Probability & Statistics Ch.2 Section 2 Question 3

· Mohammad-Ali Bandzar

Solutions for “Probability and Statistics: The Science of Uncertainty” (Second Edition). These are solutions I have come up with; I offer no guarantee of accuracy.

Question

Suppose we roll two fair six-sided dice, and let Y be the sum of the two numbers showing.

a) Compute P(Y=y)P(Y = y) for every real number y. b) Write a formula for P(YB)P(Y \in B), for any subset B of the real numbers.

Solution

(a) Compute P(Y=y)P(Y = y) for every real number y

When rolling 2 dice, there are 36 equally likely outcomes. Note that diagonals in the outcome grid (starting from any number in the leftmost column and going right one then up one) will all have the same sum. For example, (5,1) has the same sum as (4,2), (3,3), (2,4), (1,5).

  • P(Y=2)=136P(Y=2)=\frac{1}{36}
  • P(Y=3)=236P(Y=3)=\frac{2}{36}
  • P(Y=4)=336P(Y=4)=\frac{3}{36}
  • P(Y=5)=436P(Y=5)=\frac{4}{36}
  • P(Y=6)=536P(Y=6)=\frac{5}{36}
  • P(Y=7)=636P(Y=7)=\frac{6}{36}
  • P(Y=8)=536P(Y=8)=\frac{5}{36}
  • P(Y=9)=436P(Y=9)=\frac{4}{36}
  • P(Y=10)=336P(Y=10)=\frac{3}{36}
  • P(Y=11)=236P(Y=11)=\frac{2}{36}
  • P(Y=12)=136P(Y=12)=\frac{1}{36}

The probability of any other sum is zero: P(Y=y)=0P(Y=y)=0 for y{2,3,4,5,6,7,8,9,10,11,12}y\notin \{2,3,4,5,6,7,8,9,10,11,12\}

(b) Write a formula for P(YB)P(Y \in B), for any subset B of the real numbers

We sum up all probabilities multiplied by the indicator function to check if that outcome is in B:

P(YB)=136IB(2)+236IB(3)+336IB(4)+436IB(5)+536IB(6)+636IB(7)+536IB(8)+436IB(9)+336IB(10)+236IB(11)+136IB(12)P(Y\in 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:

P(YB)=136(IB(2)+IB(12))+236(IB(3)+IB(11))+336(IB(4)+IB(10))+436(IB(5)+IB(9))+536(IB(6)+IB(8))+636IB(7)P(Y\in 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)