Recent content by Eero

I think, things are very clear,actually. Lets denote the number of candies in a bag by N, the number of red candies by m and the probability that the first red will appear at k-th draw by P(k). Because we have here a sampling without replacement, the following holds...

Nice one bpet! I would not come into this as soon. Do you have a clue how to determine E(Xi*Xj) now, to calculate Var(X)?

After a messy, lengthy calculations (not the indicator method) an unexpectedly simple formula for the E(x) occurred: E(x)=p*(p+n*q) ; q=1-p I was shocked! Indeed, there must be a simple probabilistic approach that replaces involved calculations and hard analysis. Maybe really the indicator...

Seems interesting, but hard problem. I suspect that the indicator method does not work in case of p \neq 1/2. It is easier to consider a fair coin with p=1/2 at the beginning.

I am on the same frequency as you !

Hi! What is your opinion about this article: [PLAIN]"www.bme.hu/ptee2000/papers/fetea.pdf"[/URL]

Ok, i will try to analyze your input (drawing 6 out of 49 numbers (10000 times)) with my expectation approach. That means we set n=49 and r=6*10000=60000 in the expectation formula. Below is a piece of the formula outputs: j....E 1215 0.5456 1216 0.5495 1217 0.5530 1218...

In given case n=49 and r=188 the most frequent number will occur 8 times(in average) Do you want to know the probability of this happening?

Considering expectations may shed some light on this problem solution The probability that a specified number will occur exactly j times in r drawings follows the binomial distribution: p(j,r)=b(j;r,1/n) (j is number of successes, r is number of drawings and 1/n is probability for...

\sum_{j=0}^{r-1}(-1)^{j}\\( ^{r}_{j} \\)(r-j)^r = r! This expression is easy to prove by considering the r-th derivative of (1-e^x)^r at x=0