Also for the nerds who give a shit about how to answer this:
a) The sample space, which we'll denote S, can be denoted as the Cartesian product of the discrete sets {0,1} and {g,f,s}, single-dimensional spaces with respect to the discrete random variable X which is one of 0 or 1 and corresponds to the "has insurance" dummy, and the discrete categorical random variable Y which can be one of g,f, or s, corresponding to the patient's condition. There are 6 elements corresponding to distinct possible outcomes in this product set (denoted as ordered pairs, e.g. (1,g)) for each sampled individual in this event space. (If this is homework for a university class and you have a strict grader you may have to actually list all 6 ordered pairs.)
b) The event A is the subset of S for which Y=s: {(0,s),(1,s)}.
c) The event B is the subset of S for which X=0: {(0,g),(0,f),(0,s)}.
d) The event (B^c U A) is the union of the complement of B and A. This is the set for which patients are either insured (since "not B" would have to be true) OR in serious condition (since "A" would have to be true). A equals {(0,s),(1,s)} and B's complement equals {(1,g),(1,f,),(1,s)}. The union of these sets contains every scenario in both of these events, and since (1,s) belongs to both sets it gets counted twice in probability calculations unless you subtract the intersection from the sum of the probabilities of each event. Thus (B^c U A) equals {(0,s),(1,g),(1,f),(1,s)}.
Also for the nerds who give a shit about how to answer this:
a) The sample space, which we'll denote S, can be denoted as the Cartesian product of the discrete sets {0,1} and {g,f,s}, single-dimensional spaces with respect to the discrete random variable X which is one of 0 or 1 and corresponds to the "has insurance" dummy, and the discrete categorical random variable Y which can be one of g,f, or s, corresponding to the patient's condition. There are 6 elements corresponding to distinct possible outcomes in this product set (denoted as ordered pairs, e.g. (1,g)) for each sampled individual in this event space. (If this is homework for a university class and you have a strict grader you may have to actually list all 6 ordered pairs.)
b) The event A is the subset of S for which Y=s: {(0,s),(1,s)}.
c) The event B is the subset of S for which X=0: {(0,g),(0,f),(0,s)}.
d) The event (B^c U A) is the union of the complement of B and A. This is the set for which patients are either insured (since "not B" would have to be true) OR in serious condition (since "A" would have to be true). A equals {(0,s),(1,s)} and B's complement equals {(1,g),(1,f,),(1,s)}. The union of these sets contains every scenario in both of these events, and since (1,s) belongs to both sets it gets counted twice in probability calculations unless you subtract the intersection from the sum of the probabilities of each event. Thus (B^c U A) equals {(0,s),(1,g),(1,f),(1,s)}.
Kk you're in charge of rational production after the revolution
Lol there's no way I'm living past the boogaloo phase of the revolution if it even succeeds in our lifetimes.