probability of choosing an object at least once in $3$ drawings So there are $4$ objects in total, and I want to know the probability of choosing object A at least once in $3$ drawings. What I did was add  $3C1 + 3C2 + 3C3$ and got $7$ and, as a final answer, I got $7/64$, since there is a $4^3$ samples space. Why doesn't this method work?

You’re not taking into account the different possibilities for the choices that aren’t A. It’s true that if you pick A exactly once, there are $\binom31=3$ ways to pick which of the $3$ choices is the A, but in each case there are $3\cdot3=9$ possible ways to make the other two choices: each can be any of the other $3$ outcomes. That’s a total of $3\cdot9=27$ different sequences of $3$ choices having exactly one A. Similarly, if exactly $2$ of the choices are A, there are $\binom32=3$ ways to choose which $2$ choices they are, but the remaining choice can be any of the other $3$ possible outcomes. Thus, there are $3\cdot3=9$ different sequences of $3$ choices having exactly two As. Finally, there is a single sequence having A every time, so there are altogether $27+9+1=37$ sequences having at least one A, and the probability of such a sequence is therefore $\frac{37}{64}$.
