Conditional probability for a random variable

Question

If I had a random variable, $X$, such that $P(X≥3) = m$ and $P(X\geq9) = n$ and wanted to show that $P(X<3 | X<9) = \frac{m-1}{n-1}$

I let $P(X<3) = 1 - P(X≥3) = 1 - m$

Then:

$P(X<3 | X<9) = \frac{P(X<3 \cap X<9)}{P(X<9)}$

$P(X<3 | X<9) = \frac{P(X<3) * P(X<9)}{P(X<9)}$

I'm left with the following:

$P(X<3 | X<9) = P(X<3)$

...which does not give me the proof required.

Does anyone know how I can prove the above statement?

Answer 1

You seem to be assuming that $(X<3)$ and $(X<9)$ are independent. But this is almost never the case. Indeed if you know that $X<3$ then necessarily $X<9$, right? This is the opposite of independence, one event making the other automatically true.

Because of that, the event $(X<3$ and $X<9)$ is defined redundantly, and simplifies to just $(X<3)$. This is what your teacher used.