If the probability distribution of a discrete | vedclass

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

Question

(B) The sum of probabilities is $\sum_{k=0}^{\infty} P(X=k) = 1$. Given $P(X=k) = \frac{3k+1}{2^{k+c}}$,we have $\frac{1}{2^c} \sum_{k=0}^{\infty} (3k

Vedclass · Accepted Answer

$\frac{c}{4}$

Vedclass · Answer

(B) The sum of probabilities is $\sum_{k=0}^{\infty} P(X=k) = 1$. Given $P(X=k) = \frac{3k+1}{2^{k+c}}$,we have $\frac{1}{2^c} \sum_{k=0}^{\infty} (3k+1) \left(\frac{1}{2}\right)^k = 1$. Let $S = \sum_{k=0}^{\infty} (3k+1) x^k$ where $x = \frac{1}{2}$. $S = 3 \sum_{k=0}^{\infty} k x^k + \sum_{k=0}^{\infty} x^k = 3 \frac{x}{(1-x)^2} + \frac{1}{1-x}$. Substituting $x = \frac{1}{2}$: $S = 3 \frac{1/2}{(1/2)^2} + \frac{1}{1/2} = 3(2) + 2 = 8$. Thus,$\frac{1}{2^c} (8) = 1 \implies 2^3 = 2^c \implies c = 3$. We need to find $P(X \leq 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)$. $P(X=k) = \frac{3k+1}{2^{k+3}}$. $P(X=0) = \frac{1}{8}, P(X=1) = \frac{4}{16} = \frac{2}{8}, P(X=2) = \frac{7}{32}, P(X=3) = \frac{10}{64} = \frac{5}{32}$. Sum $= \frac{4}{32} + \frac{8}{32} + \frac{7}{32} + \frac{5}{32} = \frac{24}{32} = \frac{3}{4}$. Since $c=3$,the option $\frac{c}{4} = \frac{3}{4}$ matches option $B$.

$X = x$	$0$	$1$	$2$	$3$	$4$
$P(X = x)$	$k$	$2k$	$4k$	$2k$	$k$

$X = x$	$1$	$2$	$3$	$4$
$P(X = x)$	$1/10$	$2/10$	$3/10$	$4/10$

Similar Questions

The p.m.f of a random variable $X$ is given by $P(X) = \frac{2x}{n(n+1)}$ for $x = 1, 2, 3, \ldots, n$,and $0$ otherwise. Then $E(X) = $

The probability distribution of a discrete random variable $X$ is given by the following table:
$X = x$ $0$ $1$ $2$ $3$ $4$
$P(X = x)$ $k$ $2k$ $4k$ $2k$ $k$

Then the value of $P(X \leq 2)$ is:

Suppose that a book of $600$ pages contains $40$ print mistakes. Assume that these errors are randomly distributed throughout the book and the number of errors per page follows a Poisson distribution. The probability that all the $10$ pages selected at random have no print mistakes is

The probability distribution of a random variable $X$ is given by:
$X = x$ $1$ $2$ $3$ $4$
$P(X = x)$ $1/10$ $2/10$ $3/10$ $4/10$

Then the cumulative distribution function (c.d.f.) $F(x)$ of $X$ is given by:

$A$ random variable $X$ takes values $0, 1, 2, 3, \ldots$ with probability $P(X=x) = K(x+1)\left(\frac{1}{5}\right)^x$,where $K$ is a constant. Then $P(X=0)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module