Recent content by Arenholt

For the given scenario, the formula indeed doesn't hold up. However, I had the doubt if there can only be one bucket with exactly n-balls or are multiple buckets each with n-balls allowed? Because if multiple buckets are allowed then we need to take cases to get the total probability.

The probability calculated here is just for one bucket. For all buckets, $P_{total}=P_{m1}+P_{m2}+...P_{M}=M\times P$(since the probabilities for each bucket are identical). You can think about it intuitively as: if you don't focus on a particular bucket, the chances of you finding exactly...

This is exactly what I am trying to wrap my head around. I thought of this but how is $f^{-1}(Y)\setminus f^{-1}(B)$ equal to $f^{-1}(Y\setminus B)$.

Hi everyone, didn't know where to post question on sigma algebra so here it is:- What I've tried till now: Let $$C\in G$$ 1) For $$C=X, f^{-1}(B)=X$$ which will be true for $$B=Y$$ (by definition) 2) For closure under complementation, to show $$C^{c}\in G. So, C^{c}=X\setminus C=X\setminus...