# Theoretical Probability: How To Get It, Examples, Exercises

The theoretical (or Laplace) probability of an event E occurring that belongs to a sample space S, in which all events have the same probability of occurrence, is defined in mathematical notation as:P (E) = n (E) / N (S)

Where P (E) is the probability, given as the quotient between the total number of possible outcomes of event E, which we call n (E), divided by the total number N (S) of possible outcomes in the sample space S.

The theoretical probability is a real number between 0 and 1, but it is often expressed as a percentage, in which case the probability will be a value between 0% and 100%.

Calculating the probability of an event occurring is very important in many fields, such as stock trading, insurance companies, gambling, and many more.

## How to get the theoretical probability?

An illustrative case is the case of raffles or lotteries. Suppose 1,000 tickets are issued to raffle asmartphone. As the drawing is done randomly, any of the tickets has an equal chance of being a winner.

To find the probability that a person who buys a ticket with the number 81 is a winner, the following calculation of theoretical probability:

P (1) = 1 / 1,000 = 0.001 = 0.1%

The above result is interpreted as follows: if the draw were repeated infinitely many times, every 1,000 times ticket 81 would be selected, on average, once.

If for some reason someone acquires all the tickets it is certain that they will win the prize. The probability of winning the prize if you have all the tickets is calculated as follows:

P (1,000) = 1,000 / 1,000 = 1 = 100%.

That is, that probability 1 or 100% means that it is totally certain that this result will occur.

If someone owns 500 tickets the chances of winning or losing are the same. The theoretical probability of winning the prize in this case is calculated as follows:

P (500) = 500 / 1,000 = ½ = 0.5 = 50%.

He who does not buy any ticket has no chance of winning and his theoretical probability is determined as follows:

P (0) = 0 / 1,000 = 0 = 0%

## Examples

### Example 1

You have a coin with face on one side and shield or stamp on the other. When the coin is tossed, what is the theoretical probability that it will come up heads?

P (face) = n (face) / N ( face + shield ) = ½ = 0.5 = 50%

The result is interpreted as follows: if a huge number of tosses were made, on average for every 2 tosses one of them would come up heads.

In percentage terms, the interpretation of the result is that by making an infinitely large number of tosses, on average out of 100 of them 50 would result in heads.

### Example 2

In a box there are 3 blue marbles, 2 red marbles and 1 green. What is the theoretical probability that when you take a marble out of the box it will be red?

The probability that it comes out red is:

P (red) = Number of favorable cases / Number of possible cases

That is to say:

P (red) = Number of red marbles / Total number of marbles

Finally, the probability that a red marble is drawn is:

P (red) = 2/6 = ⅓ = 0.3333 = 33.33%

While the probability that when drawing a green marble is:

P (green) = ⅙ = 0.1666 = 16.66%

Finally, the theoretical probability of obtaining a blue marble in a blind extraction is:

P (blue) = 3/6 = ½ = 0.5 = 50%

That is, for every 2 attempts the result will be blue in one of them and another color in another attempt, under the premise that the extracted marble is replaced and that the number of trials is very, very large.

## Exercises

### Exercise 1

Determine the probability that rolling a die will yield a value less than or equal to 4.

#### Solution

To calculate the probability of this event occurring, the definition of theoretical probability will be applied:

P (≤4) = Number of favorable cases / Number of possible cases

P (≤5) = 5/6 = = 83.33%

### Exercise 2

Find the probability that in two Consecutive rolls of a normal six-sided die roll 5 the 2 times.

#### Solution

To answer this exercise, it is convenient to make a table to show all the possibilities. The first digit indicates the result of the first die and the second the result of the other.

To calculate the theoretical probability we need to know the total number of possible cases, in this case, as can be seen from the previous table, there are 36 possibilities.

Also observing the table, it can be deduced that the number of cases favorable to the event that in the two consecutive launches comes out 5 is only 1, highlighted with color, therefore the probability that this event occurs is:

P (5 x 5) = 1/36.

This result could also have been arrived at using one of the properties of theoretical probability, which states that the combined probability of two independent events is the product of their individual probabilities.

In this case, the probability that the first toss will roll 5 is ⅙. The second toss is completely independent of the first, therefore the probability that 5 is rolled in the second is also ⅙. So the combined probability is:

P (5 × 5) = P (5) P (5) = (1/6) (1/6) = 1/36.

### Exercise 3

Find the probability that a number less than 2 is rolled on the first toss and a number greater than 2 is rolled on the second.

#### Solution

Again, a table of possible events must be constructed, where those in which the first throw was less than 2 and in the second greater than 2 are underlined.

In total there are 4 possibilities out of a total of 36. In other words, the probability of this event is:

P (<2;> 2) = 4/36 = 1/9 = 0.1111 = 11.11%

Using the probability theorem that states:

The probability of occurrence of two independent events is equal to the product of the individual probabilities.

The same result is obtained:

P (<2) P (> 2) = (1/6) (4/6) = 4/36 = 0.1111 = 11.11%

The value obtained with this procedure coincides with the previous result, by means of the theoretical or classical definition of probability.

### Exercise 4

What is the probability that when rolling two dice the sum of the values ​​is 7.

#### Solution

To find the solution in this case, a table of possibilities has been drawn up, which is indicated in color the cases that satisfy the condition that the sum of the values ​​is 7.

Looking at the table, 6 possible cases can be counted, so the probability is:

P (I + II: 7) = 6/36 = 1/6 = 0.1666 = 16.66%

## References

1. Canavos, G. 1988. Probability and Statistics: Applications and methods. McGraw Hill.
2. Devore, J. 2012. Probability and Statistics for Engineering and Science. 8th. Edition. Cengage.
3. Lipschutz, S. 1991. Schaum Series: Probability. McGraw Hill.
4. Obregón, I. 1989. Theory of probability. Editorial Limusa.
5. Walpole, R. 2007. Probability and Statistics for Engineering and Sciences. Pearson.