Probability

Probability theory is one of the most important branches of mathematics. The goal of probability is to examine random phenomena. While this may sound complicated, it can be better understood by looking at the definition of probability.
probability_formula_simple

Probability is the likelihood that something will happen. The image above shows the simplest equation for probability. Say, for example, you have a jar of bank notes.

Bank Note Amount
50 pounds 1
20 pounds 1
5 pounds 3
Total Amount 5

The probability of getting a 50 pound note is 1 divided by the total possible outcomes, 4, giving us 0.25 or 25%

Random Variable and Sample Space

To understand more complex notions of probability, it is necessary to understand the notation that is often used in probability theory. The table below summarizes some of the most important elements.

Notation Description Example
Random Variable A A variable whose outcomes are unknown A coin toss
Probability P() The likelihood of something occurring P(A)
Sample Space S={} All possible outcomes of a random variable
\cap AND The probability of event A AND B occurring P(A \cap B)
\cup OR The probability of event A OR B occurring P(A \cup B)

Independent Events

Two events are independent when the probability of the first event does not affect the probability of the second. This shouldn’t be confused with mutually exclusive events, which are events that cannot physically happen at the same time.

random_variable_probabilities

Let’s look at an example in the image above. Event A, selecting a red circle, is mutually exclusive from event B, selecting a blue circle. This is because it is not possible to select a red and blue circle at the same time.

If we pick a circle, but it back with the rest, then pick again, these two events are also independent. This is because picking a red circle, with probability 0.5, does not change in the next round because we’ve put the red circle back in and therefore have the same amount of circles as before.

independent_events_example dependent_probability_example3

 

Dependent Events

Dependent events, on the other hand, are event whose probability is affected by the outcome of another event. Let’s take the example from before but this time we do not replace the circle we pick from the group.

dependent_probability_example2

While the two events A and B are still mutually exclusive, they are now dependent events. If in the first pick we select a red circle, that leaves us with only 5 circles: 2 red and 3 blue. The probability of picking a red circle for a second time is now only 2/5.

dependent_probability_example

On the other hand, if you pick a blue circle the first time, that still leaves us with 5 circles, only now it would be: 3 red, 2 blue. The probability of picking a red circle for the second time would now be 3/5. In fact, whenever you’re picking without replacing, the events you’re studying will be dependent.

Definition Events Example
With Replacement When you’re selecting something and then putting it back Independent Selecting cards out of a deck with replacement
Without Replacement When you’re selecting something and then not putting it back Dependent Selecting cards out of a deck without replacement

Conditional Probability

Card selections are a classic example of events that have conditional probability. If something is conditional, it means that it is dependent on something else happening. For example, the probability that you will bring an umbrella with you is conditional on, or dependent on, the probability that it will rain outside.

conditional_probability_formula

independent_probability_formula

The formula above gives us the conditional probability of event A and B. The table below explains each element in the formula.

Description Example
P(A|B) The probability that event A will occur given that event B has already happened The probability of selecting a king given that a spade has already been selected
P(A \cap B) The probability of A and B occurring The probability of selecting a king and spade
P(B) The probability of B The probability of selecting a spade

In order to calculate the probabilities, you should know that a deck of cards has:

  • 4 kings
  • 13 spades

Probability
P(K \cap S) 1/52 = 0.019
P(S) 13/52 = 0.25
P(K|S) 0.077

Bayes’ Theorem

Notice that the formula for conditional probability only applies to dependent events that are not mutually exclusive. If two events are independent, then the probability that A occurs is not affected by the fact that event B has occurred. On top of that, if two events are mutually exclusive, there is no way for event A and B to occur together, which gives us a conditional probability of 0.

Mutually Exclusive Not Mutually Exclusive
Independent P(A|B) = 0

P(B|A) = 0

P(A|B) = P(A)

P(B|A) = P(B)

Dependent P(A|B) = 0

P(B|A) = 0

P(A|B) = \frac{P(B|A)P(A)}{P(B)}

P(B|A) = \frac{P(A|B)P(B)}{P(A)}

Another way of finding the conditional probability of events is with Bayes’ Theorem, which gives us the conditional probability of event A given B when we know the probability of event B given A.

bayes_theorem

Let’s try our King and Spade example with this formula, given that the probability of picking a Spade given a King has been picked is 0.25.

Probability
P(K) 1/52 = 0.077
P(S|K) 0.25
P(S) 13/52 = 0.25
P(K|S) 0.077

 

Problem 1

Find the conditional probability of P(A|B) given the following information. Please keep in mind this is fictitious data.

Description Probability
A Fatal illnesses are quite rare 0.005 = 0.5%
B Coughs are very common, resulting from a number of illnesses 0.3 = 30%
B|A Coughs that occur due to fatal illnesses are also quite rare 0.001 = 1%

Solution 1

We can calculate P(A|B), or the probability of having a fatal illness given that we have a cough using Bayes’ formula.

Description Probability
A Fatal illnesses are quite rare 0.005 = 0.5%
B Coughs are very common, resulting from a number of illnesses 0.3 = 30%
B|A Coughs that occur due to fatal illnesses are also quite rare 0.001 = 1%
A|B A fatal illness given that we have a cough \frac{P(A)P(B|A)}{P(B)} =

\frac{0.005*0.3}{0.001} = 0.002%

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Danica

Located in Prague and studying to become a Statistician, I enjoy reading, writing, and exploring new places.