Probability & Statistics Ch.2 Section 3 Question 4

· 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 flipping two fair coins. Let X = 1 if the first coin is heads, and X = 0 if the first coin is tails. Let Y = 1 if the two coins show the same thing (both heads or both tails), with Y = 0 otherwise. Let Z=X+YZ = X + Y, and W=XYW = XY.

a) What is the probability function of Z? b) What is the probability function of W?

Solution

(a) What is the probability function of Z?

Valid outcomes for Z = 0:

  • Tails, Heads (X=0, Y=0): Z=0+0=0Z = 0 + 0 = 0
pZ(0)=12×12=14p_Z(0) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}

Valid outcomes for Z = 1:

  • Heads, Tails (X=1, Y=0): Z=1+0=1Z = 1 + 0 = 1
  • Tails, Tails (X=0, Y=1): Z=0+1=1Z = 0 + 1 = 1
pZ(1)=14+14=12p_Z(1) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}

Valid outcomes for Z = 2:

  • Heads, Heads (X=1, Y=1): Z=1+1=2Z = 1 + 1 = 2
pZ(2)=14p_Z(2) = \frac{1}{4}

pZ(z)=0p_Z(z)=0 for all z∉{0,1,2}z\notin\{0,1,2\}

(b) What is the probability function of W?

Valid outcomes for W = 0:

  • Heads, Tails (X=1, Y=0): W=1×0=0W = 1 \times 0 = 0
  • Tails, Tails (X=0, Y=1): W=0×1=0W = 0 \times 1 = 0
  • Tails, Heads (X=0, Y=0): W=0×0=0W = 0 \times 0 = 0
pW(0)=14+14+14=34p_W(0) = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}

Valid outcomes for W = 1:

  • Heads, Heads (X=1, Y=1): W=1×1=1W = 1 \times 1 = 1
pW(1)=14p_W(1) = \frac{1}{4}

pW(w)=0p_W(w)=0 for all w∉{0,1}w\notin\{0,1\}