An urn contains $5$ red and $2$ black balls. Two balls are randomly drawn. Let $X$ represent the number of black balls. What are the possible values of $X?$ Is $X$ a random variable?
(N/A) The two balls selected can be represented as $BB$,$BR$,$RB$,and $RR$,where $B$ represents a black ball and $R$ represents a red ball.
$X$ represents the number of black balls.
$X(BB) = 2$
$X(BR) = 1$
$X(RB) = 1$
$X(RR) = 0$
Therefore,the possible values of $X$ are $0, 1,$ and $2$.
Yes,$X$ is a random variable because it is a real-valued function whose domain is the sample space of a random experiment.
$X$ represents the number of black balls.
$X(BB) = 2$
$X(BR) = 1$
$X(RB) = 1$
$X(RR) = 0$
Therefore,the possible values of $X$ are $0, 1,$ and $2$.
Yes,$X$ is a random variable because it is a real-valued function whose domain is the sample space of a random experiment.
Explore More
Similar Questions
The probability that an individual suffers a bad reaction from an injection is $0.001$. The probability that out of $2000$ individuals exactly three will suffer a bad reaction is:
DifficultTS EAMCET 2011
View SolutionIf a discrete random variable $X$ is defined as follows:
$P(X=x) = \begin{cases} \frac{k(x+1)}{5^x}, & x=0, 1, 2, \ldots \\ 0, & \text{otherwise} \end{cases}$
then $k=$
$P(X=x) = \begin{cases} \frac{k(x+1)}{5^x}, & x=0, 1, 2, \ldots \\ 0, & \text{otherwise} \end{cases}$
then $k=$
If a Poisson variate $X$ satisfies $P(X=2) = P(X=3)$, then $P(X=5) =$
Which of the following cannot be a valid assignment of probabilities for outcomes of sample space $S = \{\omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}, \omega_{5}, \omega_{6}, \omega_{7}\}$?
Outcome $\omega_{1}$ $\omega_{2}$ $\omega_{3}$ $\omega_{4}$ $\omega_{5}$ $\omega_{6}$ $\omega_{7}$ Probability $-0.1$ $0.2$ $0.3$ $0.4$ $-0.2$ $0.1$ $0.3$
| Outcome | $\omega_{1}$ | $\omega_{2}$ | $\omega_{3}$ | $\omega_{4}$ | $\omega_{5}$ | $\omega_{6}$ | $\omega_{7}$ |
| Probability | $-0.1$ | $0.2$ | $0.3$ | $0.4$ | $-0.2$ | $0.1$ | $0.3$ |
Easy
View SolutionThe 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) = $
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo