The random variable X has a probability distribution P(X) of the following form, where k is some number : P(X) = k, if x=0 2k, if x=1 3k, if x=2 0, otherwise 1. Determine the value of k. 2. Find P(X < 2), P(X ≤ 2), P(X ≥ 2).

Probability distribution: x_i 0 1 2 P(x_i) k 2k 3k 1. P(X = 0) + P(X = 1) + P(X = 2) = 1 k + 2k + 3k = 1 6k = 1 k=1 6 2. P(X < 2) = P(X = 0) + P(X = 1) =k+2k=3k=3 1 6 =1 2 P(X 2) = P(X = 2) = 3k =3 1 6 =1 2

The random variable X has a probability distribution P(X) of the …

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The random variable $X$ has a probability distribution $P(X)$ of the following form, where k is some number :$ \text{P}(\text{X}) = \begin{cases} \text{k, }\overline{\text{if}\ \text{x}=0} \\ \overline{ 2\text{k, }\text{if}\ \text{x}=1}\\3\text{k, }\text{if}\ \text{x}=2\\0,\ \text{otherwise} \end{cases}$

Determine the value of $k.$
Find $P(X < 2), P(X ≤ 2), P(X ≥ 2).$

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Answer

Probability distribution:

$x_i$	$0$	$1$	$2$
$P(x_i)$	$k$	$2k$	$3k$

$P(X = 0) + P(X = 1) + P(X = 2) = 1$

$\Rightarrow k + 2k + 3k = 1 $
$\Rightarrow 6k = 1$
$\Rightarrow\ \text{k}=\frac{1}{6}$

$P(X < 2) = P(X = 0) + P(X = 1)$

$=\text{k}+2\text{k}=3\text{k}=3\times\frac{1}{6}=\frac{1}{2}$
$P(X \geq 2) = P(X = 2) = 3k $
$=3\times\frac{1}{6}=\frac{1}{2}$

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