Q80 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Mathematics
If the probability function of a random variable $X$ is given by $P(X=n)=\frac{k(n+1)}{3 n}$ for $n \in \mathbf{N} \cup\{0\}$ where $k$ is a constant, then $P(X < 2)=$
- Correct answerA$\frac{20}{27}$
- B$\frac{20}{81}$
- C$\frac{2}{27}$
- D$\frac{8}{81}$
Solution
Given, probability function of random variable $X$ is
$P(X=n)=\frac{k(n+1)}{3^n} \text { for } n \in \mathbf{N} \cup\{0\}$
$\therefore \quad \sum_{n=0}^{\infty} P(X=n)=1$
$\Rightarrow \quad k \sum_{n=0}^{\infty} \frac{(n+1)}{3^n}=1 \Rightarrow k\left[1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+\ldots.\right]=1$
$\because$ Sum of infinite AGP
$a+(a+d) r+(a+2 d) r^2+\ldots . .=\frac{a}{1-r}+\frac{d r}{(1-r)^2}$
$\therefore k\left[\frac{1}{1-1 / 3}+\frac{1 / 3}{(1-1 / 3)^2}\right]=1 \Rightarrow k\left[\frac{3}{2}+\frac{3}{4}\right]=1 \Rightarrow k=\frac{4}{9}$
$\therefore \quad P(X < 2)=P(X=0)+P(X=1)$
$=k\left[1+\frac{2}{3}\right]=\frac{4}{9} \times \frac{5}{3}=\frac{20}{27}$
Hence, the correct answer is option: 'A'.