# How do you determine the theoretical probability of rolling a five on a die?

Feb 13, 2015

The probability of any event is based on given probabilities of elementary events and a composition of our event as a set of elementary events.

If we know the probabilities of elementary events and a composition of our event, the probability of our event is a sum of probabilities of all elementary events that comprise it.

In case of a die, the elementary events are numbers rolled on this die. The so-called "fair" die has all its 6 numbers equally probable. Since the total probability always equals to 1, the probability of each elementary event (rolling each number from 1 to 6) equals to
$P \left\{1\right\} = P \left\{2\right\} = P \left\{3\right\} = P \left\{4\right\} = P \left\{5\right\} = P \left\{6\right\} = \frac{1}{6}$.

An event offered in the problem, "rolling a five on a die", contains only one elementary event - number 5, whose probability we know as being equal to $\frac{1}{6}$. Therefore, the probability of rolling five on a die equals to
$P \left\{5\right\} = \frac{1}{6}$.

Just as an example, an event "Rolling a number that is less than five" contains 4 different elementary events - rolling 1, 2, 3 or 4. Each of them has a probability $\frac{1}{6}$. The sum of all elementary events comprising our event, that is the probability of our event, is
$P \left\{< 5\right\} = P \left\{1 \mathmr{and} 2 \mathmr{and} 3 \mathmr{and} 4\right\} = \frac{4}{6} = \frac{2}{3}$.