Recent content by GreenLeaf

@DrClaude - Thank you for such an elaborate explanation and to have done it so promptly as well! This forum is amazing!

@DrClaude - Ok, that makes sense. So you're saying that, in the solution the statement with b > 0 is only true when r >= 1 is guaranteed? Also - would you happen to know the answer to the 3 questions I posted as quoted below:

@DrClaude - I don't see how you got to the conclusion. How did you get there?

@DrClaude - Yes I got 1 for the LHS, but the RHS became quite complicated. For the RHS, I got \frac{r^2+br-r-b}{r^2+br-r} So what do you do with that?

@DrClaude - Yea I tried multiplying both sides by \frac{r+b}{r} , and got 0 > -b, which is false. So are you saying that the solution in the book is actually incorrect?

Hm. I see your point. The thing is - this is part of a solution to a probability problem. Here's the problem: A drawer contains red socks and black socks. When two socks are drawn at random, the probability that both are red is 1/2. (a) How small can the number of socks in the drawer be? (b)...

Hi all, I've been trying to figure out why this statement is true: \frac{r}{r+b} > \frac{r-1}{r+b-1} \quad \text{for all } b>0 I can't seem to reason it out. I've plugged in a few values greater than 0 and yes, it works out. But I don't understand how I can look at this and find it true...