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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.

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    Please read [this post](http://meta.math.stackexchange.com/a/9960) and the others there for information on writing a good question for this site. In particular, people will be more willing to help if you [edit] your question to include some motivation, and an explanation of your own attempts. – 2017-02-26

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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.

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    Thank you, I'm trying to help a friend and probability isn't my strong suit. I'll be sure to pass on the advice! – 2017-02-26