Engineers presented 3-point, single-link robotic system

I cannot even understand what the question is asking!!!!!!

Engineers collaborated on research involving a robot-sensor system in an unknown environment. As an example, the engineers presented the three-point, single-link robotic system shown in the accompanying figure. Each point (A,B, or C) in the physical space of the system has either an "obstacle" status or a "free" status. There are two single links in the system: A-B and B-C. A link has a free status if and only if both points in the link are "free". Otherwise the link has an "obstacle" status. Of interest is the random variable Y, the total number of links in the system that are "free".

a)list the possible values of Y for the system
b)the researchers stated that the probability of any point in the system having a "free" status is 0.5. Assuming the three points operate independently, find the probability distribution for y.

And then the problem gives these three boxes in a L shape:

C
B A

any any help at all would be appreciated!!!!!!!! I do not even know where to begin

Re: random variable word problem

You have a random variable \(\displaystyle Y\) which takes on the values \(\displaystyle 0,1,2\). The key here is that the points are independent, but the links depend very much on \(\displaystyle B\). Let \(\displaystyle A,B,C\) take on values of \(\displaystyle 1\) if free and \(\displaystyle 0\) if not free. Then \(\displaystyle P(Y=0) = P(B=0)+P(B=1,A=0,C=0)\). Likewise, \(\displaystyle P(Y=1) = P(A=1,B=1,C=0)+P(A=0,B=1,C=1)\). Finally, \(\displaystyle P(Y=2)=P(A=1,B=1,C=1)\).

Re: random variable word problem

Thanks. I think I under when y=1 and 2, although I am still not sure as to why
P(y=0)= P(b=0)+ P(b=1, a=0, c=0)
Is it even possible for B to be free? And why would you add the two together?

Re: random variable word problem

Point B is free with probability 1/2. You add together probabilities of disjoint events. If both links are not free, there are two distinct ways that can happen. FIrst, if B is not free, it doesn't matter what the states of A and C are. Second, if B is free, then A and C are not free.