If X is a random variable with probability di | vedclass

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

Question

(B) For a probability distribution, the sum of all probabilities must be $1$: $\sum_{k=0}^{\infty} P(X=k) = 1$. Substituting the given expression: $c

Vedclass · Accepted Answer

$\frac{1}{18}$

Vedclass · Answer

(B) For a probability distribution, the sum of all probabilities must be $1$: $\sum_{k=0}^{\infty} P(X=k) = 1$. Substituting the given expression: $c \sum_{k=0}^{\infty} \frac{2k+3}{3^k} = 1$. Let $S = \sum_{k=0}^{\infty} \frac{2k+3}{3^k} = 3 \sum_{k=0}^{\infty} \frac{1}{3^k} + 2 \sum_{k=0}^{\infty} \frac{k}{3^k}$. The first part is a geometric series: $3 	imes \frac{1}{1 - 1/3} = 3 	imes \frac{3}{2} = \frac{9}{2}$. The second part is an arithmetico-geometric series: $\sum_{k=0}^{\infty} k x^k = \frac{x}{(1-x)^2}$. For $x = 1/3$, this is $\frac{1/3}{(1-1/3)^2} = \frac{1/3}{4/9} = \frac{3}{4}$. So, $S = \frac{9}{2} + 2 	imes \frac{3}{4} = \frac{9}{2} + \frac{3}{2} = 6$. Since $c 	imes S = 1$, we have $c = 1/6$. Now, $P(X=3) = \frac{(2(3)+3)c}{3^3} = \frac{9c}{27} = \frac{c}{3}$. Substituting $c = 1/6$, we get $P(X=3) = \frac{1/6}{3} = \frac{1}{18}$.

$X = x$	$-2$	$-1$	$0$	$1$	$2$
$P(X = x)$	$0.2$	$0.3$	$0.1$	$0.15$	$0.25$

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 $P(X=3) =$

Similar Questions

$A$ random variable $X$ takes values $-1, 0, 1, 2$ with probabilities $\frac{1+3p}{4}, \frac{1-p}{4}, \frac{1+2p}{4}, \frac{1-4p}{4}$ respectively,where $p$ varies over $\mathbb{R}$. Then the minimum and maximum values of the mean of $X$ are respectively.

$A$ person throws an unbiased die. If the number shown is even,he gains an amount equal to the number shown. If the number is odd,he loses an amount equal to the number shown. Then his expectation is ₹.

For the probability distribution of $X$ given below,the variance of $X$ is:
$X = x$ $-2$ $-1$ $0$ $1$ $2$
$P(X = x)$ $0.2$ $0.3$ $0.1$ $0.15$ $0.25$

The range of a random variable $X$ is $\{1, 2, 3, \ldots\}$ and $P(X=x) = \frac{c^x}{x!}$ for $x = 1, 2, 3, \ldots$. Then the value of $c$ is

If the probability distribution of a random variable $X$ is as follows,then the mean of $X$ is:
$X = x_i$ $-1$ $0$ $1$ $2$
$P(X = x_i)$ $k^3$ $2k^3 + k$ $4k - 10k^2$ $4k - 1$

Vedclass Test Series

Exam Paper Generator

Online Exam Module