So we are discussing probability and so far, I’ve just used some simple examples where I used the rule:

Probability = \[

\frac{{\mathrm{Number}}\hspace{0.33em}{\mathrm{of}}\hspace{0.33em}{\mathrm{favorable}}\hspace{0.33em}{\mathrm{outcomes}}}{{\mathrm{Total}}\hspace{0.33em}{\mathrm{number}}\hspace{0.33em}{\mathrm{of}}\hspace{0.33em}{\mathrm{possible}}\hspace{0.33em}{\mathrm{outcome}}{s}}

\]

Now I would like to be able to show more complex examples , but first a definition:

Event – a collection of one or more outcomes of an experiment.

An experiment is flipping a coin, rolling a die, picking a card, etc. The outcome is what happens, that is the result of the experiment. To save writing, we use the following notation:

P(A) is the probability of event A. Event A can be flipping heads, rolling a 2, picking the ace of hearts. Other letters can be used to represent other events.

So for example, what is the probability of rolling a die and getting a 1 or a 6? Well one way is to note that the number of favorable outcomes is 2 and the number of possibilities is 6, so

P(rolling a 1 or a 6) = 2/6 = 1/3

Another way is by using the probability addition rule:

P(A or B) = P(A) + P(B)

Event A can be rolling a 1 and event B can be rolling a 6. We know that the probability of rolling any single number is 1/6 so,

P(A or B) = 1/6 + 1/6 = 2/6 = 1/3

The addition rule only works for mutually exclusive events. Rolling a 1 means that rolling a 6 is impossible and vice versa.

What if I asked what is the probability of not rolling a 1 or a 6? Well there is something called the complement rule that is useful:

P(~A) = 1 – P(A), where ~A means not A

If A is rolling a 1 or a 6, then ~A is rolling any other number. But since we now know what the probability of throwing a 1 or a 6 is, we can use this to find the probability of not throwing a 1 or a 6:

P(not throwing a 1 or a 6) = 1 – P(rolling a 1 or a 6) = 1 – 1/3 = 2/3

In my next post, I’ll explore questions like what is the probability of tossing 3 heads in a row.