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$X$ is a random variable taking the values $3, 4 $and $12$ with probabilities $\large\frac{1}{3},\frac{1}{4}$ and $\large\frac{4}{7}.$ Then $E(X)$ is

\[\begin{array}{1 1}(1)5&(2)6\\(3)7&(4)3\end{array}\]

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$E(x)= 3 \bigg(\large\frac{1}{3} \bigg)$$ +4 \bigg( \large\frac{1}{4} \bigg)$$ + 7 \bigg( \large\frac{4}{7} \bigg)$
$\qquad= 1+1+4=6$
Hence 2 is the correct answer.
answered May 23, 2014 by meena.p
 

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