MCQ
If the probability that the random variable $X$ takes the value x is given by $\mathrm{P}(\mathrm{X}=\mathrm{x})=\mathrm{k}(\mathrm{x}+1) 3^{-\mathrm{x}}$, $\mathrm{x}=0,1,2,3 \ldots \ldots$, where k is a constant, then $\mathrm{P}(\mathrm{X} \geq 3)$ is equal to
  • A
    $\frac{7}{27}$
  • B
    $\frac{4}{9}$
  • C
    $\frac{8}{27}$
  • D
    $\frac{1}{9}$

Answer

D. $\frac{1}{9}$
$\sum_{x=0}^{\infty} k(x+1) 3^{-x}=1$
$\Rightarrow \frac{1}{\mathrm{k}}=1+\frac{2}{3}+\frac{3}{3^{2}}+\frac{4}{3^{3}}+ \quad\quad...(i)$
$\frac{1}{3 \mathrm{k}}=\frac{1}{3}+\frac{2}{3^{2}}+\frac{3}{3^{3}}+\quad\quad\quad \ldots(ii)$
(i)- (ii) $\Rightarrow \frac{1}{\mathrm{k}}-\frac{1}{3 \mathrm{k}}=1+\frac{1}{3}+\frac{1}{3^{2}}+\ldots$
$\Rightarrow \mathrm{k}=\frac{4}{9}$
$P(x \geq 3)=1-P(x=0)-P(x=1)-P(x=2)$
$=1-\mathrm{k}\left(1+\frac{2}{3}+\frac{3}{9}\right)=\frac{1}{9}$

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