Chapter 2 Section 2 Question 3
Question
Suppose we roll two fair six-sided dice, and let Y be the sum of the two numbers showing.
1. Compute $P(Y = y)$ for every real number y.
2. Write a formula for$P(Y \in\textbf{B} )$, for any subset $\textbf{B}$ of the real numbers.
Author: Mohammad-Ali Bandzar| Date:Oct 14 2020
(a) Compute $P(Y = y)$ for every real number y.
All possible outcomes from rolling 2 dice: please note that diagonals starting from any number in the left most column and going right one then up one will all have the same sum. So if we look at (5,1) it will have the same sum as (4,2),(3,3),(2,4),(1,5).
The probability of a sum of 2 is $P(Y=2)=\frac{1}{36}$
The probability of a sum of 3 is $P(Y=3)=\frac{2}{36}$
The probability of a sum of 4 is $P(Y=4)=\frac{3}{36}$
The probability of a sum of 5 is $P(Y=5)=\frac{4}{36}$
The probability of a sum of 6 is $P(Y=6)=\frac{5}{36}$
The probability of a sum of 7 is $P(Y=7)=\frac{6}{36}$
The probability of a sum of 8 is $P(Y=8)=\frac{5}{36}$
The probability of a sum of 9 is $P(Y=9)=\frac{4}{36}$
The probability of a sum of 10 is $P(Y=10)=\frac{3}{36}$
The probability of a sum of 11 is $P(Y=11)=\frac{2}{36}$
The probability of a sum of 12 is $P(Y=12)=\frac{1}{36}$
The probability of any other sum is zero $P(Y=y)=0$ for $x\notin \{2,3,4,5,6,7,8,9,10,11,12\}$
(b) Write a formula for$P(Y \in\textbf{B} )$, for any subset $\textbf{B}$ of the real numbers.
This question just like the last one just wants us to sum up all probabilities multiplied by the indicator function to check if that outcome is in $\textbf{B}$ $$P(Y\in \textbf{B})=\frac{1}{36}\textit{I}_B(2)+\frac{2}{36}\textit{I}_B(3)+\frac{3}{36}\textit{I}_B(4)+\frac{4}{36}\textit{I}_B(5)+\frac{5}{36}\textit{I}_B(6)+\frac{6}{36}\textit{I}_B(7)+\frac{5}{36}\textit{I}_B(8)+\frac{4}{36}\textit{I}_B(9)+\frac{3}{36}\textit{I}_B(10)+\frac{2}{36}\textit{I}_B(11)+\frac{1}{36}\textit{I}_B(12)$$ We can factor to simplify this to the following: $$P(Y\in \textbf{B})=\frac{1}{36}(\textit{I}_B(2)+\textit{I}_B(12))+\frac{2}{36}(\textit{I}_B(3)+\textit{I}_B(11))+\frac{3}{36}(\textit{I}_B(4)+\textit{I}_B(10))+\frac{4}{36}(\textit{I}_B(5)+\textit{I}_B(9))+\frac{5}{36}(\textit{I}_B(6)+\textit{I}_B(8))+\frac{6}{36}\textit{I}_B(7)$$