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Q)

A random variable $X$ has the following probability mass function as follows:\[\]$\begin{array} {llllllll} \textbf{X:}& -2& 3 & 1 \\ \textbf{P(X=x):}& \large\frac{\lambda}{6}& \large\frac{\lambda}{4} &\large\frac{\lambda}{12} \end{array}$\[\] Then the value of $\lambda$ is

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

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A)
$\large\frac{\lambda}{6} +\frac{\lambda}{4} +\frac{\lambda}{12}$$=1$
$\large\frac{4 \lambda + 6 \lambda+ 2 \lambda}{24}$$=1$
$\qquad =\large\frac{12 \lambda}{24}$$=1$
$ \lambda=2$
Hence 2 is the correct answer.
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