If the probability function of a random varia | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the probability mass function $P(X=n) = \frac{k(n+1)}{3^n}$ for $n \in \{0, 1, 2, \dots\}$.Since the sum of all probabilities must be $1$, w

Vedclass · Accepted Answer

$\frac{20}{27}$

Vedclass · Answer

(A) Given the probability mass function $P(X=n) = \frac{k(n+1)}{3^n}$ for $n \in \{0, 1, 2, \dots\}$.Since the sum of all probabilities must be $1$, we have $\sum_{n=0}^{\infty} P(X=n) = 1$.$k \sum_{n=0}^{\infty} \frac{n+1}{3^n} = k \left( \frac{1}{3^0} + \frac{2}{3^1} + \frac{3}{3^2} + \frac{4}{3^3} + \dots \right) = 1$.This is an arithmetico-geometric series with $a=1$, $d=1$, and $r=\frac{1}{3}$.The sum of the series is $S = \frac{a}{1-r} + \frac{dr}{(1-r)^2} = \frac{1}{1-1/3} + \frac{1 \cdot (1/3)}{(1-1/3)^2} = \frac{3}{2} + \frac{1/3}{4/9} = \frac{3}{2} + \frac{3}{4} = \frac{9}{4}$.Thus, $k \cdot \frac{9}{4} = 1 \implies k = \frac{4}{9}$.We need to find $P(X $P(X=0) = k \cdot \frac{0+1}{3^0} = k \cdot 1 = \frac{4}{9}$.$P(X=1) = k \cdot \frac{1+1}{3^1} = k \cdot \frac{2}{3} = \frac{4}{9} \cdot \frac{2}{3} = \frac{8}{27}$.$P(X < 2) = \frac{4}{9} + \frac{8}{27} = \frac{12+8}{27} = \frac{20}{27}$.

$X = k$	$-2$	$-1$	$0$	$1$	$2$
$P(X = k)$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{3}$	$\frac{1}{6}$	$\frac{1}{6}$

If the probability function of a random variable $X$ is given by $P(X=n) = \frac{k(n+1)}{3^n}$ for $n \in \mathbb{N} \cup \{0\}$ where $k$ is a constant, then $P(X < 2) = $

Similar Questions

If $X$ is a random variable with probability distribution $P(X=k) = \frac{(2k+3)c}{3^k}$, $k=0, 1, 2, \ldots, \infty$, then $P(X=3) =$

If the p.m.f. of a random variable $X$ is given by the following table,then the standard deviation of $X$ is (given $p+q=1$):

$x$ $0$ $1$ $2$

$P(X=x)$ $q^2$ $2pq$ $p^2$

The variance of the random variable $X$ having the following distribution is:
$X = k$ $-2$ $-1$ $0$ $1$ $2$
$P(X = k)$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{3}$ $\frac{1}{6}$ $\frac{1}{6}$

Let a sample space be $S = \{\omega_{1}, \omega_{2}, \ldots, \omega_{6}\}$. Which of the following assignments of probabilities to each outcome is valid?

Outcome $\omega_1$ $\omega_2$ $\omega_3$ $\omega_4$ $\omega_5$ $\omega_6$

$(b)$ $1$ $0$ $0$ $0$ $0$ $0$

The probability mass function of a random variable $X$ is given by $P(X=x) = \frac{{}^{5}C_{x}}{2^{5}}$ for $x = 0, 1, 2, 3, 4, 5$ and $0$ otherwise. Then,$P(X \leq 2)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module