Counting problem in conditional probability exercise

I want to solve the following problem:

A real estate agent has 8 master keys to open several new homes. Only one master key will open any of the houses. If 40% of these are usually left open, what is the probability that the real estate consultant will open any of the houses?

So, we can define the events:

A:The real estate consultant enters the house B:The house is open C:The house is closed D: The key is correct

And note that

$$P(A)=P(B) \cup P(C \cap D)$$ And for the independence

$$P(A)=P(B) \cup P(C) \cap P(D)$$

Note that $$P(A)=0.4$$ and $$P(C)=1-P(A)=0.6$$

But my question is how to do the counting to find $$P(D)$$, this is because it is not specified how many keys the consultant is trying, because for example if it were for three keys the count is easy

$$\frac{\binom{1}{1} \binom{7}{2}}{\binom{8}{3}}$$

Any suggestions?, I'll apreciatte it

• Please add the self-study tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. Please make these changes as just posting your homework & hoping someone will do it for you is grounds for closing. Mar 8 at 13:52