Conditional probability refresher

Probability is always…

A number between 1 (certain) and 0 (impossible):

$1 \leq P \geq 0$

The probability experiment

This is process that can be repeated and in which the outcome will be uncertain, for example rolling a dice, tossing a coin, or selecting a card from a pack.

The sample space

This is the set of all possible outcomes of a probability experiment, written in set notation which is denoted by Ω or S.

When rolling a dice the sample space is:

$\Omega = \{1, 2, 3, 4, 5, 6\}$

When tossing a coin the sample space is:

$\Omega = \{H, T\}$

Note how tossing two coins can result in different sample space scenarios, depending on whether you care about the order or not.

For tossing a coin where order is important the sample space is:

$\Omega = \{HH, HT, TH, TT\}$

Whereas for tossing a coin where order is not important the sample space is, because in this scenario HT and TH both amount to ‘one of each’ which is all we care about:

$\Omega = \{HH, HT, TT\}$

Discrete vs Continuous sample spaces

Discrete means I can count it, even if there are a LOT of values. For example shoe sizes are discrete – I could be a 6 or a 6.5 or a 7… there are only so many options.

Continuous means I can’t count it, it could be almost anything. Foot sizes are continuous – my foot length could be 24cm, but the next person might be 24.1 and the next 24.10456 → ∞.

Events

An event is a subset of a sample space – usually represented by some capital letter, for example P(A) represents the probability of event A occurring – where A might represent ‘getting tails on a coin toss. There are many different types of events:

Elementary event

One item selected from a sample space, where the probability is calculated as:

$P(A) = \frac{1}{\Omega}$

For example, we might ask, if a dice is thrown once what is the probability of a 5?

$P(A) = \frac{1}{6} = 0.167$

Impossible event

No items can be selected from the sample space. The probability = 0.

Certain event

All items are selected from the sample space. The probability = 1.

Independent events

Two or more events occur, where the outcome of the first event does not affect the outcome of the next event, and so on – think “and”, which is calculated by:

$P(A \cap B) = P(A) \times P(B)$

For example, a card is selected from a deck of 52 cards, it is then put back in the deck and a second card is selected. What is the probability of choosing a king and then a nine?

$P(king) = \frac{4}{52}$

$P(nine) = \frac{4}{52}$

$P(king \cap nine) = \left(\frac{4}{52}\right) \times \left(\frac{4}{52}\right) = \frac{1}{169} = 0.006$

Dependent events

Two or more events occur, where the outcome of the 1st event affects the outcome of the next event, and so on – again think “and”, but this one is calculated as follows – where P(B|A) read as “the probability that B happens, given that A has already occurred”:

$P(A \cap B) = P(A) \times P(B|A)$

For example, a card is selected from a deck of 52 cards, it is kept to one side and a second card is selected. What is the probability of choosing a king and then a nine?

$P(\text{king}) = \frac{4}{52}$

$P(\text{nine} \mid \text{king}) = \frac{4}{51}$

$P(\text{king} \cap \text{nine}) = \left(\frac{4}{52}\right) \times \left(\frac{4}{51}\right) = \frac{4}{663} = 0.006$

In the above scenario the size sample space when calculating the probability of getting a nine is now 51, since the first card drawn has been removed from the pack.

Mutually exclusive events

Two or more events, where the outcome cannot be can only be one of the identified events – think “or”, which is calculated by:

$P(A \cup B) = P(A) + P(B)$

For example, a dice is thrown once, what is the probability of a 2 or a 5? (Clearly, in a single role of a die you could never get both a 2 and a 5!)

$P(A - \text{throwing a 2)} = \frac{1}{6}$

$P(B - \text{throwing a 5)} = \frac{1}{6}$

$P(A \cup B) = \frac{1}{6} + \frac{1}{6} = \frac{1}{3} = 0.333$

Non-mutually exclusive events

Two or more events, where the outcome of the first may have an impact on the second – think “or”, which is calculated by:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

For example, a card is selected from a deck of 52 cards, and we want to know what is the probability of choosing a king or a club? (What we want to avoid here is counting the probability of the king of clubs twice!)

$P(A - \text{drawing a king)} = \frac{4}{52}$