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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!

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    I think you are correct. – 2017-01-26

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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, $$P(A)=\frac{|A|}{|\Omega_1|}=\frac{25\cdot 24 \cdot 3 \cdot 9\cdot 8 \cdot 7 \cdot 4}{26\cdot 25\cdot 24 \cdot 10\cdot 9\cdot 8\cdot 7} = \frac{3}{65}$$

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    Think you made a mistake somewhere, I calculate that as $54,925/1,134$? – 2017-01-26
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    @Bob I think he swapped $|\Omega|$ and $|A|$ – 2017-01-26
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    @Bob sorry, my mistake. – 2017-01-26
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    Wait, your solution assumes repetition is allowed doesn't it? Without repetition, shouldn't the total number of plates be $((26*25*24)*3)((10*9*8*7)*4)$ shouldn't it? I made an edit to the problem description, I think it might've not been clear on my part. – 2017-01-26
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    Repetition is allowed within the whole set of plates, there was no restriction about it. Total number of plates without repetition wold be $26\cdot 25\cdot 24\cdot 10\cdot 9\cdot 8\cdot 7$ – 2017-01-26
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    I was totally thinking about this differently, as in the no repetition was part of the plate creation. Suddenly my confusion makes more sense. – 2017-01-26
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    @Bob I've edited my answer – 2017-01-26
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It seems like you got the correct answer but your explanation sounds flawed - after selecting one A, you cannot "not care" about the other two letters - you have to guarantee that they are different from A and each other.

For exactly one A, you simply ask yourself: How many possible valid license plates can I construct? There are three possible schemes for one A:

AXY XAY and XYA

where X denotes a non-A letter and Y is a non-A non-X letter. You cannot "not care" about the X - it has to be a non-A letter since otherwise it could possibly violate the "exactly one A" rule.

The A is fixed, you can only choose it in one way. The X can be selected in 25 different ways, the Y in 24 (neither A nor X) so you have $1 \times 25 \times 24 = 600$ possible ways of choosing three letters of which two are non-A and one is A. Now multiply this number by the total number of distinct orderings, $3$. You get $3 \times 600 = 1800$, which is the number of possible arrangements for all valid letter choices. Dividing this by the total number of orderings, $26 \times 25 \times 24$, gives us the probability of a random ordering conforming to the one A, two non-As rule. So the final probability is:

$$\frac{3 \times 25 \times 24}{26 \times 25 \times 24} = \frac{3}{26} \quad \text{or about} \quad 11.54\%$$

For the other case, just try the same strategy as above. (How many valid choices of numbers with exactly one 1 are there? How many ways can I arrange them? What is the total number of arrangements?) The final probability for it having exactly one A and exactly one 1 is the product of both probabilities.

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    The question says no repetition is allowed if I'm not mistaken. – 2017-01-26
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    I was thinking by 'repetition not allowed' that was a part of the plate creation. So I don't have to worry about plates with more than 1 A, because they won't exist. That's why I said the other 2 won't matter, because they definitely won't be A. – 2017-01-26
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    I'm now not actually sure which it is, having re-looked at the problem, the description isn't clear. Thanks for your explanation though, it was very helpful! – 2017-01-26
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    @Bob Either it is part of the plate creation or it isn't, in both cases only the denominator changes, not the numerator. In the case where only non-repeating plates are created, you got the correct solution since the terms cancel out nicely. In the other case you get a large fraction which is not guaranteed to yield a clean result. – 2017-01-26
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    @ThreeFx, Aren't there $3!=6$ possible orderings for one A? AXY, AYX, XAY, YAX, XYA, YXA. And perhaps, there's ${3\choose{1}}$, because, you can place A(think $r=1$ ball) in one of three slots(think $n=3$ cells). And $(25)_{2}$ for the other two slots. Please correct me, if that's wrong. :) – 2017-01-26
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    @Quasar That is correct, but for some values of X and Y, AXY and AYX are the same, which means swapping X and Y does not increase the number of permutations. – 2017-01-26
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    @Quasar Using Haskell, we can see that: ``[('a', x, y) | x <- ['b'..'z'], y <- ['b'..'z'], x /= y]`` contains the same elements as ``[('a', y, x) | x <- ['b'..'z'], y <- ['b'..'z'], x /= y]``. So changing the order does not yield any new permutations. – 2017-01-26
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    @ThreeFx, just a bit confused. We distinguish between a plate, starting with ABD and ADB right? – 2017-01-26
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    @Quasar Yes, but you can get both from the structure AXY. For ABD choose X = B and Y = D. For ADB choose X = D and Y = B. Now by reversing X and Y in both cases, we get what AYX would have yielded. Since it is exactly what AXY yielded, we do not have to distinguish between AXY and AYX. – 2017-01-26
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    @Quasar If we disregard the A for a moment, let's look at $XY$ and $YX$, where $X \not= Y$. You habe 25 choices for $X$ and 24 for $Y$, totaling $25 \times 24$. Looking at $YX$, you have $24 \times 25$ choices. But every choice $YX$ is actually covered by the corresponding $XY$ case. Let's fix $X = B$ for a moment and look at the case $Y = C$. $XY$ yields $BC$, while $YX$ yields $CB$. It is correct that $YX$ is different from $XY$ in this case, but we can still construct it for $X = C$. Since $X$ takes on all possible values, you can always transform specific choices for $YX$ into $XY$. – 2017-01-26
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    @ThreeFx, thanks for the elaborate reply. Yes, I understand that the chronological order of events does not matter, when using the multiplication rule. For example, $4$ shirts multiply into $3$ ties to yield ${4}\times{3}=12$ shirt-ties, whether we choose the shirts followed by ties or vice versa. – 2017-01-26