Probability & Statistics Ch.2 Section 3 Question 5

· 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

Consider rolling two fair six-sided dice. Let W be the product of the numbers showing. What is the probability function of W?

Solution

Valid outcomes for each product value:

  • W = 1: (1,1) → pW(1)=136p_W(1) = \frac{1}{36}
  • W = 2: (1,2), (2,1) → pW(2)=236p_W(2) = \frac{2}{36}
  • W = 3: (1,3), (3,1) → pW(3)=236p_W(3) = \frac{2}{36}
  • W = 4: (1,4), (4,1), (2,2) → pW(4)=336p_W(4) = \frac{3}{36}
  • W = 5: (1,5), (5,1) → pW(5)=236p_W(5) = \frac{2}{36}
  • W = 6: (1,6), (6,1), (2,3), (3,2) → pW(6)=436p_W(6) = \frac{4}{36}
  • W = 8: (2,4), (4,2) → pW(8)=236p_W(8) = \frac{2}{36}
  • W = 9: (3,3) → pW(9)=136p_W(9) = \frac{1}{36}
  • W = 10: (2,5), (5,2) → pW(10)=236p_W(10) = \frac{2}{36}
  • W = 12: (2,6), (6,2), (3,4), (4,3) → pW(12)=436p_W(12) = \frac{4}{36}
  • W = 15: (3,5), (5,3) → pW(15)=236p_W(15) = \frac{2}{36}
  • W = 16: (4,4) → pW(16)=136p_W(16) = \frac{1}{36}
  • W = 18: (3,6), (6,3) → pW(18)=236p_W(18) = \frac{2}{36}
  • W = 20: (4,5), (5,4) → pW(20)=236p_W(20) = \frac{2}{36}
  • W = 24: (4,6), (6,4) → pW(24)=236p_W(24) = \frac{2}{36}
  • W = 25: (5,5) → pW(25)=136p_W(25) = \frac{1}{36}
  • W = 30: (5,6), (6,5) → pW(30)=236p_W(30) = \frac{2}{36}
  • W = 36: (6,6) → pW(36)=136p_W(36) = \frac{1}{36}

pW(w)=0p_W(w)=0 for all w∉{1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,30,36}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.