A random variable X takes values 0, 1, 2, 3, | vedclass

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

Question

(C) The sum of all probabilities in a probability distribution must be equal to $1$. Thus,$\sum_{x=0}^{\infty} P(X=x) = 1$. Substituting the given exp

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(C) The sum of all probabilities in a probability distribution must be equal to $1$. Thus,$\sum_{x=0}^{\infty} P(X=x) = 1$. Substituting the given expression: $\sum_{x=0}^{\infty} k(x+1)\left(\frac{1}{2}\right)^x = 1$. $k \sum_{x=0}^{\infty} (x+1)r^x = 1$,where $r = \frac{1}{2}$. We know that $\sum_{x=0}^{\infty} (x+1)r^x = 1 + 2r + 3r^2 + \dots = (1-r)^{-2}$. For $r = \frac{1}{2}$,the sum is $(1 - \frac{1}{2})^{-2} = (\frac{1}{2})^{-2} = 4$. Therefore,$k(4) = 1$,which gives $k = \frac{1}{4}$. Now,we need to find $P(X=1)$. $P(X=1) = k(1+1)\left(\frac{1}{2}\right)^1 = k(2)(\frac{1}{2}) = k$. Since $k = \frac{1}{4}$,$P(X=1) = \frac{1}{4}$.

$X=x$	$1$	$2$	$3$	$4$
$P(X=x)$	$0.1$	$0.2$	$0.3$	$0.4$

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X)$	$0$	$k$	$2k$	$3k$	$3k^2$	$k^2$	$2k^2$	$7k^2+k$

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X = x)$	$0.15$	$0.23$	$k$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

$X$	$1$	$2$	$3$	$4$
$P(X)$	$\frac{1}{10}$	$\frac{1}{5}$	$\frac{3}{10}$	$\frac{2}{5}$

$A$ random variable $X$ takes values $0, 1, 2, 3, \dots$ with probabilities $P(X=x) = k(x+1)\left(\frac{1}{2}\right)^x$. If $k$ is a constant,then $P(X=1) = $

Similar Questions

$A$ random variable $X$ has the following probability distribution:
$X=x$ $1$ $2$ $3$ $4$
$P(X=x)$ $0.1$ $0.2$ $0.3$ $0.4$

The mean and standard deviation of $X$ are respectively:

$A$ random variable $X$ has the following probability distribution:

$X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$

$P(X)$ $0$ $k$ $2k$ $3k$ $3k^2$ $k^2$ $2k^2$ $7k^2+k$

Determine $P(0 < X < 3)$. (in $/10$)

$A$ random variable $X$ has the following probability distribution:
$X = x$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$
$P(X = x)$ $0.15$ $0.23$ $k$ $0.10$ $0.20$ $0.08$ $0.07$ $0.05$

For the events $E = \{x : x \text{ is a prime number}\}$ and $F = \{x : x < 4\}$,then $P(E \cup F) = $

If $X$ is a Poisson variate with $P(X=0)=0.8$,then the variance of $X$ is

For the given probability distribution,find $E(X^2)$.
$X$ $1$ $2$ $3$ $4$
$P(X)$ $\frac{1}{10}$ $\frac{1}{5}$ $\frac{3}{10}$ $\frac{2}{5}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module