## (Basic Statistics for Citizen Data Scientist)

# Basic Probability Concepts

**Definition 1**: Typically in the field of statistics we study data that results from **experiments**. An experiment can be considered to be a series of **trials**, each with a particular **outcome**. An **event** is a collection of outcomes corresponding to some result in the experiment. The number of outcomes in event *E* (i.e. the number of elements in set *E*) is written as |*E*|. The set of all possible outcomes is called the **sample space**, often designed *S*. An event is then simply a subset of the sample space. The **probability** *P*(*E*) of the event *E* is |*E*| / |*S*|, assuming *S* is not empty.

**Example 1**: Consider the simple experiment of tossing a coin twice. What is the probability that the coin comes up heads both time?

The sample space *S* = {HH, HT, TH, TT} and the required event *E* = {HH}. Thus the probability that the coin is heads both times is *P*(*E*) = |*E*| / |*S*| = ¼, or 25%.

**Observation**: We now state the fundamental properties of probability, using the usual set notation.

**Property 1**:

- 0 ≤
*P*(*A*) ≤ 1 *P*(Ø) = 0*P*(*S*) = 1*P*(*A′*) = 1 –*P*(*A*), where*A′ = S – A**P*(*A*∪*B*) =*P*(*A*) +*P*(*B*) –*P*(*A*∩*B*)

Proof: Simple consequences of Definition 1.

**Example 2**: Consider the experiment of drawing one card from a standard deck of 52 cards. What is the probability of drawing either a spade or face card?

There are 13 spades and 12 face cards, but 3 of these face cards are also spades, which we should not count twice. Thus, there are 13 spades and 9 non-spade face cards for a total of 22 cards out of 52. The probability is therefore 22/52. We now show how to calculate the result using Property 1e.

Let *A* = the event that a spade is drawn and *B* = the event that a face card (King, Queen or Jack) is drawn. *P*(*A*) = 13/52, *P(B)* = 12/52 and *P*(*A ∩ B*)= 3/52. Thus the probability of drawing either a spade or face card is *P*(*A *∪* B*)* = P*(*A*)* + P*(*B*)* – P*(*A ∩ B*) = 13/52 + 12/52 – 3/52 = 22/52.

**Definition 2**: The probability that an event *A* occurs assuming that event *B* occurs is called the **conditional probability** of *A* **given** *B* and is denoted *P*(*A|B*).

**Observation**: By Definitions 1 and 2

**Property 2**:

*P*(*A|B*) ∙*P*(*B*) =*P*(*A*∩*B*) =*P*(*B|A*) ∙*P*(*A*)*P*(*A|B*) =*P*(*B|A*) ∙*P*(*A*) /*P*(*B*) called**Bayes’ Theorem***P*(*A*) =*P*(*A|B*) ∙*P*(*B*) +*P*(*A|B′*) ∙*P*(*B′*) called the**Law of Total Probability**

Proof: The first assertion is a restatement of the last observation. The second assertion is a consequence of two applications of the first since

We now prove the third assertion. Since *A* = (*A*∩*B*) ∪ (*A*∩*B′*), by Properties 1b and 1e,

Now by Property 2a and 2b,

which proves the third assertion.

**Example 3**: Consider the experiment of picking two balls at random without replacement from a bag which contains 3 reds and 2 blacks. What is the probability that both balls are red?

Let *A* = a red ball is taken on the first draw and *B* = a red ball is taken on the second draw. The probability that the first draw is red is *P*(*A*) = 3/5. The probability that the second draw is red given the first draw is red is *P*(*B|A*) = 2/4 = ½. From Property 2a, we see that the probability that both draws are red is

**Definition 3**: Two events *A* and *B* are **independent** if *P*(*A*∩*B*)* = P*(*A*)* ∙ P*(*B*)

**Property 3**: Two events *A* and *B* are independent if and only if *P*(*A*) =* P*(*A|B*)

Proof: *A* and *B* are independent if and only if *P*(*A∩B*)* = P*(*A*)* ∙ P*(*B*), which by Property 2a is true if and only if *P*(*A|B*)* ∙ P*(*B*)* = P*(*A*)* ∙ P*(*B*), which in turn is true if and only if *P*(*A|B*)* = P*(*A*)*.*

**Observation**: *A* and *B* are independent if *B*’s occurring (or not occurring) has no influence on *A*’s occurring, i.e. it doesn’t increase or decrease the probability of *A* occurring. By Property 3, *A* and *B* are independent if any only if *P*(*B|A*)* = P*(*B*)*,* and so it also follows that if *A* and *B* are independent then *A*’s occurring has no influence on* B*’s occurring either.

**Example 4**: Repeat the experiment from Example 3, but this time we put the ball picked on the first draw back in the bag before drawing a second ball (i.e. **sampling with replacement**).

Since *P*(*B|A*) = 3/5 = *P*(*B*)*, A* and *B* are independent, it follows that

*P*(*A*∩*B*)* = P*(*A*)* ∙ P*(*B*) = 3/5 ∙ 3/5 = 36%.

**Example 5**: You have two bags, one containing 3 red and 2 black balls, the other containing 1 red, 1 blue and 2 black balls. You pick a bag at random and then pick a ball from that bag at random. What is the probability that the ball picked is red?

Let *A* = event that the first bag is picked and let *B* = event that a red ball is drawn. By Property 2c,

*P*(*B*)* = P*(*B|A*)* ∙ P*(*A*) +* P*(*B|A′*)* ∙ P*(*A′*) = .6(.5) + .25(.5) = 42.5%.

**Example 6**: Suppose you role a die 12 times. What is the probability that the number 1 will not appear on any of the throws? What is the probability that the number 1 will appear on at least one of the 12 throws?

The 12 throws represent 12 independent events. The probability of throwing a 1 on any single trial is 1/6 and so the probability of not throwing a 1 on any single trial is 1 – 1/6 = 5/6 (by Property 1d). Thus the probability of not throwing a 1 on any of the 12 throws is (5/6)^{12} = 11.2% (by Definition 3).

The probability that the number 1 will appear at least once is simply 1 – 11.2% = 88.8% (by Property 1d). This is equivalent to 1 – (1 – 1/6)^{12}.

## Statistics for Beginners – Basic Probability Concepts

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