If today's typical math or STEM undergrad is as sheltered and brainwormed as I was in 2012 2013 when I first took probability theory, the implications are going to take at least a year to sink in. The fact that health insurance is prominently featured in the question suggests it's very likely from a course or textbook for entry-level actuaries, or possibly even an easy practice question for the first actuarial exam. If it weren't for the ACA's implementation being such a hobbled Rube-Goldberg clusterfuck of a compromise from the initial betrayed promises of a universal single-payer system, it would have taken much longer to piece together that life/health actuaries were getting paid to do the devil's work.
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)}.
If today's typical math or STEM undergrad is as sheltered and brainwormed as I was in
20122013 when I first took probability theory, the implications are going to take at least a year to sink in. The fact that health insurance is prominently featured in the question suggests it's very likely from a course or textbook for entry-level actuaries, or possibly even an easy practice question for the first actuarial exam. If it weren't for the ACA's implementation being such a hobbled Rube-Goldberg clusterfuck of a compromise from the initial betrayed promises of a universal single-payer system, it would have taken much longer to piece together that life/health actuaries were getting paid to do the devil's work.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.