Problem

Alvie Singer lives at 0 in the accompanying diagram and has four friends who live at A,...

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?

Step-by-Step Solution

Solution 1

a) Alvie chooses the four friends to be visited, each with equal probability

Once at a friend's house, Alvie will either return home or else proceed to one of the two adjacent houses.

That is if he is at \(B\), he proceeds to \(\mathrm{O}\) or \(\mathrm{A}\) or \(\mathrm{C}\)

Probability of each of these three events is \(\frac{1}{3}\)

\(X\) is defined as the number of times that Alvie visits a friend

The possible values that \(X\) can take are \(\{1,2,3,4, \ldots . .\}\)

\(X\) takes value 1 if Alvie visits a friend once

This happens when Alvie returns home after visiting a friend

The probability that he returns home from a friend's house \(=\frac{1}{3}\)

So we have \(P(X=1)=\frac{1}{3}\)

| x | 1 | 2 | 3 | dots dots dots. | x | |

| :--- | :---: | :---: | :---: | :---: | :---: | :---: |

| p(x) | (1)/(3) | (2)/(9) | (4)/(27) | dots dots dots.. | ((2)/(3))^(x-1)*(1)/(3) | dots dots dots.. |

(b) The random variable Y is defined as:

`

If Alvie visits a friend once he has to traverse through two line segments

For example if he visits only \(B\), he has to traverse through the line segment \(O B\) and then again as he has to go home immediately after this visit, he has to traverse through the line segment BO.

So if Alvie visits a friend once, the number of line segments that he has to traverse \(=2\)

Probability that he traverses 2 line segments is therefore equal to the probability that the number of times he visits a friend is 1

In other words,

$$ \begin{aligned} P(Y=2) &=P(X=1) \\ &=\frac{1}{3} \end{aligned} $$

If the number of times that Alive visits a friend is 2 , he has to traverse through three line segments

For example if he visits only \(B\), and then he visits \(A\), he has to traverse through the line segment \(O B\) and then \(B A\), and again as he has to go home immediately after this visit, he has to traverse through the line segment \(A O\).

So if the number of times that Alive visits a friend is 2 , the number of line segments that he has to traverse \(=3\)

Probability that he traverses 3 line segments is therefore equal to the probability that the number of times he visits a friend is 2

In other words,

$$ \begin{aligned} P(Y=3) &=P(X=2) \\ &=\frac{1}{3}\left(\frac{2}{3}\right) \end{aligned} $$

| y | 2 | 3 | dots dots dots. | y |

| :--- | :--- | :--- | :--- | :--- |

| p(y) | (2)/(9) | (4)/(27) | dots dots dots dots+dots dots | |

| ((2)/(3))^(y-2) | (1)/(3) | dots dots dots. | | |

Since the visits are independent, we get:

$$ \begin{aligned} p(1) &=\left(\frac{1}{2} \times \frac{1}{3}\right)+\left(\frac{1}{2} \times \frac{2}{3} \times \frac{1}{3}\right)+\left(\frac{1}{2} \times \frac{2}{3} \times \frac{1}{3}\right)+\left(\frac{1}{2} \times \frac{2}{3} \times \frac{2}{3} \times \frac{1}{3}\right) \\ &=\left(\frac{1}{2} \times \frac{1}{3}\right)\left(1+\frac{2}{3}\right)+\left(\frac{1}{2} \times \frac{2}{3} \times \frac{1}{3}\right)\left(1+\frac{2}{3}\right) \\ &=\left(\frac{1}{2}\right)\left(1+\frac{2}{3}\right)\left(\frac{1}{3}\right)+\left(\frac{1}{2} \times \frac{2}{3}\right)\left(1+\frac{2}{3}\right)\left(\frac{1}{3}\right) \\ &=\left(\frac{1}{2}\right)\left(\frac{5}{3}\right)\left(\frac{1}{3}\right)+\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)\left(\frac{5}{3}\right)\left(\frac{1}{3}\right) \end{aligned} $$

The first term in the above expression corresponds to the probability of first visiting a female friend and the second term corresponds to the probability of first visiting a male friend.

So proceeding in the similar way, we can write:

\(\begin{aligned} p(2) &=\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)^{2}\left(\frac{5}{3}\right)\left(\frac{1}{3}\right)+\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)^{3}\left(\frac{5}{3}\right)\left(\frac{1}{3}\right) \\ &=\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)^{2}\left(\frac{5}{3}\right)\left(\frac{1}{3}\right)+\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)^{2}\left(\frac{2}{3}\right)\left(\frac{5}{3}\right)\left(\frac{1}{3}\right) \\ p(3) &=\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)^{4}\left(\frac{5}{3}\right)\left(\frac{1}{3}\right)+\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)^{4}\left(\frac{2}{3}\right)\left(\frac{5}{3}\right)\left(\frac{1}{3}\right) \\ \text { Generalizing this series, we get the pmf of } Z \text { is obtained as follows: } \\ p(z) &=\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)^{2 z-2}\left(\frac{5}{3}\right)\left(\frac{1}{3}\right)+\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)^{2 z-2}\left(\frac{2}{3}\right)\left(\frac{5}{3}\right)\left(\frac{1}{3}\right) \\ &=\left(\frac{2}{3}\right)^{2 z-2}\left[\left(\frac{1}{2}\right)\left(\frac{5}{3}\right)\left(\frac{1}{3}\right)+\left(\frac{2}{3}\right)\left(\frac{1}{2}\right)\left(\frac{5}{3}\right)\left(\frac{1}{3}\right)\right] \\ &=\left(\frac{2}{3}\right)^{2 z-2}\left[\left(\frac{5}{18}\right)+\left(\frac{10}{54}\right)\right] \\ &=\left(\frac{2}{3}\right)^{2 z-2}\left(\frac{25}{54}\right) \end{aligned}\)

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