As your random experiment, consider rolling a cubical die (*die* is singular; *dice* is the plural of *die*):

The sample space has six outcomes: {1, 2, 3, 4, 5, 6}. If the die has been constructed in a perfectly symmetrical manner, then we can expect that no outcome will be preferred over the other. Thus, for a large number of rolls, the relative occurrence of each outcome should be roughly one-sixth (1/6). In other words, the probability of each outcome will be 1/6. Note that we have used the assumption of equal likelihood in saying that the probability of each outcome will be 1/6. This assumption itself will have to be tested empirically for its truth.

Thus, for a fair die,

\[\begin{align}&P\left( 1 \right) = P\left( 2 \right) = P\left( 3 \right)\\ \qquad\quad &\qquad= P\left( 4 \right) = P\left( 5 \right) = P\left( 6 \right) = \frac{1}{6}\end{align}\]

The sum of the six probabilities is 1.

A die need not be fair. In a loaded die, one face is made heavier – this can be done in a number of ways. When the die is rolled, the heavier face will then always tend to come to rest at the bottom, so the face opposite the heavier face will tend to show up a larger number of times compared to the other faces - its relative occurrence will increase.

The following table shows the number of occurrences and the relative occurrence of each outcome for a loaded die which was rolled 600 times:

Outcome |
1 |
2 |
3 |
4 |
5 |
6 |

Number of Occurrences |
100 |
80 |
150 |
120 |
100 |
50 |

Relative Occurrence |
1/6 |
2/15 |
1/4 |
1/5 |
1/6 |
1/12 |

The relative occurrence of Face-6 is the lowest, while that of Face-3 is the highest. This suggests that Face-6 is the heaviest face of the die, and Face-3 is opposite Face-6.

The relative occurrence values in the last row of the table above are a good measure of the probability of each face, since the number of repetitions of the experiment is sufficiently large (600). A larger number of repetitions would have been even better, but we can make do with these values for the probabilities:

\[\begin{align}&P\left( 1 \right) = \frac{1}{{12}}, & P\left( 2 \right) = \frac{1}{4}, \;\;\;\; P\left( 3 \right) = \frac{1}{6}\\&P\left( 4 \right) = \frac{2}{{15}}, & P\left( 5 \right) = \frac{1}{5}, \;\;\;\;P\left( 6 \right) = \frac{1}{6}\end{align}\]

Once again, note that the sum of the probabilities is 1. This is because the sum of all the relative occurrences must be 1.

Now, suppose that our random experiment is rolling a fair die. Let us define an *event* *E* as follows:

*E*: A number greater than 4 shows up

Note that there are two possible outcomes in *E*: either 5 shows up, or 6 shows up. What is the probability of event *E* occurring? In other words, what will be the relative occurrence of event *E* for a very larger number of rolls?

For a large number of rolls, since the relative occurrence of the outcome “5” is 1/6, and the relative occurrence of the outcome “6” is also 1/6, the relative occurrence of event *E* is

\[\frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}\]

Thus, we can say that \(P\left( E \right) = \frac{1}{3}\): the probability of event *E* occurring is \(\frac{1}{3}\), or in other words, for a large number of rolls, event *E* will occur one-thirds of the time.

Now, let us define another event *F* as follows:

*F*: 1 does not show up

What is the probability of *F* occurring, or \(P\left( F \right)\)? Observe that *F* contains the following outcomes: {2, 3, 4, 5, 6}. Thus, for a large number of rolls, the relative occurrence of event *F* will be the sum of the relative occurrences of all these outcomes. Thus,

\[P\left( F \right) = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{5}{6}\]

We note that the sample space has a *total* of 6 outcomes, of which 5 are *favorable* to event *F* – that is, if any of the 5 favorable outcomes occurs, we can say that the event *F* as occurred. Thus, we can think of the probability of *F* occurring as the ratio of the number of favorable outcomes to the total number of outcomes:

\[P\left( F \right) = \frac{{{\rm{No}}{\rm{. of \,favorable\,\, outcomes}}}}{{{\rm{No}}{\rm{. of \,total\,\, outcomes}}}} = \frac{5}{6}\]

Note that doing this is possible *only because* all the outcomes are equally likely, and so the relative occurrence of each outcome for a large number of rolls (the probability of each outcome) is equal to 1/6.

For an unfair die, we cannot follow a similar approach of taking the ratio of the number of favorable outcomes to the number of total outcomes. For example, suppose that the relative occurrences of the various faces for a large number of rolls (the probabilities) are:

\[\begin{align}&P\left( 1 \right) = \frac{1}{{12}}, & P\left( 2 \right) = \frac{1}{4}, \;\;\;P\left( 3 \right) = \frac{1}{6}\\&P\left( 4 \right) = \frac{2}{{15}}, & P\left( 5 \right) = \frac{1}{5}, \;\;\;P\left( 6 \right) = \frac{1}{6}\end{align}\]

You can verify that the sum of the 6 probability values above is equal to 1. Now, if we take the event *F* to be

*F*: 1 does not show up,

what is the probability of event *F* occurring? Clearly, it will be the sum of the relative occurrence values or probability values of the outcomes {2, 2, 4, 5, 6}, which in this case will be \(\frac{{11}}{{12}}\). Thus, even though the expression

\[\frac{{{\rm{No}}{\rm{. of\, favorable \,\,outcomes}}}}{{{\rm{No}}{\rm{. of\, total \,\, outcomes}}}}\]

gives us a value of \(\frac{5}{6}\), the outcomes in this case are not equally likely, and so the actual probability value cannot be calculated using this expression; it must be calculated using the actual relative occurrence values (for a large number of rolls).

**Example 1:** A fair die is rolled. The event *E* is defined as follows:

*E*: A multiple of 3 shows up

What is the probability of *E* occurring?

**Solution:** The event *E* consists of two possible outcomes: 3 or 6. Thus, the probability of *E* occurring is:

\[\begin{align}&P\left( E \right) = \frac{{{\rm{No}}{\rm{. of \,favorable\,\, outcomes}}}}{{{\rm{No}}{\rm{. of \,total \,\,outcomes}}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\;\;= \frac{2}{6} = \frac{1}{3}\end{align}\]