Probability Rules

I object to the idea that math is a bunch of "rules" but in this case the IB has truly boiled it down to a bunch of rules…

### Setting the Stage

To get started I'm going to define a simple situation and use that throughout this page. I'm going to define three events related to rolling a standard 6-sided die.

Event A - Rolling a 1,2 or 3
Event B - Rolling an even number
Event C - Rolling a 5 or 6

In a Venn diagram this looks something like: From this we can also say:

(1)
\begin{align} P(A)=\frac{3}{6}, P(B)=\frac{3}{6}, P(C)=\frac{2}{6}, \end{align}

Notice that the sum of these probabilities do not add to 1!

### A OR B - Combined Events

Probability questions are often phrased in terms of "A or B." The key word is OR. In this case we are interested in if one or the other occurs.

The probability of Event A or Event C happening is pretty straight forward to figure out. From the Venn diagram we can see that 5 different outcomes out of 6 possible outcomes meet this criteria. In math symbols we can say:

(2)
\begin{align} P(A\cup C) = P(A)+P(C) = \frac{5}{6} \end{align}

That works great, maybe just like you learned in middle school, but this only works because the events are Mutually Exclusive - meaning that A and C (in this case) can not both happen at the same time. However if we look at the probability of B or C occurring it's a bit more complicated.

From the Venn diagram we can see that 4 out of 6 outcomes would satisfy the conditions of B or C. Meaning the probability is $\frac{4}{6}$.

If we simply add $P(B)$ and $P(C)$ we'd get $\frac{5}{6}$ which is wrong. This is because we have effectively double counted the intersection or the event when they both occur. So we can sum $P(B)$ and $P(C)$ but we must subtract the intersection:

(3)
\begin{align} P(B \cup C) = P(B) + P(C) - \underbrace{P(B \cap C)}_{intersection} = \frac{4}{6} \end{align}

In general, and on your Data Booklet, the equation for combined events is :

(4)
\begin{align} P(A \cup B) = P(A) + P(B) - P(A \cap B) \end{align}

In many simple cases $P(A \cap B)=0$.

### Conditional Probability

We are often interested in multiple events occurring. So we can ask the question what is the chance that event B occurs given that event A has occurred.

Using the example at the top this can be thought of as "knowing" that we rolled a 1,2 or 3 and then asking what is the chance that it was an even number. Clearly the probability is $\frac{1}{3}$.

In math symbols this type of question is written as $P(B|A)$ and is read as "B given A" and has the meaning of "what is the probability of B occurring given than A has occurred?" We can also express this as:

(5)
\begin{align} P(B|A)=\frac{P(A \cap B)}{P(A)} \end{align}

The truth of this equation is not obvious. I might add in an explanation at some later point.

Example 1

Use the formula above to calculate $P(B|A)$.

#### Independent vs. Dependent Events

If the occurrence of event A does not affect the probability of event B then the events are said to be independent. So we can say:

(7)
\begin{align} P(A and B) = P(A \cap B)=P(A)P(B) \end{align}

An example of this is rolling a die twice - the outcome of the first roll does not affect the second. This is pretty straight forward and matches much of what folks learned before IB math.

If the outcome of one event affects the second then the events are not independent (i.e. dependent). A simple example is pulling two cards from a deck of cards at random without replacing the first card.

### IB Examples

A few examples of IB problems - minus any really context.

Example 2

Consider the independent events A and B. Given that $P(B)=2P(A)$, and $P(A \cup B) = 0.52$ find $P(B)$.

Example 3

Consider the events A and B, where $P(A)=\frac{2}{5}$, $P(B')=\frac{1}{4}$ and $P(A \cup B)=\frac{7}{8}$.

a) Write down $P(B)$.
b) Find $P(A \cap B)$.
c) Find $P(A|B)$.