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 Oct11 comment By convention $P[X=x] = 0$ for all x. How would you explain pdf $f(x) =3x^2$ (where x is between 0 and 1) when x =0.9 ok, when $x =0.9$ then $f(0.9)= 3(0.9)^2 which \neq 0$. That's what I'm trying to ask Oct11 asked By convention $P[X=x] = 0$ for all x. How would you explain pdf $f(x) =3x^2$ (where x is between 0 and 1) when x =0.9 Oct10 revised How can gender and class classification be dependent? edited title Oct10 comment How can gender and class classification be dependent? Ross, thank you for your answer! I upvoted everyone who contributed, honestly I still don't quite get it. but it's pretty late tonight. I'll think more on each answer when i wake up tomorrow morning. Oct10 comment How can gender and class classification be dependent? oh no =( this is so convoluted. I started a post before on how I can identify a probability problem as a conditional probability problem(math.stackexchange.com/questions/205103/…). it seems like you're suggesting they're the same. this is still confusing =( Oct10 comment How can gender and class classification be dependent? why would you think of this in terms of conditional probability? I'm thinking it's simple union and interects Oct9 revised How can gender and class classification be dependent? deleted 7 characters in body Oct9 comment How can gender and class classification be dependent? If, however, the first die landed on 6, we would be unhappy because we would no longer have a chance of getting a total of 6. In other words, our chance of getting a total of 6 depends on the outcome of the first die; thus, E1 and F cannot be independent. So is there an explanation like this for class and gender? Oct9 revised How can gender and class classification be dependent? deleted 46 characters in body Oct9 revised How can gender and class classification be dependent? deleted 13 characters in body Oct9 comment How can gender and class classification be dependent? @QiaochuYuan Suppose that we toss 2 fair dice. Let E1 denote the event that the sum of the dice is 6 and F denote the event that the first die equals 4. Hence, E1 and F are not independent. Realistically, the reason for this is clear because if we are interested in the possibility of throwing a 6 (with 2 dice), we shall be quite happy if the first die lands on 4 (or, indeed, on any of the numbers 1, 2, 3, 4, and 5), for then we shall still have a possibility of getting a total of 6. Oct9 revised How can gender and class classification be dependent? deleted 4 characters in body Oct9 asked How can gender and class classification be dependent? Oct9 accepted Why does negative binomial random variable uses ${n-1 \choose r-1}$ instead of ${n \choose r}$ as coefficient? Oct7 accepted Why can we use poisson distribution in this case? the n is very small!! Oct7 comment Probability given a mean and standard deviation of a random variable @DilipSarwate had the distribution been normal, i'd have been right? Oct7 comment Probability given a mean and standard deviation of a random variable @DilipSarwate sorry, i take that back. I think i had assumed the distribution to be normal... Oct7 answered Probability given a mean and standard deviation of a random variable Oct7 comment Why can we use poisson distribution in this case? the n is very small!! are you sure $$$\frac{n(n-1)\ldots(n-k+1)}{n^n} \approx 1$$$? because (by looking at how it's derived) the denominator is $k!$ Oct6 asked Why can we use poisson distribution in this case? the n is very small!!