In a given game, when defeating an enemy, the player has a $16\%$ chance of getting the item [A] and a $32\%$ chance of getting the item [B]. Considering that in a phase the player will face 3 equal enemies, what are the chances of the player to get:
Thank you very much!
Call getting no item $C$. It is clear that $$\mathbb{P}(C) = 1 - \mathbb{P}(A) - \mathbb{P}(B) = 0.52.$$
Now, the only outcome that results in no items is $$C,C,C.$$ Since the individual draws are independent, we have that $$\mathbb{P}(C,C,C) = \mathbb{P}(C)\mathbb{P}(C)\mathbb{P}(C) = 0.52^3.$$ Now, the probability of getting any three items is $$\mathbb{P}(A\cup B,A\cup B,A\cup B) = \mathbb{P}(A\cup B)\mathbb{P}(A\cup B)\mathbb{P}(A\cup B).$$ Because $A$ and $B$ are mutually exclusive, we have that $\mathbb{P}(A\cup B) = \mathbb{P}(A)+\mathbb{P}(B) = 0.48.$ Hence, $$\mathbb{P}(A\cup B,A\cup B,A\cup B) = 0.48^3.$$
Now, getting $1$ item means that we get one of the following outcomes $$A\cup B,C,C\\C,A\cup B,C\\C,C,A\cup B.$$ I'll leave it to you to convince yourself that these events have the same probability which is $$\mathbb{P}(A\cup B,C,C) = 0.48\cdot0.52^2.$$ Therefore, the probability of getting exactly one item is $$3\left(0.48\cdot0.52^2\right).$$ Now, the last probability is the trickiest. First, we note there are several ways to get exactly one $A$ and one $B$ and one $C$. Yet, each will have probability $$\mathbb{P}(A,B,C) = \mathbb{P}(A)\mathbb{P}(B)\mathbb{P}(C) = 0.16\cdot0.32\cdot0.52.$$ Therefore, we just need to figure out how many different (equally likely and mutually exclusive) events yield one of each event. I will leave it to you to convince yourself that there are $3\cdot2\cdot1 = 3! = 6$ events that give us one of each type. Thus the probability of exactly one item $A$ and one item $B$ is $$6\left(0.16\cdot0.32\cdot0.52\right).$$