# 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

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

## 1 Answer

$\large\frac{\lambda}{6} +\frac{\lambda}{4} +\frac{\lambda}{12}$$=1 \large\frac{4 \lambda + 6 \lambda+ 2 \lambda}{24}$$=1$

1 answer