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What is Probability?


2023-09-13 |    0

You've probably come across the concept of probability before. Let's tackle some straightforward problems to refresh your memory.

Problem 1: What is the probability of choosing a yellow ball from a bin containing 5 yellow balls and 15 blue balls? The correct answer is 5 / (5 + 15) = 0.25. To find this, we divide the number of favorable outcomes (in this case, there are 5 ways to choose a yellow ball) by the total number of outcomes (the total number of balls in the bin).

Problem 2: In a bin, there are 2 red, 2 blue, and 4 white balls. What is the probability (P) of choosing either a red or a blue ball? Again, we need to find the ratio of favorable outcomes (F) to total outcomes (T). In this case, a favorable outcome means choosing a red or a blue ball. So, we have F = 2 + 2 = 4 and T = 2 + 2 + 4 = 8, which gives us P = 4 / 8 = 0.5.

Problem 3: What is the probability of rolling a prime and odd number on a 6-sided die?

 

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Let's list all possible outcomes when rolling a die: 1, 2, 3, 4, 5, 6. Now, recall that prime numbers have exactly two positive integer divisors. So, the prime numbers in our list are 2, 3, and 5, while the odd numbers are 1, 3, 5. A favorable outcome here is choosing a prime number that is also odd. In other words, favorable outcomes are elements that belong to both {2, 3, 5} and {1, 3, 5}. Thus, we have two favorable outcomes: 3 and 5. So, F = 2, T = 6, and P = 2/6 = 1/3.

As you may have noticed, P is a function where the argument is an event X. In Problem 1, event X is selecting a yellow ball, and we get P(X) = 0.25. In Problems 2 and 3, we have P(X) = 0.5 and P(X) = 1/3, respectively.

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In Problem 2, event X is selecting a red OR a blue ball. We can write X = Y OR Z, where Y is selecting a red ball and Z is selecting a blue ball. So we have P(Y OR Z) = 0.5. Similarly, in Problem 3, we have P(Y AND Z) = 1/3. This demonstrates that we can use OR and AND operations for events and then calculate probabilities. For instance, we may want to find something like P((X AND Y) OR (X AND Z)).

This brings us to the precise definition of an event (allowing us to use OR and AND operations for events). Surprisingly, we can define an event as simply a subset of a set. In mathematics, defining something using sets is a classical approach. Remember that the power set Pow(A) of a set A contains all subsets of A, including the empty set {} and A itself. For example, if A = {1,2}, then Pow(A) = {{}, {1}, {2}, {1,2}}. In probability theory, we can call Pow(A) the event space or probability space, and the elements of Pow(A) are events. In Problem 3, the probability space is Pow({1,2,3,4,5,6}), and the event of rolling an odd prime number is the subset {3,5}.

Now, for events, we can define OR and AND as simply the union and intersection of sets. It makes more sense to write P((X ∩ Y) (X ∩ Z)) instead of P((X AND Y) OR (X AND Z)). But what about probability? Now we can provide its precise definition. Recall that the cardinality |X| of a set X is the number of its elements. For instance, if X = {3,5,7}, then |X| = 3. Given a probability space Pow(A) and an event X (which must be a subset of A), we can define:

P(X) = |X| / |A|

Understanding this definition simplifies all calculations involving probability. For example, we can apply the following purely set-theoretic result to derive a formula for probability theory:

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P(X  Y) =

= |X  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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