When you roll 2 dice there are 36 equally likely options, starting at double one, $(1,1)$, and going all the way to double six, $(6,6)$. Of these, there are 11 equally likely ways to roll a $5$, $(1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3),(5,4),(5,6)$, so the odds of rolling a 5 is $\frac{11}{36}$.
Similarly there are 18 equally likely ways to roll an odd number (I'll let you think about which combinations these are), so the odds of rolling an odd number are $\frac{18}{36}=\frac{1}{2}$.
At this point you may think the odds of rolling a $5$ or and odd number should be $\frac{11}{36}+\frac{18}{36}=\frac{29}{36}$, however this is not the case. Note that six of the rolls contain both a $5$ and are odd:
$(2,5), (4,5), (6,5), (5,2), (5,4),(5,6)$
If we just add the two probabilities, we'll count these situations twice, even though they're no more likely to occur than any other combination! In reality there are only $11+18-6=23$ dice rolls which contain a $5$ or are odd (see if you can list them), and so the odds of rolling either a $5$ or an odd combination is $\frac{11}{36}+\frac{18}{36}-\frac{6}{36}=\frac{23}{36}$.
This is a simple application of the "principle of inclusion and exclusion", which you may want to look up.
This is a fairly standard application of the binomial distribution, which is in fashion in games these days. It gives the probability of getting exactly $k$ successes in $n$ trials as $$
\binom n kp^k(1-p)^{n-k},
$$ where $p$ is the probability of a single trial succeeding.
In your case, the number of trials $n=X$, the number of dice that you’re rolling, and the probability of success $p$ is the probability of rolling $Z$ or higher, namely, $\frac{7-Z}6$.
Best Answer
The easiest way to figure out odd for events with the words "at least" is often to figure out the contrary. So here the contrary is : "you have no 1 on any of your 3 dice" and this probality is easy to figure out. It is : $\left(\frac{5}{6}\right)^3=\frac{125}{216}$
So the probality of your event is : $1-\frac{125}{216}=\frac{91}{216}$