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 Sep24 awarded Autobiographer Jul10 awarded Supporter Oct24 revised Self-study resources for basic probability? Added 'reference-request' tag Oct14 awarded Teacher Oct12 comment A Simple probability question You're surely welcome. Thank you, too. Oct12 comment A Simple probability question Ah! Of course! I forgot about that. If you feel inclined, you may mark my answer as accepted, but you don't have to. Oct12 awarded Commentator Oct12 comment A Simple probability question You're welcome! Be sure to upvote the question since you found it helpful. Thanks! Oct12 comment A Simple probability question @Kevinsb, that would work; that works out to about 31311329, so you could say that there is about a 1 in 31,311,329 chance of getting 15 questions right. Oct11 revised annuities and interest Added 'finance' tag Oct11 suggested approved edit on annuities and interest Oct11 comment annuities and interest Please do not use "homework" as the only tag for a question. Oct11 revised What is a blow-up? Added 'terminology' tag Oct11 comment Differentiation using definition of derivatives By providing the hint, are you implying you know the answer (and want to see if others can figure it out), or is it (rather) that you're working on homework or doing self-learning? Oct11 suggested approved edit on What is a blow-up? Oct11 comment Relation between a random variable and its conditional expectation @lezebulon Can you clarify what you are representing with the letter $G$? Is $G$ an event (such as "$G$ is the event that $X > 0$")? Or, if $G$ is a random variable, by "for any $G$ possible", do you mean "for any selection of random variable $G$", or do you mean "allowing $G$ to be any of the values in its support" (thus distinguishing $E(X|G)$ from $E(X|G=g)$)? Oct11 comment Relation between a random variable and its conditional expectation The conditional probability part doesn't always work. You can only split $P(X = 0 \textrm{ and } Y \neq 0)$ into $P(Y \neq 0 | X = 0) P(X = 0)$ if $P(X = 0) \neq 0$, because, by definition, $P(Y \neq 0 | X = 0) \equiv \frac{P(X = 0 \textrm{ and } Y = 0)}{P(X = 0)}$, so $P(X = 0) = 0$ results in division by zero, making $P(Y \neq 0 | X = 0)$ undefined. Oct11 revised Explanation of term “closed under” in the definition of a sigma algebra? Made title more specific, added 'terminology' tag Oct11 suggested approved edit on Explanation of term “closed under” in the definition of a sigma algebra? Oct11 revised sigma algebra generated by random variable Include the close-brackets in the subscript on the last two MathJaX equations of the first paragraph, capitalize Borel