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Let x be a random variable such that the probability function of a distribution is given by \mathrm{P}(\mathrm{X}=0)=\frac{1}{2}, \mathrm{P}(\mathrm{X}=\mathrm{j})=\frac{1}{3^{\mathrm{j}}}(\mathrm{j}=1,2,3, \ldots, \infty)
Then the mean of the distribution and \mathrm{P}(\mathrm{X} is positive and even) respectively are
Option: 1 \frac{3}{8}\, and\, \frac{1}{8}
Option: 2 \frac{3}{8}\, and\, \frac{1}{8}
Option: 3 \frac{3}{8}\, and\, \frac{1}{8}
Option: 4 \frac{3}{8}\, and\, \frac{1}{8}
Option: 5 \frac{3}{4}\,\, and \,\,\frac{1}{8}
Option: 6 \frac{3}{4}\,\, and \,\,\frac{1}{8}
Option: 7 \frac{3}{4}\,\, and \,\,\frac{1}{8}
Option: 8 \frac{3}{4}\,\, and \,\,\frac{1}{8}
Option: 9 \frac{3}{4}\,\, and \,\,\frac{1}{9}
Option: 10 \frac{3}{4}\,\, and \,\,\frac{1}{9}
Option: 11 \frac{3}{4}\,\, and \,\,\frac{1}{9}
Option: 12 \frac{3}{4}\,\, and \,\,\frac{1}{9}
Option: 13 \frac{3}{4}\,\, and \,\,\frac{1}{16}
Option: 14 \frac{3}{4}\,\, and \,\,\frac{1}{16}
Option: 15 \frac{3}{4}\,\, and \,\,\frac{1}{16}
Option: 16 \frac{3}{4}\,\, and \,\,\frac{1}{16}

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Mean\, of\, distribution= \sum xi\, pi
= 0\cdot \left ( \frac{1}{2} \right )+1\left ( \frac{1}{3} \right )+2\left ( \frac{1}{3} \right )^{2}+\cdots
Let\, S= \frac{1}{3}+\frac{2}{3^{2}}+\frac{3}{3^{3}}+\frac{4}{3^{4}}+\cdots
       \frac{S}{3}= \frac{1}{3^{2}}+\frac{2}{3^{3}}+\frac{3}{3^{4}}+\cdots

Subtracting
\frac{2S}{3}= \frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\cdots
\frac{2S}{3}= \frac{1}{2}\Rightarrow S= \frac{3}{4}

Now,P\left ( x\, is\, even \right )= P\left ( x= 2 \right )+P\left ( x= 4 \right )+\cdots
                                   = \frac{1}{3^{2}}+\frac{1}{3^{4}}+\frac{1}{3^{6}}+\cdots
                                   = \frac{1}{8}

Posted by

Kuldeep Maurya

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