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

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(D) 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 $p_0 + p_1 + p_2 + p_3 = 1$.

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(D) 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 $p_0 + p_1 + p_2 + p_3 = 1$. Substituting the values,$p_0 + p_1 + 0.3 + 2p_1 = 1 \Rightarrow p_0 + 3p_1 = 0.7 \dots (i)$. The mean is given by $E(X) = \sum x_i p_i = 0(p_0) + 1(p_1) + 2(p_2) + 3(p_3) = 1.3$. Substituting the values,$0 + p_1 + 2(0.3) + 3(2p_1) = 1.3$. $p_1 + 0.6 + 6p_1 = 1.3 \Rightarrow 7p_1 = 0.7 \Rightarrow p_1 = 0.1$. From equation $(i)$,$p_0 + 3(0.1) = 0.7 \Rightarrow p_0 + 0.3 = 0.7 \Rightarrow p_0 = 0.4$. Thus,$P(X=0) = 0.4$.

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X)$	$0$	$k$	$2k$	$2k$	$3k$	$k^2$	$2k^2$	$7k^2 + k$

Outcome	Probability
$\omega_{1}$	$\frac{1}{12}$
$\omega_{2}$	$\frac{1}{12}$
$\omega_{3}$	$\frac{1}{6}$
$\omega_{4}$	$\frac{1}{6}$
$\omega_{5}$	$\frac{1}{6}$
$\omega_{6}$	$\frac{3}{2}$

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

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