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

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(D) We know that the sum of probabilities for all possible values of a random variable is $1$. Given $P(X=0) = 0.4$. Since $P(X=0) + P(X=1) + P(X=2) =

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$0.9$

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(D) We know that the sum of probabilities for all possible values of a random variable is $1$. Given $P(X=0) = 0.4$. Since $P(X=0) + P(X=1) + P(X=2) = 1$,we have $0.4 + P(X=1) + P(X=2) = 1$. This implies $P(X=1) + P(X=2) = 0.6$. Given $P(X=1) = P(X=2)$,let $P(X=1) = P(X=2) = p$. Then $p + p = 0.6 \Rightarrow 2p = 0.6 \Rightarrow p = 0.3$. Thus,$P(X=1) = 0.3$ and $P(X=2) = 0.3$. The mean of a random variable $X$ is given by $E(X) = \sum x_i P(X=x_i)$. $E(X) = (0 	imes P(X=0)) + (1 	imes P(X=1)) + (2 	imes P(X=2))$. $E(X) = (0 	imes 0.4) + (1 	imes 0.3) + (2 	imes 0.3)$. $E(X) = 0 + 0.3 + 0.6 = 0.9$.

$A$ random variable $X$ takes the values $0, 1$ and $2$. If $P(X=1)=P(X=2)$ and $P(X=0)=0.4$,then the mean of the random variable $X$ is

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