Let's look at this from the ground up.
A sample space is the set of all possible outcomes of a random process.
An event is a subset of that sample space.
Let E be the event and let S be the sample space.
Then the Equally Likely Probability Formula is:
P(E) = the number of outcomes in E/ the number of outcomes in S or P(E) = N(E)/N(S)
(N(R) is just the number of elements in R)
Now just to clear it up, in set theory {A,B,C} is not equal to {A,B,C,0}. Let {A,B,C} bet set 1, and let {A,B,C,0} bet set 2.
Consider,
We randomly select one element from each sample space, then our possible outcomes are:
For set 1: A or B or C
For set 2: A or B or C or 0
That means for set 2 there is a one in four chance of 0 being selected. Put that in the context of our problem and that really does not make any sense.
Now conditional probability is the the probability of event K given that event L has already occurred.
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Consider this argument:
Our sample space is {H,T} with P(H)=N(H)/N(S) or P(H) = 1/2.
Let H equal the set {M1} and let T equal the set {M2, Tu}
Where,
M1 equals Monday and Heads
M2 equals Monday and Tails
Tu equals Tails and Tuesday.
So now our possible events are sets. Set H has one element and set T has two elements, each with a 50% chance of being selected. Then P(M2) or P(Tu), given the event T, by our Equally Likely Probability Formula is P(M2) = 1/4 and P(Tu) = 1/4. Given Tails she could be in P(M2) or P(Tu), so 1/4 + 1/4 = 1/2 therefore P(M2) + P(Tu) = 1/2 = P(T) = P(H).