Probability of Primitive Root (Mod 43) Probability of Primitive Root (mod 43)
Q: If you independently select 5 random integers from the reduced residue system (mod 43), What is the probability that at least one of your integers is a primitive root (mod 43)?
So I know that if I select a random integer from the reduced residue system (mod 43), the probability that my integer is a primitive root is (# or primitive roots)/ phi(43) or 12/42. 
I thought that since the 5 random integers were independently selected the probability would be (12/42)^5, but that not correct. Can someone please help me understand how to do this!
 A: My interpretation of the question is that you want to choose $5$ different primitive roots, therefore given that interpretation the answer should be:
$$\frac{12}{42} \times \frac{11}{41}\times \frac{10}{40}\times \frac{9}{39} \times\frac{8}{38}$$
I'm assuming that there are $12$ primitive roots in $\mathbb{Z}_{43}$ as you have said. (Sorry that I didn't leave this as a comment, I can't post a comment until I have 50 reputation scores).
Remark: I just checked that the number of primitive roots modulo $n$ (if any exists) is given by $\varphi(\varphi(n))$. The reason is that if $\mathbb{Z}_n$ has a primitive root, then it's cyclic and the number of primitive roots therefore are equal to the number of generators of this cyclic group. I thought some people might find this useful. In this case, $\varphi(\varphi(43))$ is $12$.
EDIT: 
The probability that I initially calculated gives you the probability of all the 5 selected integers being primitive!
You want at least one of them to be primitive. That's a different probability. Because there are several different cases that need to be handled, namely choosing 1 primitive and 4 non-primitives, 2 primitive and 3 non-primitives, et cetera.
The probability of choosing at least one primitive root is equal to the compliment of the event that you choose none!
In that case, you will have:
$$1-(\frac{30}{42}\times \frac{29}{41}\times \frac{28}{40}\times \frac{27}{39}\times \frac{26}{38})$$
