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$\frac{1}{4}$
- B$\frac{1}{3}$
- C$\frac{1}{6}$
- D$\frac{1}{7}$
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Similar Questions
If a random variable $X$ has the following probability distribution values,then $P(X \geq 6)$ has the value:
$X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $P(X)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2 + k$
| $X$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ |
| $P(X)$ | $0$ | $k$ | $2k$ | $2k$ | $3k$ | $k^2$ | $2k^2$ | $7k^2 + k$ |
Which of the following can not 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 Probability $\omega_{1}$ $0.1$ $\omega_{2}$ $0.01$ $\omega_{3}$ $0.05$ $\omega_{4}$ $0.03$ $\omega_{5}$ $0.01$ $\omega_{6}$ $0.2$ $\omega_{7}$ $0.6$
| Outcome | Probability |
| $\omega_{1}$ | $0.1$ |
| $\omega_{2}$ | $0.01$ |
| $\omega_{3}$ | $0.05$ |
| $\omega_{4}$ | $0.03$ |
| $\omega_{5}$ | $0.01$ |
| $\omega_{6}$ | $0.2$ |
| $\omega_{7}$ | $0.6$ |
Easy
View Solution$A$ random variable $X$ has the following probability distribution:
$X$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $P(X=x)$ $0.15$ $0.23$ $0.12$ $0.10$ $0.20$ $0.08$ $0.07$ $0.05$
For the events $E = \{X \text{ is a prime number}\}$ and $F = \{X < 4\}$,find $P(E \cup F)$.
| $X$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |
|---|---|---|---|---|---|---|---|---|
| $P(X=x)$ | $0.15$ | $0.23$ | $0.12$ | $0.10$ | $0.20$ | $0.08$ | $0.07$ | $0.05$ |
For the events $E = \{X \text{ is a prime number}\}$ and $F = \{X < 4\}$,find $P(E \cup F)$.
Two dice are rolled. If a random variable $X$ denotes the sum of the numbers on them and $\mu$ denotes the mean of $X$,then $\mu+P(X < 5)+P(X>9)+P(X=7)=$
An unbiased coin is tossed $5$ times. Suppose that a variable $X$ is assigned the value $k$ when $k$ consecutive heads are obtained for $k=3, 4, 5$; otherwise,$X$ takes the value $-1$. Then the expected value of $X$ is:
DifficultJEE MAIN 2020
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