Chapter 2 Section 3 Question 5
Question
Consider rolling two fair six-sided dice. Let W be the product of the numbers showing. What is the probability function of W?
Answer
Author: Mohammad-Ali Bandzar| Date:Oct 15 2020
valid outcomes for w=1
1,1 $$p\small{W}(1) = \frac{1}{36}$$ valid outcomes for w=2
1,2
2,1 $$p\small{W}(2) = \frac{2}{36}$$ valid outcomes for w=3
1,3
3,1 $$p\small{W}(3) = \frac{2}{36}$$ valid outcomes for w=4
1,4
4,1
2,2 $$p\small{W}(4) = \frac{3}{36}$$ valid outcomes for w=5
1,5
5,1 $$p\small{W}(5) = \frac{2}{36}$$ valid outcomes for w=6
1,6
6,1
2,3
3,2 $$p\small{W}(6) = \frac{4}{36}$$ valid outcomes for w=8
2,4
4,2 $$p\small{W}(8) = \frac{2}{36}$$ valid outcomes for w=9
3,3 $$p\small{W}(9) = \frac{1}{36}$$ valid outcomes for w=10
2,5
5,2 $$p\small{W}(10) = \frac{2}{36}$$ valid outcomes for w=12
2,6
6,2
3,4
4,3 $$p\small{W}(12) = \frac{4}{36}$$ valid outcomes for w=15
3,5
5,3 $$p\small{W}(15) = \frac{2}{36}$$ valid outcomes for w=16
4,4 $$p\small{W}(16) = \frac{1}{36}$$ valid outcomes for w=18
3,6
6,3 $$p\small{W}(18) = \frac{2}{36}$$ valid outcomes for w=20
4,5
5,4 $$p\small{W}(20) = \frac{2}{36}$$ valid outcomes for w=24
4,6
6,4 $$p\small{W}(24) = \frac{2}{36}$$ valid outcomes for w=25
5,5 $$p\small{W}(25) = \frac{1}{36}$$ valid outcomes for w=30
5,6
6,5 $$p\small{W}(30) = \frac{2}{36}$$ valid outcomes for w=36
6,6 $$p\small{W}(36) = \frac{1}{36}$$ Otherwise:
$p\scriptsize{W}\normalsize{(w)=0}$ for all $w\notin\{1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,30,36\}$
Note that all probabilities above add up to one.