2 Independent events probability

Probability is by far not my strong suit and I need to solve this question thoroughly. So if somebody could help me with this one:

A firm z is considering acquiring two different firms: firm A and firm B. It believes it will acquire firm A with probability 0.6 and firm B with probability 0.1. Assuming that those two events are independent, find the probability that Z acquires either firm.

• Welcome to Stackexchange. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – GNUSupporter 8964民主女神 地下教會 Oct 10 '18 at 18:53
• Try to consider every possibility: A and B, A but not B, B but not A, neither A nor B. Calculate their probabilities and figure out which add up to the requested probability. – SmileyCraft Oct 10 '18 at 18:54

Let $$A^c$$ be the event that the company will not acquire the firm A and $$B^c$$ analogously for firm B.
Then $$P(A^c \cap B^c) = P(A^c)P(B^c)=(1-0.6)(1-0.1)=0.4 \times 0.9=0.36$$ where the first equality follows from the independency.
$$\mathbf {Edit}:$$ I misread the question, but this is still a good hint for what you actually need to compute.
Let $$A$$ be the event acquiring firm A and $$B$$ be the event acquiring firm B. The probability of acquiring either firm is: $$\mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(A \cap B)$$ Since events A and B are independent $$\mathbb{P}(A \cap B) = \mathbb{P}(A) \cdot \mathbb{P}(B)$$. The rest is putting the numbers...