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(A) We know that the sum of all probabilities in a probability distribution is $1$,i.e.,$\sum P(X=x) = 1$.Given the distribution:$k^2 + 2k + k + 2k +

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$\frac{35}{36}$

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(A) We know that the sum of all probabilities in a probability distribution is $1$,i.e.,$\sum P(X=x) = 1$.Given the distribution:$k^2 + 2k + k + 2k + 5k^2 = 1$$6k^2 + 5k - 1 = 0$$(6k - 1)(k + 1) = 0$Since $P(X) \geq 0$,we must have $k > 0$,so $k = \frac{1}{6}$.Now,we need to find $P(X \geq 2)$:$P(X \geq 2) = P(X=2) + P(X=3) + P(X=4) + P(X=5)$$P(X \geq 2) = 2k + k + 2k + 5k^2 = 5k + 5k^2$Substituting $k = \frac{1}{6}$:$P(X \geq 2) = 5(\frac{1}{6}) + 5(\frac{1}{6})^2$$P(X \geq 2) = \frac{5}{6} + \frac{5}{36} = \frac{30 + 5}{36} = \frac{35}{36}$.

$X=x_i$	$-2$	$-1$	$0$	$1$	$2$
$P(X=x_i)$	$1/6$	$k$	$1/4$	$k$	$1/6$

$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$K$	$K + \frac{1}{7}$	$2K$	$\frac{2}{5}$

$A$ 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:

Similar Questions

$A$ random variable $X$ has the following probability distribution:

$X=x_i$ $-2$ $-1$ $0$ $1$ $2$

$P(X=x_i)$ $1/6$ $k$ $1/4$ $k$ $1/6$

The variance of this random variable is

For the random variable $X$ with the probability distribution given by the table:
$X = x$ $0$ $1$ $2$ $3$
$P(X = x)$ $K$ $K + \frac{1}{7}$ $2K$ $\frac{2}{5}$

The mean of $X$ is:

$A$ random variable $X$ has the following probability distribution:
$X = 1, P(X) = 0.15$
$X = 2, P(X) = 0.20$
$X = 3, P(X) = 0.25$
$X = 4, P(X) = 0.30$
$X = 5, P(X) = 0.10$
For the event $E = \{ X \text{ is a prime number} \}$ and $F = \{ X < 4 \}$, find $P(E \cup F)$.

If $P(X=x)=k\left(\frac{3}{8}\right)^{X}, x=1,2,3, \ldots$ is the probability distribution function of a discrete random variable $X$,then $k=$

$A$ coin is biased so that the head is $3$ times as likely to occur as tail. This coin is tossed until a head or three tails occur. If $X$ denotes the number of tosses of the coin,then the mean of $X$ is

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$A$ 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: