A random variable X takes the values 0, 1, 2. | vedclass

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(C) The random variable $X$ takes values $x_i \in \{0, 1, 2\}$.Given the mean $E(X) = \sum x_i P(x_i) = 1.2$.Substituting the values: $(0 	imes P(X=0

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$0.2$

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(C) The random variable $X$ takes values $x_i \in \{0, 1, 2\}$.Given the mean $E(X) = \sum x_i P(x_i) = 1.2$.Substituting the values: $(0 	imes P(X=0)) + (1 	imes P(X=1)) + (2 	imes P(X=2)) = 1.2$.Since $P(X=0) = 0.3$,we have: $0 + P(X=1) + 2P(X=2) = 1.2 \implies P(X=1) + 2P(X=2) = 1.2$ (Equation $1$).Also,the sum of probabilities is $1$: $P(X=0) + P(X=1) + P(X=2) = 1$.$0.3 + P(X=1) + P(X=2) = 1 \implies P(X=1) + P(X=2) = 0.7$ (Equation $2$).Subtracting Equation $2$ from Equation $1$: $(P(X=1) + 2P(X=2)) - (P(X=1) + P(X=2)) = 1.2 - 0.7$.$P(X=2) = 0.5$.Substituting $P(X=2) = 0.5$ into Equation $2$: $P(X=1) + 0.5 = 0.7 \implies P(X=1) = 0.2$.

$X = x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(x)$	$k$	$2k$	$3k$	$4k$	$4k$	$3k$	$2k$	$k$	$k$

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X = x)$	$K$	$3K$	$5K$	$7K$	$8K$	$K$

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

$A$ random variable $X$ takes the values $0, 1, 2$. Its mean is $1.2$. If $P(X=0)=0.3$,then $P(X=1)=$

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