I don't think anyone is posting questions with the specific aim of blocking out your questions.
Sorry, that's not my intention. I figured I'd forget to post this one if I didn't now. I did attempt your question, but I have no idea where to proceed after finding the discriminant. Your hint of using integration makes me think it is a portion of an area bounded by a curve and maybe the x-axis representing the values for which the quadratic has real roots, but I'm not sure if that's correct. Even if it is, I don't know how to proceed.
Haven't tried the Q. much, but to find the probability that one random variable is less than another, wouldn't we essentially need to use double integration (at least, this is one way of doing it)?
Are you sure the answers 6 men? If theres 6 men that means n=3. Which also means there is 9 people. If each player plays only 1 person there would be 1 person that doesnt play...?
I think when they said they play with everyone once, it's similar to a group of people all giving handshakes to everyone only once. So no-one is left out.
I now think it means everyone plays everyone once.
p² - 4q must be positive.
Sorry - I thought you were the one who also posted the previous question.
But p^2 - 4q positive is the same as the probability that 2logp > log4 + logq? Log is concave, so that won't change the probability.
That log thing would give half the probability, as it is the case where p > 0.
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.
Possibly. I'll think about it when I have more time.$ and $Q\sim \mathcal{U}(0,b)$ (uniform distribution over those intervals), then isn't $\mathbb{P}\left(P^2 > 4Q \right) = \mathbb{P}\left(\log P^2 > \log (4Q) \right)$, since one of these events occurs if and only if the other does? Even though $\log P^2$ and $\log (4Q)$ are not uniformly distributed, it shouldn't matter should it, since their pdf's will change accordingly?$)
Still haven't had time to think about it properly. But you talk about THE result. There are two different results depending on the relative values of a and b. = 1$ and $\mathbb{P}\left(B|A\right) = 1$, then $\mathbb{P}(A) =\mathbb{P}(B)$, since $\mathbb{P}(A) \mathbb{P}\left(B|A\right)= \mathbb{P}(B) \mathbb{P}\left(A|B\right) (=\mathbb{P}\left(A \cap B\right))\Rightarrow \mathbb{P}(A) \times 1 =\mathbb{P}(B)\times 1$, and the result follows.$)
No I wasn't referring to that question, by 'the result' I meant P(A) = P(B). (And these were unrelated to the "a" and "b" of the question.)
