 3U Marathon Alternatively - sub in n=1, 2, 3 and solve the resulting 3 equations simultaneously.

Re: HSC 2015 3U Marathon That video is WRONG. The second expression should be (2n-1)!/(n-1)!

Re: HSC 2015 3U Marathon What is the answer supposed to be? Because I get no solution.

Re: HSC 2015 3U Marathon It's also possible to do it by arranging the boys first, but it requires a bit more thought.

Re: HSC 2015 3U Marathon In how many ways can 5 men, 5 women, 5 girls and 3 boys be arranged in a line if all the children must be together, but no two boys can be adjacent?

Re: HSC 2015 3U Marathon Assuming you mean that the two black balls must be adjacent, you are essentially creating a 5-letter word from the letters W, W, M, M and BB. So the count would be 5!/(2!2!)

Re: 2015 permutation X2 marathon Now my question again: p and q are real numbers. p is randomly chosen on the interval -a ≤ p ≤ a q is randomly chosen on the interval 0 ≤ q ≤ b What is the probability that the equation x² + px + q = 0 has real roots ? At its core, this is...

Here is what Bostes says: Students are ranked by their HSC mark. Where students have equal HSC marks, they are then ranked by the average of their examination marks and their assessment marks, each taken to one decimal place. In the case of extension courses, marks awarded for other courses...