$5$ persons entered a lift cabin on the ground floor of a $7$-floor house. Suppose that each of them independently and with equal probability can leave the cabin at any floor beginning with the first. The probability of all the $5$ persons leaving the cabin at different floors is

  • A
    $\frac{360}{2401}$
  • B
    $\frac{5}{54}$
  • C
    $\frac{5}{18}$
  • D
    $\frac{5!}{7!}$

Explore More

Similar Questions

Let $X$ be a random variable such that the probability function of a distribution is given by $P(X=0) = \frac{1}{2}$ and $P(X=j) = \frac{1}{3^j}$ for $j = 1, 2, 3, \ldots, \infty$. Then the mean of the distribution and $P(X \text{ is positive and even})$ respectively are:

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 Probability
$\omega_{1}$ $1/8$
$\omega_{2}$ $2/3$
$\omega_{3}$ $1/3$
$\omega_{4}$ $1/3$
$\omega_{5}$ $-1/4$
$\omega_{6}$ $-1/3$

The p.d.f. of a continuous random variable $X$ is given by $f(x) = \frac{1}{2}$ if $0 < x < 2$ and $f(x) = 0$ otherwise. If $a = P(X < \frac{1}{2})$ and $b = P(X > \frac{3}{2})$,then the relation between $a$ and $b$ is:

$A$ biased coin with probability $p, 0 < p < 1,$ of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even is $\frac{2}{5},$ then $p = $

Difficult
View Solution

If the probability distribution of a random variable $X$ is as follows,then $P(X \leq 2) = $
$x_i$$0$$1$$2$$3$$4$
$P(X = x_i)$$3K$$5K$$3k^2$$4k^2 + k$$3k^2$

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo