You've probably heard about probability. Let us solve some easy-peasy problems so you could recall basic moments.
Problem 1. What is probability to choose a yellow ball if there are 5 yellow balls and 15 blue balls in a bin? The right answer is 5 / (5 + 15) = 0.25. Here we divide the number of favourable outcomes (there are 5 ways to choose a yellow ball we need) by the total number of outcomes (the number of balls in our bin).
Problem 2. There are 2 red, 2 blue, and 4 white balls in a bin. What is probability P to choose a red OR a blue ball? Again, we have to find the ratio of favourable outcomes F and total outcomes T. This time a favourable outcome is to choose a red or a blue ball. Thus we have F = 2 + 2 = 4. Then T = 2 + 2 + 4 = 8 and P = 4 / 8 = 0.5.
Problem 3. What is probability of rolling a prime AND odd number for a 6 sided dice?
Let us right down all possible outcomes when rolling a dice: 1, 2, 3, 4, 5, 6. Now recall that a prime number has exactly two positive integer divisors. Thus the prime numbers in the list above are 2, 3, and 5. Also, our odd numbers are 1, 3, 5. The favourable outcome is to choose a prime number which is odd. In other words, favourable outcomes are all elements lying both in {2, 3, 5} and {1, 3, 5}. Thus we have two favourable outcomes: 3 and 5. Now F = 2, T = 6, P = 2/6 = 1/3.
As you may note, P is a function while its arguments are some events X. Say, the event X in Problem 1 is taking a yellow ball and we get P(X) = 0.25. On the other hand, P(X) = 0.5 and P(X) = 1/3 in Problem 3 and Problem 3, respectively.
Note that in Problem 2, X is taking a red OR a blue ball. Thus we could write X = Y OR Z, where Y is taking a red ball and Z is taking a blue ball. Thus we heve P(Y OR Z) = 0.25. Similarly, we have P(Y AND Z) = 1/3 in Problem 33. This shows that one may wish to use operations OR and AND for events and then calculate P. Say, we may wish to find something like P((X AND Y) OR (X AND Z)).
This leads us to the precise meaning of an event (so we can use operations OR and AND for events). Surprisingly, we can define an event just as a subset of some set. In fact, this is a classical approach in math to define something via sets. Recall that the power set Pow(A) of a set A is the set of all subsets of A including the empty set {} and A itself. Say, if A = {1,2}, then Pow(A) = {{},{1},{2},{1,2}}. In probability theory, we may call Pow(A) the event space or probability space while elements of Pow(A) are events. The probability space in Problem 3 is Pow({1,2,3,4,5,6}) while the event of rolling an odd prime number is the subset {3,5}.
Now for events, we can define OR and AND simply as the union and intersection of sets. So it makes more sense to write P((X ∩ Y) U (X ∩ Z)) instead of P((X AND Y) OR (X AND Z)). But what about probability? Actually, now we can give its precise definition. Recall that the cardinality |X| of a set X is just the number of its elements. For example, if X = {3,5,7}, then |X| = 3. Now, given a probability space Pow(A) and event X (which must be a subset of A), we just put
P(X) = |X| / |A|
Understanding of this definition simplifies all calculations concerning probability. Say, we may apply the following purely set-theoretic result to get a formula for probability theory:
P(X U Y) =
= |X U Y| / |A| =
= (|X| + |Y| - |X ∩ Y|) / |A| =
= |X|/|A| + |Y|/|A| - |X ∩ Y|/|A| =
= P(X) + P(Y) – P(X ∩ Y)
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