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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}$.

$X = x$	$0$	$2$	$4$	$6$	$8$	$10$
$P(X = x)$	$0$	$k$	$2k$	$5k^2$	$2k^2$	$3k$

$X = x_i$	$0$	$1$	$2$	$3$	$4$
$P(X = x_i)$	$0.4$	$0.3$	$0.1$	$0.1$	$0.1$

$A$ 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

If the probability distribution of a random variable $X$ is given by the following table, then the mean of $X$ is:
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The probability distribution of a random variable $X$ is given by:
$X = x_i$ $0$ $1$ $2$ $3$ $4$
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If a discrete random variable $X$ takes the values $1, 2, 3, 4$ such that $2P(X = 1) = 3P(X = 2) = P(X = 3) = 5P(X = 4)$, then $P(X = 4) = ...$ (in $61$)

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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: