Q79 / 160 · AP EAMCET 2021 (24 Aug Shift 2) · Mathematics

Let $X$ be a random variable with probability distribution function, $P(X=x)=K\left(\frac{2}{5}\right)^x, x=1,2,3, \ldots \ldots$ Then, value of $K$ is

  • A
    $\frac{3}{5}$
  • B
    $\frac{5}{3}$
  • Correct answer
    C
    $\frac{3}{2}$
  • D
    $\frac{2}{3}$

Solution

Given, $P(X=x)=K(2 / 5)^x$ We know $\Sigma P_i=1$ $\therefore \quad K\left[\left(\frac{2}{5}\right)+\left(\frac{2}{5}\right)^2+\left(\frac{2}{5}\right)^3+\ldots\right]=1$ $\begin{aligned} & \Rightarrow \quad K\left(\frac{\frac{2}{5}}{1-2 / 5}\right)=1 \\ & \Rightarrow \quad K \times \frac{2}{3}=1 \Rightarrow K=3 / 2\end{aligned}$

Hence, the correct answer is option: 'C'.

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