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

| 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:

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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:

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