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(C) The sum of probabilities in a probability distribution must be equal to $1$.$\sum P(X=x) = P(X=0) + P(X=1) + P(X=2) = 1$$(4k - 10k^2) + (5k - 1) +

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$\frac{8}{9}$

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(C) The sum of probabilities in a probability distribution must be equal to $1$.$\sum P(X=x) = P(X=0) + P(X=1) + P(X=2) = 1$$(4k - 10k^2) + (5k - 1) + 3k^3 = 1$$3k^3 - 10k^2 + 9k - 2 = 0$By testing values,we find that $k = \frac{1}{3}$ is a root:$3(\frac{1}{27}) - 10(\frac{1}{9}) + 9(\frac{1}{3}) - 2 = \frac{1}{9} - \frac{10}{9} + 3 - 2 = -1 + 1 = 0$Factoring the polynomial,we get $(k - \frac{1}{3})(3k^2 - 9k + 6) = 0$,which simplifies to $(k - \frac{1}{3})(k - 1)(k - 2) = 0$.If $k = 1$,$P(X=0) = 4(1) - 10(1)^2 = -6$,which is impossible as probability cannot be negative.If $k = 2$,$P(X=0) = 4(2) - 10(4) = 8 - 40 = -32$,which is impossible.Thus,$k = \frac{1}{3}$ is the only valid value.We need to find $P(X $P(X Substituting $k = \frac{1}{3}$:$P(X < 2) = 9(\frac{1}{3}) - 10(\frac{1}{9}) - 1 = 3 - \frac{10}{9} - 1 = 2 - \frac{10}{9} = \frac{18 - 10}{9} = \frac{8}{9}$.

$X = x$	$0$	$1$	$2$
$P(X = x)$	$4k - 10k^2$	$5k - 1$	$3k^3$

$X$	$1$	$2$	$3$	$4$	$5$
$P(X = x)$	$k$	$\frac{k}{3}$	$\frac{k}{4}$	$\frac{k}{2}$	$\frac{k}{2}$

$A$ random variable $X$ has the following probability distribution:
$X = x$ $0$ $1$ $2$
$P(X = x)$ $4k - 10k^2$ $5k - 1$ $3k^3$

Then $P(X < 2)$ is:

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$A$ random variable $X$ has the following probability distribution:$X = x$$0$$1$$2$$P(X = x)$$4k - 10k^2$$5k - 1$$3k^3$Then $P(X < 2)$ is: