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 Jun 6 comment Is there a measure on $\mathbb{N}$? (What is the chance that a random integer from $\mathbb{N}$ is even?) @EricTowers, the counting measure is not a probability measure. Jul 7 comment solving a problem in probability Nitpicking: weight cannot be negative, hence the Gaussian assumption is unrealistic. Jun 23 comment Mutually exclusive events Then you should assume $P(E) = P(F) = 0.5$ Jun 22 comment Mutually exclusive events The event that $E$ does not occur first is (in my notaton) $A^c$. You cannot simply change the meaning of $E$ (which is an event in experiment $\mathcal E_1$). Perhaps the solution given by @DilipSarwate is close to what you are thinking: Think of the experiment in which either $E$ or $F$ occur for the first time. What is the probability that the event that occurred was $E$. Jun 22 comment What is the best choice in a win/lose game and how to calculate it Agreed. To explain buying of lottery, you need to add additional utility for participating in the game---the thrill of the possibility of winning. Jun 22 comment Mutually exclusive events Does my updated answer clarify this point? Jun 22 comment What is the best choice in a win/lose game and how to calculate it (Tongue in cheek comment): Math doesn't help because math assumes linear utilities. Assume concave utilities, and math will explain risk adverse behavior and buying of lotteries. Jun 22 comment Mutually exclusive events If $P(E) = P(F) = 1$, then $E$ and $F$ cannot be mutually exclusive because $E \cup F \subset \Omega$, thus $P(E \cup F) = P(E) + P(F) \le P(\Omega) = 1$. Jun 22 comment Mutually exclusive events $E^c = \{3,4,5,6\} \not\equiv \{3,4\} = F$ May 17 comment Lower bound on a minimum of maximum of a sequence of standard normal random variables In the RHS of all equations, $j$ should be replaced by $p$. Apr 26 comment Show that if $\{x_{n}\}$ and $\{y_{n}\}$ are Cauchy sequences in X then $d(x_{n},y_{n})$ converges in $\mathbb{R}$. In the reverse triangle inequality, the RHS should be $d(x,z)$. Oct 23 comment Roots of $f_n(x) = 1 + (1-x)^2 - (x+3)(1-x)^{n+1}$ in the interval $[0,1]$ Yes, it is easy to see that the roots form a decreasing sequence. $f_{n+1}(x) - f_n(x) > 0$. If $a_n$ is a root of $f_n$ then, $f_{n+1}(a_n) > 0$. Thus, there is a root of $f_{n+1}$ between $0$ (where $f_{n+1}$ is negative) and $a_n$ (where $f_{n+1}$ is positive). Oct 23 comment Roots of $f_n(x) = 1 + (1-x)^2 - (x+3)(1-x)^{n+1}$ in the interval $[0,1]$ @SL2:The derivative is not always positive. The derivative is $[ (n+2)x + 3n + 2 ](1-x)^n - 2(1-x)$ and its sign is determined by whether $a_n(x) = [(n+2)x + 3n + 2 ](1-x)^{n-1}$ is greater than or less than $2$. As $n \to \infty$, $a_n(x) \to 0$, for $x \in [0,1]$. Thus, the derivative, $a_n(x) - 2$, is negative for large $n$. Apr 26 comment Given P(A|B) and P(A|C), how to get or strategically approach P(A|(B & C))? Try expanding P(A,B,C) using the law of total probability in different ways. Aug 8 comment A probability game Sorry. I didn't realize the difference.