If the probability that the random variable $X$ takes the value $x$ is given by

$P(X=x)=k(x+1) 3^{-x}, x=0,1,2,3 \ldots$, where $k$ is a constant, then $P(X \geq 3)$ is equal to

<p>To find $ P(X \geq 3) $, we first determine the constant $ k $ using the total probability for $ X $.</p> <p>The probability $ P(X = x) $ is given by:</p> <p>$ P(X = x) = k(x+1) \cdot 3^{-x}, \quad x = 0, 1, 2, 3, \ldots $</p> <p>The total probability must equal 1:</p> <p>$ s = \sum_{x = 0}^{\infty} k(x+1) \cdot 3^{-x} $</p> <p>Calculating that series:</p> <p>$ s = \frac{k}{3^0} + \frac{2k}{3} + \frac{3k}{3^2} + \ldots $</p> <p>Therefore, dividing the series by 3:</p> <p>$ \frac{s}{3} = \frac{k}{3} + \frac{2k}{3^2} + \ldots $</p> <p>Subtracting these:</p> <p>$ s - \frac{s}{3} = k + \frac{k}{3} + \frac{k}{3^2} + \ldots $</p> <p>The resulting series is a geometric series:</p> <p>$ \frac{2s}{3} = k \left( 1 + \frac{1}{3} + \frac{1}{3^2} + \ldots \right) $</p> <p>The sum of the infinite geometric series is:</p> <p>$ \frac{2s}{3} = k \cdot \frac{1}{1-\frac{1}{3}} = \frac{3k}{2} $</p> <p>Equating:</p> <p>$ s = \frac{9k}{4} = 1 $</p> <p>Thus, solving for $ k $:</p> <p>$ k = \frac{4}{9} $</p> <p>Next, compute $ P(X \geq 3) $:</p> <p>$ P(X \geq 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2)) $</p> <p>Calculating these:</p> <p>$ P(X = 0) = k = \frac{4}{9} $</p> <p>$ P(X = 1) = \frac{2k}{3} = \frac{8}{27} $</p> <p>$ P(X = 2) = \frac{3k}{9} = \frac{4}{27} $</p> <p>Adding these probabilities:</p> <p>$ P(X = 0) + P(X = 1) + P(X = 2) = \frac{4}{9} + \frac{8}{27} + \frac{4}{27} = \frac{8}{9} $</p> <p>Finally, calculate $ P(X \geq 3) $:</p> <p>$ P(X \geq 3) = 1 - \frac{8}{9} = \frac{1}{9} $</p>