Finding probability $P({X}^{2}<0)$.

Question

Probability density function of random variable $X$ is ${f}_{X}(x)=\frac{1}{2\theta}$ where $-\theta <x<\theta $ . Calculate probability $P({X}^{2}<0)$.

I'm little bit confused. I thought that ${X}^{2}$ is never less than $0$ so $P(\oslash )=0$, is it true? Any idea will be appreciated.

Answer 1

I solve the first part but the second part, I'm little bit confused

You are perfectly right.

as $X^2\geq 0$, $\forall X$ the requested probability is zero

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Perhaps there is a typo....you can calculate $P(X^2<\theta)=\theta^{-\frac{1}{2}}$

(this makes sense and the question looks similar to the one you posted)