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Affinity : I am not saying this as balm, because I was about to say it before I posted this sol. I have noticed your superior (to me, that is ) adeptness in handling difficult series in previous threads together with this one.

___________________________________________________ Throw a set of N biased coins, with the probability(head) for nth coin being 1/(2n+1) (for n=1,2,...N). What is the probability of an odd number of heads. Note: no longer a binomial distribution...

Hi ND. Its the old class back. Lets see, we've got the actuarials, the mathematicians, the Year 12 wannabes .:D ,but we don't have the medicos yet. Archman and Affinity have the right answer. But it should only be a three line affair -though admittedly, I can't count up to 3 :) .

Too easy for archman. Raises the bar a touch. Throw a set of N biased coins, with the probability(head) for nth coin being 1/(2n+1) (for n=1,2,...N). What is the probability of an odd number of heads. Note: no longer a binomial distribution.

An interesting binomial probability question with a twist. A die is thrown n times. What is the probability of getting an odd number of sixes. Here's the twist : use (p-q)^n rather than the usual binomial distribution of (p+q)^n.

_________________________________________________ then w^(j1*k2pi/n)=w^(j2*k2pi/n) _________________________________________________ oops. Just realized I've been mixing up my exponents and cis. Please change statement to cis(j1*k2pi/n)=cis(j2*k2pi/n) I have edited the original...

____________________________________________________ Grey Council: I dunno why, but Archman is doing the HSC this year... wtf, how do we even stand a chance against him. ____________________________________________________ Don't worry there's at least 6 mths to go, to build up your...

Grey Council : Its a compliment. Appreciated your compliments as well.

Thanks McLake. But I myself felt a bit uncomfortable in the top of the IMPORTANT section. I'll continue to help anyway. Grey Council : Yes, I am sure you'll do well (not only in maths but in life as well)- what with a personality like yours!

Archman : that's correct. Could someone explain why?

aw c'mon give it a break will you. Lets help the students.

Here's a retake of an otherwise useful question. If w, w^2, ...,w^n are the n complex roots of 1, with w being of smallest positive argument, what is the condition that need to be satisfied for W=w^k, so that W, W^2, W^3,...,W^n are also the n roots of unity; and why?

Thanks for the break, Lazarus.:)

don't worry, nothing much has been flogged. You are right, the time has come for the mods to pull the plug off -really doesn't worry me. Thanks, mods, for your patience. :)

____________________________________________________ Sammeh: heard a lot about me? ____________________________________________________ Don't worry sammeh, its just American politese when an introduction is made.:)

a fellow mudgeeite! Haven't met but heard a lot about you. Hope you're fully recovered.

c) (nCo)^2 + (nC1)^2 + (nC2)^2+...+(nCn)^2 = 2n Cn ____________________________________________________ ryan.cck: no idea wot combinatoral argument means o.O .. we havent done binomial at skool either so yeh.. ____________________________________________________ c) could be done by...

The algebra is right anyway, by all means practise on the rest. But a combinatorial argument uses the definition of nCr as the number of ways of choosing r objects from n distinct objects or "n choose r" for short.

Some of the following have been discussed already. It might be useful to bring them all together to show that there usually is physical meaning behind the coefficient. Use a combinatorial argument to show : a) nCr = nC n-r b) nCr = n-1 C r-1 + n-1 Cr c) (nCo)^2 + (nC1)^2 +...

This is a useful thread. Should be bumped up again closer to exam time for revision.