$A$ random variable $X$ has the following probability distribution:
| $x$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |
|---|---|---|---|---|---|---|---|---|
| $P(x)$ | $0.15$ | $0.23$ | $0.12$ | $0.10$ | $0.20$ | $0.08$ | $0.07$ | $0.05$ |
For the events $E = \{x \text{ is a prime number}\}$ and $F = \{x < 4\}$,the probability $P(E \cup F)$ is:
| $x$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |
|---|---|---|---|---|---|---|---|---|
| $P(x)$ | $0.15$ | $0.23$ | $0.12$ | $0.10$ | $0.20$ | $0.08$ | $0.07$ | $0.05$ |
For the events $E = \{x \text{ is a prime number}\}$ and $F = \{x < 4\}$,the probability $P(E \cup F)$ is:
- A$0.50$
- B$0.77$
- C$0.35$
- D$0.87$
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Similar Questions
The distribution of a random variable $X$ is given below. The value of $k$ is:
$X = x$ $-2$ $-1$ $0$ $1$ $2$ $3$ $P(X = x)$ $\frac{1}{10}$ $k$ $\frac{1}{5}$ $2k$ $\frac{3}{10}$ $k$
| $X = x$ | $-2$ | $-1$ | $0$ | $1$ | $2$ | $3$ |
| $P(X = x)$ | $\frac{1}{10}$ | $k$ | $\frac{1}{5}$ | $2k$ | $\frac{3}{10}$ | $k$ |
If $X$ is a Poisson variate such that $P(X=1) = 2P(X=2)$,then $P(X=3)$ is equal to:
DifficultTS EAMCET 2013
View SolutionThe probability distribution of a discrete random variable $X$ is given below. If $E(X^2) = \Sigma x^2 P(X=x)$, then $6 E(X^2) - \operatorname{Var}(X) =$
$X=x$ $-1$ $0$ $1$ $2$ $P(X=x)$ $\frac{1}{3}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{3}$
| $X=x$ | $-1$ | $0$ | $1$ | $2$ |
|---|---|---|---|---|
| $P(X=x)$ | $\frac{1}{3}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{3}$ |
For a Poisson variate $X$,if $P(X=2)=3 P(X=3)$,then the mean of $X$ is:
DifficultTS EAMCET 2003
View Solution$A$ random variable $X$ has the following probability distribution:
$X$ $0$ $1$ $2$ $P(X)$ $\frac{25}{36}$ $k$ $\frac{1}{36}$
If the mean of the random variable $X$ is $\frac{1}{3}$,then the variance is:
| $X$ | $0$ | $1$ | $2$ |
|---|---|---|---|
| $P(X)$ | $\frac{25}{36}$ | $k$ | $\frac{1}{36}$ |
If the mean of the random variable $X$ is $\frac{1}{3}$,then the variance is:
MediumKCET 2024
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