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:
| $X = x$ | $0$ | $1$ | $2$ | $3$ | $4$ |
| $P(X = x)$ | $k$ | $2k$ | $4k$ | $2k$ | $k$ |
Then the value of $P(X \leq 2)$ is:
- A$\frac{1}{10}$
- B$\frac{7}{10}$
- C$\frac{3}{10}$
- D$\frac{9}{10}$
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The cumulative distribution function (c.d.f.) $F(x)$ of a discrete random variable $X$ is given by the following table:
$X$ $-3$ $-1$ $0$ $1$ $3$ $5$ $7$ $9$ $F(X=x)$ $0.1$ $0.3$ $0.5$ $0.65$ $0.75$ $0.85$ $0.90$ $1$
Then,find the value of $\frac{P[X=-3]}{P[X < 0]}$.
| $X$ | $-3$ | $-1$ | $0$ | $1$ | $3$ | $5$ | $7$ | $9$ |
|---|---|---|---|---|---|---|---|---|
| $F(X=x)$ | $0.1$ | $0.3$ | $0.5$ | $0.65$ | $0.75$ | $0.85$ | $0.90$ | $1$ |
Then,find the value of $\frac{P[X=-3]}{P[X < 0]}$.
$A$ bag contains $4$ white and $6$ black balls. Three balls are drawn at random from the bag. Let $X$ be the number of white balls among the drawn balls. If $\sigma^{2}$ is the variance of $X$,then $100 \sigma^{2}$ is equal to.
DifficultJEE MAIN 2022
View Solution$A$ random variable $X$ has the following distribution.
$X = x_{i}$ $-2$ $-1$ $0$ $1$ $2$ $3$ $P(X = x_{i})$ $0.1$ $k$ $0.2$ $2k$ $3k$ $k$
Then the variance of this distribution is
| $X = x_{i}$ | $-2$ | $-1$ | $0$ | $1$ | $2$ | $3$ |
| $P(X = x_{i})$ | $0.1$ | $k$ | $0.2$ | $2k$ | $3k$ | $k$ |
Then the variance of this distribution is
Two balls are drawn at random from a bag containing $5$ black balls and $3$ white balls. If the random variable $X$ denotes the number of white balls drawn, then the mean of $X$ is
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 $\omega_1$ $\omega_2$ $\omega_3$ $\omega_4$ $\omega_5$ $\omega_6$ $(a)$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$
| Outcome | $\omega_1$ | $\omega_2$ | $\omega_3$ | $\omega_4$ | $\omega_5$ | $\omega_6$ |
| $(a)$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ |
Easy
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