$A$ random variable $X$ has the following probability distribution. Then,$P(2 \leq X < 5) = $ . . . . . .
$X = x$ $1$ $2$ $3$ $4$ $5$ $6$ $P(X = x)$ $K$ $3K$ $5K$ $7K$ $8K$ $K$
| $X = x$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ |
| $P(X = x)$ | $K$ | $3K$ | $5K$ | $7K$ | $8K$ | $K$ |
- A$\frac{3}{5}$
- B$\frac{7}{25}$
- C$\frac{23}{25}$
- D$\frac{24}{25}$
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If the p.m.f. of a random variable $X$ is given by $P(X=x) = \frac{\binom{5}{x}}{2^{5}}$ for $x = 0, 1, 2, \ldots, 5$ and $0$ otherwise,then which of the following is not true?
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Medium
View Solution$A$ 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 $P(X=2)=0.3, P(X=3)=0.4, P(X=4)=0.3$,then the variance of random variable $X$ is (in $.6$)
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