Problem: Find probability of a plate having an A and a 1, no repetition allowed.
I haven't done these types of problems in a while, and originally did it assuming repetition was allowed before I noticed the fine print saying it isn't.
Plates are 3 letters followed by 4 numbers.
Just to show my thought process, originally when I assumed repetition was allowed, I had $(1-(25/26)^3)*(1-(9/10)^4)$ for getting a plate with at least an $A$ and a $1$ (1 – prob no A's AND no 1's).
For no repetition, I ended up with $(3/26)*(4/10) = 3/65$. If you get an A, the other 2 letters don't matter and there are 3 orderings. If you get a 1, the other 3 numbers don't matter and there are 4 orderings. but I'm not confident that's correct. Confirmation of that and/or help would be appreciated.
Edit: I will update and select an answer after I get some clarification on the problem in a few hours. Thank you both, you were extremely helpful!
Best Answer
There are $26^3 \cdot 10^4$ different plates, so: $$|\Omega| = 26^3 \cdot 10^4$$ or, if we assume that there are no repetitions within plates: $$|\Omega_1| = 26\cdot 25\cdot 24 \cdot 10\cdot 9\cdot 8\cdot 7$$ Now we want plates without repetitions, but with A and 1, so first we need to select two dofferent letters different than A (the order matters) and a place where we put A. Then we select 3 different numbers different than 1 and select a place, where we put 1. $$|A| = 25\cdot 24 \cdot {3\choose 1} \cdot 9\cdot 8 \cdot 7 \cdot {4\choose 1}$$ Finally the probability obtaining the license plate with A and 1 and without repetitions is : $$P(A)=\frac{|A|}{|\Omega|}=\frac{25\cdot 24 \cdot 3 \cdot 9\cdot 8 \cdot 7 \cdot 4}{26^3 \cdot 10^4}$$ or, for plates with no repetition, $$P(A)=\frac{|A|}{|\Omega_1|}=\frac{25\cdot 24 \cdot 3 \cdot 9\cdot 8 \cdot 7 \cdot 4}{26\cdot 25\cdot 24 \cdot 10\cdot 9\cdot 8\cdot 7} = \frac{3}{65}$$