Lesson Video: An Introduction to Probability and Simple Events Mathematics

Video Transcript

In this video, we will learn how to find the probability of a simple event. We will begin by recapping some of the key terminology associated with probability. Firstly, we recall that an experiment is an activity with an identifiable result. An outcome is a specific result of such an experiment. A sample space is the set of all possible outcomes from a random experiment. Finally, an event is a subset of the sample space where an event may consist of a single outcome or maybe composed of multiple outcomes.

Let’s consider the experiment of rolling a fair six-sided die and recording the number. In this case, a specific outcome would be rolling a particular number, for example, four. The sample space would be the set of all outcomes, in this case one, two, three, four, five, and six. And an event would be the subset of the sample space, for example, rolling a one or rolling a number greater than three. Informally, we may think of an event as something that happens and the probability of an event is how likely it is to happen. In everyday life, we may discuss probabilities using words like “certain,” “impossible,” “unlikely,” or “very likely.” For example, we may say it is likely that it will rain tomorrow.

In mathematics, we want to extend this idea and start describing probabilities using numbers. We begin by defining impossible events as having a probability of zero and events that are certain to happen as having a probability of one. This leads to the probability scale on which the probabilities of all events are placed between these values. Events that have a probability of 0.5, or one-half, have an even chance of occurring. This is because they are just as likely to happen as they are to not happen. Events with a probability between zero and 0.5 are less likely to occur. And events with a probability between 0.5 and one are more likely to occur. We may also use terms like highly unlikely or highly likely to describe events for which the probabilities are close to the extremes of zero and one.

We note that probabilities can be expressed as either fractions, decimals, or percentages. When all the outcomes of an experiment are equally likely, the probability of an event can be calculated by dividing the number of outcomes in that event, which we call the number of successful outcomes by the total number of outcomes in the sample space. This can be more formally defined as follows. If 𝐴 is an event in a sample space 𝑆 where each outcome is equally likely, then the probability of event 𝐴 occurring is 𝑃 of 𝐴 which is equal to 𝑛 of 𝐴 over 𝑛 of 𝑆, where 𝑃 of 𝐴 represents the probability of event 𝐴. 𝑛 of 𝐴 represents the number of elements in event 𝐴. And 𝑛 of 𝑆 represents the number of elements in the sample space 𝑆.

We will now consider a series of examples. In each example, the key will be to determine the total number of outcomes or elements in the sample space of the experiment and the number of outcomes which are considered to be successful.

If I roll a regular six-sided die, what is the probability that the score is three?

As the die is regular, this means it is unbiased, and so every outcome is equally likely. We can therefore calculate the required probability by recalling that the probability of an event is equal to the number of successful outcomes divided by the total number of outcomes. In this case, the event we’re interested in is getting a three when a regular six-sided die is rolled. The possible outcomes when rolling a die are the numbers one, two, three, four, five, and six. This means that the total number of outcomes is six. The only successful outcome is getting the number three. So there is one successful outcome. We can therefore conclude that the probability of rolling a three, written 𝑃 of three, is one out of six, or one-sixth.

In our next example, we will calculate the probability of a simple event, which consists of more than one outcome.

If I roll a regular six-sided die, what is the probability that the score is divisible by three?

In this question, we’re rolling a regular six-sided die, which means that every outcome is equally likely. There are six possible outcomes: the numbers one, two, three, four, five, and six. And we are asked to calculate the probability that the score that we land on is divisible by three. We recall that the probability of an event can be written as a fraction, where the numerator is the number of successful outcomes and the denominator is the total number of outcomes. Applying the formula to this question, we have the probability that the score is divisible by three is equal to the number of outcomes divisible by three divided by the total number of outcomes. The numbers three and six are both divisible by three. Therefore, there are two successful outcomes.

The probability that the score is divisible by three is therefore equal to two out of six, or two-sixths. And as both the numerator and denominator are divisible by two, this simplifies to one-third. When rolling a regular six-sided die, the probability that the score is divisible by three is one-third.

It is worth noting at this stage that if we are giving probabilities as fractions, it is generally good practice to give them in their simplest form. Let’s now consider an example in a different context.

A card is drawn at random from a deck of cards numbered one to 52. What is the probability that the card drawn is a prime number?

We begin by noting that as the card is to be drawn at random, every card has an equal chance of being chosen. In order to calculate the required probability, we can use the formula 𝑃 of 𝐴 is equal to 𝑛 of 𝐴 divided by 𝑛 of 𝑆, where 𝑃 of 𝐴 represents the probability of event 𝐴, 𝑛 of 𝐴 represents the number of outcomes in event 𝐴, and 𝑛 of 𝑆 represents the number of elements in the sample space 𝑆. Since there are 52 cards in the deck, 𝑛 of 𝑆 is equal to 52. The event we’re interested in in this question is drawing a prime number. This means that 𝑛 of 𝐴 will be the number of prime number cards in the deck.

We recall that a prime number has exactly two factors, the number one and itself. The prime numbers less than 20 are as shown, as all of these numbers are only divisible by one and the number itself. Continuing this list, the numbers 23, 29, 31, 37, 41, 43, and 47 are also prime. In total, we have 15 prime numbers between one and 52 inclusive. We can therefore conclude that the probability of event 𝐴 drawing a prime number is 15 out of 52. This fraction cannot be simplified as 15 and 52 have no common factors apart from one.

In our next example, we will calculate an experimental probability from survey data that has been presented in a frequency table.

The table shows the results of a survey that asked 100 people to vote for their favorite type of TV program. What is the probability that a randomly selected person prefers drama?

The table tells us that 14 of the surveyed people prefer drama programs. We are also told that 19 prefer documentaries, 14 prefer comedy programs, 16 prefer news programs, and 37 prefer sport programs. As mentioned in the question, there were a total of 100 people surveyed. A person is to be selected at random, and as such each person has an equal chance of being chosen. We can therefore calculate the required probability using the formula the probability of an event is equal to the number of successful outcomes over the number of possible outcomes. In this question, the number of successful outcomes will be equal to the number of people who prefer drama. And the number of possible outcomes will be the total number of people.

As already mentioned, we can see from the table that 14 people prefer drama and there were a total of 100 people surveyed. The probability that a randomly selected person prefers drama is therefore equal to 14 out of 100. Dividing both the numerator and denominator by two, this fraction simplifies to seven over 50, or seven fiftieths. We could also write this answer as the decimal 0.14 or 14 percent. This is the probability that a randomly selected person prefers drama.

It is worth noting at this point that we can use probabilities to make conclusions based on statistical data. For example, in this question, we have found that the probability that a randomly selected person prefers drama was 14 percent. The probability that a randomly selected person prefers sport, on the other hand, is 37 out of 100, or 37 percent. Based on this information, the person responsible for program scheduling may choose to show a greater proportion of sports programs than drama programs in order to appeal to a larger number of people. It is therefore important that whenever samples are taken, they are unbiased and representative of the population being sampled so that any decisions made are based on reliable data.

In our final example, we will work backwards from knowing a probability to determine an unknown.

A bag contains 16 white balls and an unknown number of red balls. The probability of choosing a red ball at random is one-third. How many balls are in the bag?

There are many ways of approaching this problem. We will begin by demonstrating an algebraic method in which we form and solve an equation. In order to determine the total number of balls in the bag, we will first need to calculate the number of red balls. And since this is currently unknown, we will let the number of red balls be 𝑟. Since there are 16 white balls and the bag only contains red and white balls, the total number of balls is equal to 𝑟 plus 16. We are also told in the question that the probability of choosing a red ball at random is one-third. Next, we recall that the probability of an event is equal to the number of successful outcomes divided by the total number of outcomes. In this question, the probability of selecting a red ball is equal to the number of red balls divided by the total number of balls.

Substituting the information we know into this formula, we have one-third is equal to 𝑟 over 𝑟 plus 16. Cross multiplying, we can rewrite this equation as 𝑟 plus 16 is equal to three 𝑟. We can then subtract 𝑟 from both sides such that two 𝑟 is equal to 16. And dividing through by two, we have 𝑟 is equal to eight. This means that there are eight red balls in the bag. And since there are 16 white balls, there are 24 balls in total in the bag. An alternative method here would be to notice that since one-third of the balls are red, two-thirds must be white. And since 16 balls are white, half of this, that is, eight of the balls, must be red. This once again confirms that there are 24 balls in total in the bag.

We will now finish this video by recapping the key points. When all the outcomes of an experiment are equally likely, the probability of an event can be calculated using the following formula. The probability of an event is equal to the number of successful outcomes over the total number of outcomes. More formally, this can be written as 𝑃 of 𝐴 is equal to 𝑛 of 𝐴 over 𝑛 of 𝑆, where 𝑃 of 𝐴 represents the probability of event 𝐴. 𝑛 of 𝐴 represents the number of outcomes in event 𝐴. And 𝑛 of 𝑆 represents the number of elements in the sample space 𝑆.