Given below is the distribution of a random variable $X$ $\begin{array}{ccccc}\mathbf{X}=\mathbf{x} & 1 & 2…
Given below is the distribution of a random variable $X$
$\begin{array}{ccccc}\mathbf{X}=\mathbf{x} & 1 & 2 & 3 & 4 \\ \mathbf{P}(\mathbf{X}=\mathbf{x}) & \lambda & 2 \lambda & 3 \lambda & 4 \lambda\end{array}$
If $\alpha=\mathrm{P}(X < 3)$ and $\beta=\mathrm{P}(X>2)$, then $\alpha: \beta=$
2 : 5
3 : 4
4 : 5
3 : 7
Solution
For a distribution of random variable $x$,
$
\begin{aligned}
& \alpha=P\left(X^6 < 3\right)=P\left(X^6=1\right)+P\left(X^6=2\right)=\lambda+2 \lambda=3 \lambda \\
& \text { and } \beta=P\left(X^6>2\right)=P\left(X^6=3\right)+P\left(X^6=4\right) \\
& \quad=3 \lambda+4 \lambda=7 \lambda \\
& \therefore \quad \alpha: \beta=3: 7
\end{aligned}
$