Checking counting/probability question

[half pelican]

Just wanna make sure these are correct!

[meager quest]

b and c are wrong

[half pelican]

i had a feeling

can you point out where at least pls? not entirely sure

[meager quest]

P(Z) = 1-P(~Z)
P(~Z) = 25^3 / 26^3

you can figure out the remaining problems similarly

That said, you're intuition isn't far off. You're just estimating instead of counting the way you did it

[half pelican]

hmmm

any chance you can elaborate a bit on why mine is wrong

like why do we do the Z part with the compleemnt

oh it's not counting everything

im only considering Z appearing in one of the indices but it could appear in 2 or 3

[half pelican]

is the 8 part correct at least?

c) was abysmally wrong lmfao

[meager quest]

counting ways to make license plates without a Z is easier
P(Z) = 1-P(~Z)

no, 8 part is wrong in the same way
Count(~8)/Count(total)

[half pelican]

a bit ugly but i think this is right

[buoyant crest]

The main mistake I see is that you are adding the probabilities, when you shouldn't. 1/26+1/26+1/26 = 3/26
If we exaggerated it, for example instead of 3 letters having 26
26/26=100%
or if we had 52 letters
52/26 =200%
Which is not true, even with large quantities there is the possibility of not obtaining a Z.

[half pelican]

that makes sense ye

on a different note, if you dont mind

this is easy enough to see algebraicallyu but is there a way to explain it by counting sets

[meager quest]

algebraic is what you should do for the proof.

try to do the 8s like I did the Zs

6/10 is wrong

[buoyant crest]

If you are asking for an example.
As if you have 52 cards and you take 5, and after those 5 you take two, it would be equal to
You take 2 from 52 cards, and from the remaining 50, you take another 3.
Or You take 3 from 52 cards, and from the remaining 49, you take another 2.

In all cases you end up with a group of 3 cards and one of 2

[half pelican]

that makes sense ty