Let's work through the problem step by step. ### Given: The probability mass function (PMF) of \(X\): \[ P(X = x) = \left\{ \begin{array}{ll} \frac{1}{3}\left(\frac{2}{3}\right)^x & x = 0, 1, 2, \ldots \\ 0 & \text{otherwise} \end{array} \right. \] We are to **find the distribution of \( Y = \frac{X}{X+1} \)**. --- ### 1. **Find the range of \( Y \):** For \( x = 0, 1, 2, \ldots \): \[ Y = \frac{X}{X+1} \] So, \[ Y(x) = \frac{x}{x+1} \] for \(x = 0, 1, 2, \ldots\) - For \(x = 0\): \(Y = 0\) - For \(x = 1\): \(Y = \frac{1}{2}\) - For \(x = 2\): \(Y = \frac{2}{3}\) - For \(x = 3\): \(Y = \frac{3}{4}\) - etc. So the possible values of \(Y\) are: \[ \left\{0, \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \ldots, \frac{x}{x+1}, \ldots\right\} \] --- ### 2. **Find the PMF of \( Y \):** Let \( y = \frac{x}{x+1} \implies x = \frac{y}{1-y} \). But \(x\) must be a non-negative integer, so \(y\) can only take the values given above. The probability: \[ P(Y = y) = P\left(X = x\right) \] where \( x = \frac{y}{1-y} \) must be an integer. So for each \(x = 0,1,2,\ldots\), \(Y = \frac{x}{x+1}\) and \[ P\left(Y = \frac{x}{x+1}\right) = P(X = x) = \frac{1}{3}\left(\frac{2}{3}\right)^x \] --- ### 3. **Final Answer (Distribution of \(Y\)):** \[ \boxed{ P\left(Y = \frac{x}{x+1}\right) = \frac{1}{3}\left(\frac{2}{3}\right)^x, \quad x = 0, 1, 2, \ldots } \] or, in terms of \(y\): \[ P(Y = y) = \frac{1}{3} \left(\frac{2}{3}\right)^{\frac{y}{1-y}}, \quad y = 0, \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \ldots \] --- **Summary Table:** | \( Y \) | \( P(Y) \) | |------------------|--------------------------------------------| | \( 0 \) | \( \frac{1}{3} \) | | \( \frac{1}{2} \)| \( \frac{2}{9} \) | | \( \frac{2}{3} \)| \( \frac{4}{27} \) | | \( \frac{3}{4} \)| \( \frac{8}{81} \) | | \( \vdots \) | \( \frac{1}{3}\left(\frac{2}{3}\right)^x \)| This is the required distribution of \(Y\).