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 followin...
The only sensible question I can imagine here would be "compute the probability that $X>9$ given that $X≥3$". Or maybe you could do "compute the probability that $X<3$ given that $X≤9$", something like that.
If they garbled the question, there's little point trying to guess what they intended. sometimes you can reverse engineer the offered solution, but if things are really garbled badly, that might fail as well. You could just set your own question and solve that.
First of all you are making the serious mistake of assuming (X<3) and (X<9) are independent in your current derivation, which is very rarely the case. If you make this assumption you will always end up with $P(A | B) = P(A)$
Second of all the prompt doesn't make much sense in the current setting. Can you show us the original material?
Ok based on your teacher's answer I think the corrected question should be "If $X$ is such that $P(X≥3) = m$ and $P(X\geq 9) = n$, show that $P(X<3 | X<9) = \frac{m-1}{n-1}$".
So there were not one but two typos in the question.
Discussion on question by George Orwe…
Discussion on question by George Orwe…