Alvie Singer lives at 0 in the accompanying diagram and has four friends who live at A, B, C, and D. One day Alvie decides to go visiting, so he tosses a fair coin twice to decide which of the four to visit. Once at a friend’s house, he will either return home or else proceed to one of the two adjacent houses (such as 0, A, or C when at B), with each of the three possibilities having probability . In this way, Alvie continues to visit friends until he returns home. a. Let X = the number of times that Alvie visits a friend. Derive the pmf of X. b. ?Let Y= the number of straight-line segments that Alvie traverses (including those leading to and from 0). What is the pmf of Y? c. ?Suppose that female friends live at A and C and male friends at B and D. If Z = the number f visits to female friends, what is the pmf of Z?

Answer: Step 1 of 1 Alvie Singer lives at 0 in the accompanying diagram and has four friends who live at A, B, C, and D. One day Alvie decides to go visiting, so he tosses a fair coin twice to decide which of the four to visit. Once at a friend’s house, he will either return home or else proceed to one of the two adjacent houses (such as 0, A, or C when at B), with each of the three possibilities having probability. a) Alvie visits 1 friend = (1/3) Alvie visits 2 friends = (2/3)(1/3) Alvie visits 3 friends = (2/3) (1/3) 3 Alvie visits 4 friends = (2/3) (1/3) Now, Alvie visits x friends = (2/3) (1/3) Where X is the number of number of times Alvie visits a friend. Thus, b) Here, Alvie will travel on x+1 straight line segments. He travels x segments to visit x friends and then travel one more segment to return home. Thus Y=X+1. X = Y - 1 The probability mass function for Y is: c) If two of the locations are female and two locations are male. Suppose that female friends live at A and C and male friends at B and D. If Z = the number of visits to female friends Let Z is the number of female friends Alvie will visit. If he visits a male first and then goes home, Z=0 Probability of (1/4)(1/3) + (1/4)(1/3) = 1/6 For visiting one female: Probability of visiting one female P(Z = 1) = P(FH) + P(FMH) + P(MFH) + P(MFMH) = (1/2)(1/3) + (1/2)(2/3)(1/3) + (1/2)(2/3)(1/3) + (1/2)(2/3)(2/3)(1/3) = 0.4629. Therefore, the probability of visiting one female is 0.4629.