What is the Probability that 1 out of 7 letters is chosen out of 2 draws? So let's say we have the letters AAABBCD,
We write them down on cards and shuffle them,
Then we randomly pick 2 cards.
I would like to know the probability of at least 1 of those cards being A. So either the 1st or 2nd or both being A.
I'm trying to wrap my head around this.
In my mind it's the probability of choosing A twice added to the probability of choosing A first and the probability of choosing it second, so (3/7 * 1/3) + (3/7 * 2/3) + (4/7 * 1/2). 
Would this be correct?
Thank you in advance.

Yes, your calculations are correct.
Note: it could have perhaps been easier to have approached the problem indirectly instead of directly by instead calculating the probability of choosing no $A$'s.
Picking no $A$'s occurs with probability $\frac{4}{7}\cdot \frac{3}{6}=\frac{2}{7}$ implying picking at least one $A$ occurs with probability $1-\frac{2}{7}=\frac{5}{7}$.
Alternatively calculated, picking no $A$'s occurs with probability $\binom{4}{2}/\binom{7}{2}$ which again simplifies to $\frac{2}{7}$.
As a general suggestion, keep an eye out for whether calculating directly or calculating indirectly will require more arithmetic or complication and pick the easier route.
