$A$ box contains $8$ batteries, of which $3$ are defective. If a person randomly selects $2$ batteries from this box, find the probability distribution of the number of defective batteries $X$.
- A
$X = x$ $0$ $1$ $2$ $P(X = x)$ $\frac{10}{28}$ $\frac{15}{28}$ $\frac{3}{28}$ - B
$X = x$ $1$ $2$ $3$ $P(X = x)$ $\frac{10}{28}$ $\frac{15}{28}$ $\frac{3}{28}$ - C
$X = x$ $0$ $1$ $2$ $P(X = x)$ $\frac{15}{28}$ $\frac{10}{28}$ $\frac{3}{28}$ - D
$X = x$ $1$ $2$ $3$ $P(X = x)$ $\frac{15}{28}$ $\frac{10}{28}$ $\frac{3}{28}$
Explore More
Similar Questions
$A$ random variable $X$ has the following probability distribution:
$X = x_i$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $P(X = x_i)$ $10k$ $9k$ $8k$ $8k$ $6k$ $5k$ $4k$ $3k$ $k$
where $k$ is a real number. If $A = \{ x_i : x_i \text{ is a prime number} \}$ and $B = \{ x_i : x_i > 5 \}$ are two events, then $P(A \cup B) = $
| $X = x_i$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ |
| $P(X = x_i)$ | $10k$ | $9k$ | $8k$ | $8k$ | $6k$ | $5k$ | $4k$ | $3k$ | $k$ |
where $k$ is a real number. If $A = \{ x_i : x_i \text{ is a prime number} \}$ and $B = \{ x_i : x_i > 5 \}$ are two events, then $P(A \cup B) = $
$A$ 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 SolutionFor the probability distribution of a discrete random variable $X$ as given below,the mean of $X$ is:
$X = x$ $-2$ $-1$ $0$ $1$ $2$ $3$ $P(X = x)$ $\frac{1}{10}$ $K + \frac{2}{10}$ $K + \frac{3}{10}$ $K + \frac{3}{10}$ $K + \frac{4}{10}$ $K + \frac{2}{10}$
| $X = x$ | $-2$ | $-1$ | $0$ | $1$ | $2$ | $3$ |
|---|---|---|---|---|---|---|
| $P(X = x)$ | $\frac{1}{10}$ | $K + \frac{2}{10}$ | $K + \frac{3}{10}$ | $K + \frac{3}{10}$ | $K + \frac{4}{10}$ | $K + \frac{2}{10}$ |
If a drunkard person tries to take a step,then it will be a forward or backward step with probabilities $\frac{1}{4}$ and $\frac{1}{2}$ respectively,or he will remain in his 'as it is' position. If he tries to take a step $5$ times,then find the probability that he will be one step away from the initial position.
If $X$ is a random variable such that $P(X=-2)=P(X=-1)=P(X=2)=P(X=1)=\frac{1}{6}$ and $P(X=0)=\frac{1}{3}$,then the mean of $X$ is
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