At a telephone enquiry system,the number of phone calls regarding relevant enquiries follows a Poisson distribution with an average of $5$ phone calls during $10$-minute time intervals. The probability that there is at most one phone call during a $10$-minute time period is:
- A$6e^{-5}$
- B$5e^{-5}$
- C$e^{-5}$
- D$4e^{-5}$
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If the probability distribution of a random variable $X$ is as follows,then $P(X \leq 2) = $ $x_i$ $0$ $1$ $2$ $3$ $4$ $P(X = x_i)$ $3K$ $5K$ $3k^2$ $4k^2 + k$ $3k^2$
| $x_i$ | $0$ | $1$ | $2$ | $3$ | $4$ |
|---|---|---|---|---|---|
| $P(X = x_i)$ | $3K$ | $5K$ | $3k^2$ | $4k^2 + k$ | $3k^2$ |
If the probability that an individual will suffer a bad reaction from an injection is $0.001$,then the probability that out of $2000$ individuals,exactly $3$ individuals suffer a bad reaction is
$A$ random variable $X$ takes the values $0, 1, 2, 3, \dots$ with probability $P(X=x) = k(x+1)\left(\frac{1}{5}\right)^x$,where $k$ is a constant. Then $P(X=0)$ is
$A$ random variable $X$ has the following probability distribution:
| $X$ | $1$ | $2$ | $3$ | $4$ | $5$ |
|---|---|---|---|---|---|
| $P(X)$ | $k^2$ | $2k$ | $k$ | $2k$ | $5k^2$ |
Then $P(X \geq 2)$ is equal to:
| $X$ | $1$ | $2$ | $3$ | $4$ | $5$ |
|---|---|---|---|---|---|
| $P(X)$ | $k^2$ | $2k$ | $k$ | $2k$ | $5k^2$ |
Then $P(X \geq 2)$ is equal to:
If the probability distribution function of a random variable $X$ is given as follows:
$X=x_i$ $-2$ $-1$ $0$ $1$ $2$ $P(X=x_i)$ $0.2$ $0.3$ $0.15$ $0.25$ $0.1$
Then $F(0)$ is equal to:
| $X=x_i$ | $-2$ | $-1$ | $0$ | $1$ | $2$ |
|---|---|---|---|---|---|
| $P(X=x_i)$ | $0.2$ | $0.3$ | $0.15$ | $0.25$ | $0.1$ |
Then $F(0)$ is equal to:
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