# Tag Info

You can make a table. The names of the rows are the outcomes of the first dice (d1). The names of the columns are the outcomes of the second dice. In total we have 36 possible outcomes. And the cells marked with $\color{red}x$ are the outcomes where the second dice has a greater outcome than the first dice. $\begin{array}[ht]{|p{2cm}|||p{0.5cm}|p{0.5cm}|p{... 3 The probability of a tied result on a fair die:$\mathsf P(A=B) ~=~ \tfrac 1 6$By symmetry:$\mathsf P(A>B)~=~\mathsf P(B>A)$. By total probability:$~\color{blue}{\mathsf P(B>A)}~\color{blue}{=~ \tfrac 12(1-\mathsf P(A=B)) ~\\=~ \dfrac 5{12}}$Note: if you wanted a tie or greater:$~\mathsf P(B\geq A) ~$$=~ \mathsf P(A=B)+\mathsf P(B>A)... 3 The straight forward, but somewhat tedious solution is to simply list the outcomes: First throw 1 then throw 1, no score First throw 1 then throw 2, score First throw 1 then throw 3, score etc. Then it's just a matter of counting scores vs no scores (every outcome is equally probable). I get 15 outcomes that scores and 21 that doesn't (there's 36 ... 2 It's a general fact that if \{A_n\} is a sequence of measurable sets, then \bigcup_{n=1}^{\infty}A_n is also measurable, with$$ \mu\Big(\bigcup_{n=1}^{\infty}A_n\Big)\leq \sum_{n=1}^{\infty}\mu(A_n)$$This property is sometimes called countable subadditivity. One way to prove it is as follows: let B_1=A_1, and for n>1 let B_n=A_n\setminus (\... 1 I assume m refers to probability$$A \subset B \to P(A) \le P(B)$$In our case we have$$P(B) \le P(\bigcup_j A_j)$$We also have$$P(\bigcup_j A_j) \le \sum_j P(A_j) by countable subadditivity.