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

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(B) Let $P(X=0) = p_0, P(X=1) = p_1, P(X=2) = p_2, P(X=3) = p_3$.Given $p_2 = 0.3$ and $p_3 = 2p_1$.The sum of probabilities is $p_0 + p_1 + p_2 + p_3

Vedclass · Accepted Answer

$0.4$

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(B) Let $P(X=0) = p_0, P(X=1) = p_1, P(X=2) = p_2, P(X=3) = p_3$.Given $p_2 = 0.3$ and $p_3 = 2p_1$.The sum of probabilities is $p_0 + p_1 + p_2 + p_3 = 1$.$p_0 + p_1 + 0.3 + 2p_1 = 1 \implies p_0 + 3p_1 = 0.7 \implies p_0 = 0.7 - 3p_1$.The mean is $E(X) = \sum x_i p_i = 0(p_0) + 1(p_1) + 2(p_2) + 3(p_3) = 1.3$.$p_1 + 2(0.3) + 3(2p_1) = 1.3$.$p_1 + 0.6 + 6p_1 = 1.3$.$7p_1 = 0.7 \implies p_1 = 0.1$.Substituting $p_1$ in $p_0 = 0.7 - 3p_1$:$p_0 = 0.7 - 3(0.1) = 0.7 - 0.3 = 0.4$.

$X$	$1$	$2$	$3$	$4$	$5$
$P(X)$	$K$	$2K$	$2K$	$3K$	$K$

$A$ random variable $X$ takes the values $0, 1, 2, 3$ and its mean is $1.3$. If $P(X = 3) = 2P(X = 1)$ and $P(X = 2) = 0.3$, then $P(X = 0)$ is

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