State which of the following is not the probability distribution of a random variable. Give reasons for your answer.
$Z$ $3$ $2$ $1$ $0$ $-1$ $P(Z)$ $0.3$ $0.2$ $0.4$ $0.1$ $0.05$
| $Z$ | $3$ | $2$ | $1$ | $0$ | $-1$ |
| $P(Z)$ | $0.3$ | $0.2$ | $0.4$ | $0.1$ | $0.05$ |
(N/A) For a probability distribution of a random variable,two conditions must be satisfied:
$1$. Each probability $P(Z_i)$ must be $\ge 0$.
$2$. The sum of all probabilities $\sum P(Z_i)$ must be equal to $1$.
In the given table,the sum of probabilities is:
$\sum P(Z) = 0.3 + 0.2 + 0.4 + 0.1 + 0.05 = 1.05$.
Since the sum of probabilities is $1.05$,which is not equal to $1$,the given table does not represent a probability distribution of a random variable.
$1$. Each probability $P(Z_i)$ must be $\ge 0$.
$2$. The sum of all probabilities $\sum P(Z_i)$ must be equal to $1$.
In the given table,the sum of probabilities is:
$\sum P(Z) = 0.3 + 0.2 + 0.4 + 0.1 + 0.05 = 1.05$.
Since the sum of probabilities is $1.05$,which is not equal to $1$,the given table does not represent a probability distribution of a random variable.
Explore More
Similar Questions
The probability distribution of a random variable $X$ is given below:
$X = x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $P(X = x)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2 + k$
Then, $P(0 < X < 4)$ is equal to:
| $X = x$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ |
| $P(X = x)$ | $0$ | $k$ | $2k$ | $2k$ | $3k$ | $k^2$ | $2k^2$ | $7k^2 + k$ |
Then, $P(0 < X < 4)$ is equal to:
$A$ random variable $X$ has the following probability distribution:
$X = x$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $P(X = x)$ $0.15$ $0.23$ $0.10$ $0.12$ $0.20$ $0.08$ $0.07$ $0.05$
For the event $E = \{ X \text{ is a prime number} \}$,$F = \{ X < 4 \}$,then $P(E \cup F)$ is
| $X = x$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |
|---|---|---|---|---|---|---|---|---|
| $P(X = x)$ | $0.15$ | $0.23$ | $0.10$ | $0.12$ | $0.20$ | $0.08$ | $0.07$ | $0.05$ |
For the event $E = \{ X \text{ is a prime number} \}$,$F = \{ X < 4 \}$,then $P(E \cup F)$ is
If the probability that the random variable $X$ takes values $x$ is given by $P(X = x) = k(x + 1)3^{-x}$ for $x = 0, 1, 2, 3, \ldots$,where $k$ is a constant,then $P(X \geq 2)$ is equal to
DifficultJEE MAIN 2023
View Solution$A$ random variable $X$ assumes values $1, 2, 3, \ldots, n$ with equal probabilities. If $\operatorname{Var}(X) = E(X)$,then $n$ is:
The distribution of a random variable $X$ is given below. The value of $k$ is:
$X = x$ $-2$ $-1$ $0$ $1$ $2$ $3$ $P(X = x)$ $\frac{1}{10}$ $k$ $\frac{1}{5}$ $2k$ $\frac{3}{10}$ $k$
| $X = x$ | $-2$ | $-1$ | $0$ | $1$ | $2$ | $3$ |
| $P(X = x)$ | $\frac{1}{10}$ | $k$ | $\frac{1}{5}$ | $2k$ | $\frac{3}{10}$ | $k$ |
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