1. Tardigrade
  2. Question
  3. Mathematics
  4. A random variable X takes values 0, 1, 2, 3,... with probability p(X=x)=k(x+1)( (1/5) )x where k is constant, then P X=0) is

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. A random variable $X$ takes values $0, 1, 2, 3,...$ with probability $ p(X=x)=k(x+1){{\left( \frac{1}{5} \right)}^{x}} $ where $k$ is constant, then $ P\{X=0) $ is

KEAMKEAM 2007Probability - Part 2

Solution:


$ \therefore $ $ P(X=0)=kp(X=1)=2k{{\left( \frac{1}{5} \right)}^{1}} $ $ p(X=2)=3k{{\left( \frac{1}{5} \right)}^{1}},.... $
Since, $ P(X=0)+P(X=1)+P(X=2) $ $ +.....=1 $ $ \begin{align} & \underline{\begin{align} & \therefore \,\,\,\,\,\,\,\,\,k+2k\left( \frac{1}{5} \right)+3k{{\left( \frac{1}{5} \right)}^{2}}+.....=1 \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\frac{k}{5}+2k{{\left( \frac{1}{5} \right)}^{2}}+......=\frac{1}{5} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,- \\ \end{align}} \\ & \,\,\,\,\,\,\,\,\,\,\,k+k\frac{1}{5}+k{{\left( \frac{1}{5} \right)}^{2}}+.....=\frac{4}{5} \\ \end{align} $
$ \Rightarrow $ $ \frac{k}{1-\frac{1}{5}}=\frac{4}{5} $
$ \Rightarrow $ $ k=\frac{16}{25} $
$ \therefore $ $ P(X=0)=\frac{16}{25}(0+1){{\left( \frac{1}{5} \right)}^{0}} $