If we take 5 shots with a basketball, what's the probability that you make 2 given you make at least 1? Probability of making a shot is $\displaystyle \frac{3}{5}$.
So $2$ shots would be $\displaystyle \dbinom{5}{2} \left(\frac{3}{5}\right)^2 \left(\frac{2}{5}\right)^3$ right? 
 A: To find the probability in question, you have to first check what the probability of making at least $1$ shot is, and also the probability of making exactly $2$ shots, and then divide the latter by the former. That is, the probability is $$\frac{\text{(probability of landing 2 shots)}}{\text{(probability of landing at least 1 shot)}}$$

Probability of landing at least $1$ shot:
This is the same as $1$ minus the probability of missing all $5$ shots. The answer, therefore, is $$1 - \left(\frac{2}{5}\right)^5$$

Probability of landing exactly $2$ shots:
This is the answer that you had, i.e. $$\dbinom{5}{2} \left(\frac{3}{5}\right)^2 \left(\frac{2}{5}\right)^3$$

Final answer:
$$\frac{\dbinom{5}{2} \left(\frac{3}{5}\right)^2 \left(\frac{2}{5}\right)^3}{1-\left(\frac{2}{5}\right)^5} = \boxed{\frac{240}{1031}}$$
A: It is probability you make 2 and and least 1, which is just the probability you make 2, divided by the probability of making at least 1, which is 1 minus the probability of making none. So you have to divide your answer by that second quantity.
A: The other answers are OK, but you should keep in mind that it is supposed the probability of making two different shots are independent of each other. In real life, for most of the players, if you managed to make one shot, it influences the probability of making the following shots. For example, if the probability of making a single shot is 0.6, the probability of making 5 independent shots would be 0.07777, but the probability of making 5 consecutive shots might be much higher, depending on the player.
