Hi, I'm an amateur so I'm sorry if this is something well known and uninteresting. Is 1 + 1 = 1 in probability theory?:
Consider tossing a coin and throwing a dice. Let the set of all possible outcomes for the coin be C. which implies p(C) = 1. Let the set of all possible outcomes for the dice be D, which implies p(D) = 1. Now p(C∪D) which is the probability that either the events D or C occur is also 1.
Here's the interesting bit: C and D are disjoint sets and therefore p(C∪D)= p(C) + P(D) which implies 1 = 1 + 1.
But then I started having doubts because I made some unproved assumptions such as p(C) and p(D) and so on, are actually defined in such a situation as this and whether C and D are truly disjoint.
Help!
The question is indeed about your sample space. If you throw either dice or a coin but you do not know (or do not specify) which then the sample space is
{H, T, 1, 2, 3, 4, 5, 6}
so that P(C) = 1 and P(D) = 1 are both false.
If you throw both a dice and a coin then the sample space is
{H, T} × {1, 2, 3, 4, 5, 6}
in which case the events C and D are simply not defined.
If you just throw a coin then certainly P(C) = 1. If, in another experiment, you throw a dice then, too, P(D) = 1. But in this case the event C∪D is undefined because the events C and D do not belong to the same space.
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